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| description
stringlengths 31
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| public_tests
dict | private_tests
dict | solution_type
stringclasses 2
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stringclasses 5
values | solution
stringlengths 1
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1189_D1. Add on a Tree
|
Note that this is the first problem of the two similar problems. You can hack this problem only if you solve both problems.
You are given a tree with n nodes. In the beginning, 0 is written on all edges. In one operation, you can choose any 2 distinct leaves u, v and any real number x and add x to values written on all edges on the simple path between u and v.
For example, on the picture below you can see the result of applying two operations to the graph: adding 2 on the path from 7 to 6, and then adding -0.5 on the path from 4 to 5.
<image>
Is it true that for any configuration of real numbers written on edges, we can achieve it with a finite number of operations?
Leaf is a node of a tree of degree 1. Simple path is a path that doesn't contain any node twice.
Input
The first line contains a single integer n (2 ≤ n ≤ 10^5) — the number of nodes.
Each of the next n-1 lines contains two integers u and v (1 ≤ u, v ≤ n, u ≠ v), meaning that there is an edge between nodes u and v. It is guaranteed that these edges form a tree.
Output
If there is a configuration of real numbers written on edges of the tree that we can't achieve by performing the operations, output "NO".
Otherwise, output "YES".
You can print each letter in any case (upper or lower).
Examples
Input
2
1 2
Output
YES
Input
3
1 2
2 3
Output
NO
Input
5
1 2
1 3
1 4
2 5
Output
NO
Input
6
1 2
1 3
1 4
2 5
2 6
Output
YES
Note
In the first example, we can add any real x to the value written on the only edge (1, 2).
<image>
In the second example, one of configurations that we can't reach is 0 written on (1, 2) and 1 written on (2, 3).
<image>
Below you can see graphs from examples 3, 4:
<image> <image>
|
{
"input": [
"2\n1 2\n",
"3\n1 2\n2 3\n",
"5\n1 2\n1 3\n1 4\n2 5\n",
"6\n1 2\n1 3\n1 4\n2 5\n2 6\n"
],
"output": [
"YES",
"NO",
"NO",
"YES"
]
}
|
{
"input": [
"50\n16 4\n17 9\n31 19\n22 10\n8 1\n40 30\n3 31\n20 29\n47 27\n22 25\n32 34\n12 15\n40 32\n10 33\n47 12\n6 24\n46 41\n14 23\n12 35\n31 42\n46 28\n31 20\n46 37\n1 39\n29 49\n37 47\n40 6\n42 36\n47 2\n24 46\n2 13\n8 45\n41 3\n32 17\n4 7\n47 26\n28 8\n41 50\n34 44\n33 21\n25 5\n16 40\n3 14\n8 18\n28 11\n32 22\n2 38\n3 48\n44 43\n",
"10\n8 1\n1 2\n8 9\n8 5\n1 3\n1 10\n1 6\n1 7\n8 4\n",
"5\n5 1\n5 4\n4 3\n1 2\n",
"7\n1 2\n2 3\n1 4\n1 5\n3 6\n3 7\n",
"3\n1 3\n2 3\n",
"60\n26 6\n59 30\n31 12\n31 3\n38 23\n59 29\n53 9\n38 56\n53 54\n29 21\n17 55\n59 38\n26 16\n24 59\n24 25\n17 35\n24 41\n30 15\n31 27\n8 44\n26 5\n26 48\n8 32\n53 17\n3 34\n3 51\n30 28\n47 10\n53 60\n36 42\n24 53\n59 22\n53 40\n26 52\n36 4\n59 8\n29 37\n36 20\n17 47\n53 18\n3 50\n30 2\n17 7\n8 58\n59 1\n31 11\n24 26\n24 43\n53 57\n59 45\n47 13\n26 46\n17 33\n30 31\n26 39\n26 19\n24 36\n8 49\n38 14\n",
"7\n1 2\n2 3\n3 4\n3 5\n1 6\n1 7\n",
"20\n19 16\n19 18\n20 7\n9 4\n6 17\n14 2\n9 15\n2 13\n5 11\n19 12\n12 20\n16 9\n11 8\n19 5\n3 1\n19 14\n5 3\n12 10\n19 6\n",
"7\n1 2\n1 3\n2 4\n2 5\n3 6\n3 7\n",
"10\n9 5\n7 1\n9 10\n7 2\n5 4\n9 6\n2 9\n10 8\n1 3\n",
"4\n2 4\n2 3\n2 1\n",
"4\n1 4\n3 2\n1 3\n",
"3\n1 2\n1 3\n",
"5\n1 2\n1 5\n1 3\n1 4\n",
"20\n14 9\n12 13\n10 15\n2 1\n20 19\n16 6\n16 3\n17 14\n3 5\n2 11\n3 10\n15 8\n14 2\n6 4\n3 20\n5 18\n1 7\n1 16\n4 12\n",
"20\n7 5\n14 13\n17 6\n3 8\n16 12\n18 9\n3 18\n14 1\n17 3\n15 2\n17 4\n9 11\n2 7\n15 17\n3 20\n16 10\n17 14\n2 16\n1 19\n",
"8\n1 2\n2 3\n3 4\n1 7\n1 8\n4 5\n4 6\n",
"5\n5 1\n5 2\n5 3\n5 4\n",
"50\n49 6\n43 7\n1 27\n19 35\n15 37\n16 12\n19 21\n16 28\n49 9\n48 39\n13 1\n2 48\n9 50\n44 3\n41 32\n48 31\n49 33\n6 11\n13 20\n49 22\n13 41\n48 29\n13 46\n15 47\n34 2\n49 13\n48 14\n34 24\n16 36\n13 40\n49 34\n49 17\n43 25\n11 23\n10 15\n19 26\n34 44\n16 42\n19 18\n46 8\n29 38\n1 45\n12 43\n13 16\n46 30\n15 5\n49 10\n11 19\n32 4\n",
"20\n13 1\n18 2\n3 7\n18 5\n20 16\n3 12\n18 9\n3 10\n18 11\n13 6\n3 18\n20 15\n20 17\n3 13\n3 4\n13 14\n3 20\n18 8\n3 19\n",
"10\n8 2\n5 6\n1 8\n2 9\n1 4\n8 10\n10 5\n2 7\n2 3\n"
],
"output": [
"NO",
"YES",
"NO",
"NO",
"NO",
"YES",
"NO",
"NO",
"NO",
"NO",
"YES",
"NO",
"NO",
"YES",
"NO",
"NO",
"NO",
"YES",
"NO",
"YES",
"NO"
]
}
|
IN-CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
int32_t main() {
cin.sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
;
long long n;
cin >> n;
vector<vector<long long>> graph(n + 1);
for (long long i = 0; i < n - 1; i++) {
long long u, v;
cin >> u >> v;
graph[u].push_back(v);
graph[v].push_back(u);
}
for (long long i = 0; i < graph.size(); i++) {
if (graph[i].size() > 2) {
cout << "YES"
<< "\n";
return 0;
}
}
cout << "NO"
<< "\n";
}
|
1189_D1. Add on a Tree
|
Note that this is the first problem of the two similar problems. You can hack this problem only if you solve both problems.
You are given a tree with n nodes. In the beginning, 0 is written on all edges. In one operation, you can choose any 2 distinct leaves u, v and any real number x and add x to values written on all edges on the simple path between u and v.
For example, on the picture below you can see the result of applying two operations to the graph: adding 2 on the path from 7 to 6, and then adding -0.5 on the path from 4 to 5.
<image>
Is it true that for any configuration of real numbers written on edges, we can achieve it with a finite number of operations?
Leaf is a node of a tree of degree 1. Simple path is a path that doesn't contain any node twice.
Input
The first line contains a single integer n (2 ≤ n ≤ 10^5) — the number of nodes.
Each of the next n-1 lines contains two integers u and v (1 ≤ u, v ≤ n, u ≠ v), meaning that there is an edge between nodes u and v. It is guaranteed that these edges form a tree.
Output
If there is a configuration of real numbers written on edges of the tree that we can't achieve by performing the operations, output "NO".
Otherwise, output "YES".
You can print each letter in any case (upper or lower).
Examples
Input
2
1 2
Output
YES
Input
3
1 2
2 3
Output
NO
Input
5
1 2
1 3
1 4
2 5
Output
NO
Input
6
1 2
1 3
1 4
2 5
2 6
Output
YES
Note
In the first example, we can add any real x to the value written on the only edge (1, 2).
<image>
In the second example, one of configurations that we can't reach is 0 written on (1, 2) and 1 written on (2, 3).
<image>
Below you can see graphs from examples 3, 4:
<image> <image>
|
{
"input": [
"2\n1 2\n",
"3\n1 2\n2 3\n",
"5\n1 2\n1 3\n1 4\n2 5\n",
"6\n1 2\n1 3\n1 4\n2 5\n2 6\n"
],
"output": [
"YES",
"NO",
"NO",
"YES"
]
}
|
{
"input": [
"50\n16 4\n17 9\n31 19\n22 10\n8 1\n40 30\n3 31\n20 29\n47 27\n22 25\n32 34\n12 15\n40 32\n10 33\n47 12\n6 24\n46 41\n14 23\n12 35\n31 42\n46 28\n31 20\n46 37\n1 39\n29 49\n37 47\n40 6\n42 36\n47 2\n24 46\n2 13\n8 45\n41 3\n32 17\n4 7\n47 26\n28 8\n41 50\n34 44\n33 21\n25 5\n16 40\n3 14\n8 18\n28 11\n32 22\n2 38\n3 48\n44 43\n",
"10\n8 1\n1 2\n8 9\n8 5\n1 3\n1 10\n1 6\n1 7\n8 4\n",
"5\n5 1\n5 4\n4 3\n1 2\n",
"7\n1 2\n2 3\n1 4\n1 5\n3 6\n3 7\n",
"3\n1 3\n2 3\n",
"60\n26 6\n59 30\n31 12\n31 3\n38 23\n59 29\n53 9\n38 56\n53 54\n29 21\n17 55\n59 38\n26 16\n24 59\n24 25\n17 35\n24 41\n30 15\n31 27\n8 44\n26 5\n26 48\n8 32\n53 17\n3 34\n3 51\n30 28\n47 10\n53 60\n36 42\n24 53\n59 22\n53 40\n26 52\n36 4\n59 8\n29 37\n36 20\n17 47\n53 18\n3 50\n30 2\n17 7\n8 58\n59 1\n31 11\n24 26\n24 43\n53 57\n59 45\n47 13\n26 46\n17 33\n30 31\n26 39\n26 19\n24 36\n8 49\n38 14\n",
"7\n1 2\n2 3\n3 4\n3 5\n1 6\n1 7\n",
"20\n19 16\n19 18\n20 7\n9 4\n6 17\n14 2\n9 15\n2 13\n5 11\n19 12\n12 20\n16 9\n11 8\n19 5\n3 1\n19 14\n5 3\n12 10\n19 6\n",
"7\n1 2\n1 3\n2 4\n2 5\n3 6\n3 7\n",
"10\n9 5\n7 1\n9 10\n7 2\n5 4\n9 6\n2 9\n10 8\n1 3\n",
"4\n2 4\n2 3\n2 1\n",
"4\n1 4\n3 2\n1 3\n",
"3\n1 2\n1 3\n",
"5\n1 2\n1 5\n1 3\n1 4\n",
"20\n14 9\n12 13\n10 15\n2 1\n20 19\n16 6\n16 3\n17 14\n3 5\n2 11\n3 10\n15 8\n14 2\n6 4\n3 20\n5 18\n1 7\n1 16\n4 12\n",
"20\n7 5\n14 13\n17 6\n3 8\n16 12\n18 9\n3 18\n14 1\n17 3\n15 2\n17 4\n9 11\n2 7\n15 17\n3 20\n16 10\n17 14\n2 16\n1 19\n",
"8\n1 2\n2 3\n3 4\n1 7\n1 8\n4 5\n4 6\n",
"5\n5 1\n5 2\n5 3\n5 4\n",
"50\n49 6\n43 7\n1 27\n19 35\n15 37\n16 12\n19 21\n16 28\n49 9\n48 39\n13 1\n2 48\n9 50\n44 3\n41 32\n48 31\n49 33\n6 11\n13 20\n49 22\n13 41\n48 29\n13 46\n15 47\n34 2\n49 13\n48 14\n34 24\n16 36\n13 40\n49 34\n49 17\n43 25\n11 23\n10 15\n19 26\n34 44\n16 42\n19 18\n46 8\n29 38\n1 45\n12 43\n13 16\n46 30\n15 5\n49 10\n11 19\n32 4\n",
"20\n13 1\n18 2\n3 7\n18 5\n20 16\n3 12\n18 9\n3 10\n18 11\n13 6\n3 18\n20 15\n20 17\n3 13\n3 4\n13 14\n3 20\n18 8\n3 19\n",
"10\n8 2\n5 6\n1 8\n2 9\n1 4\n8 10\n10 5\n2 7\n2 3\n"
],
"output": [
"NO",
"YES",
"NO",
"NO",
"NO",
"YES",
"NO",
"NO",
"NO",
"NO",
"YES",
"NO",
"NO",
"YES",
"NO",
"NO",
"NO",
"YES",
"NO",
"YES",
"NO"
]
}
|
IN-CORRECT
|
cpp
|
#include<bits/stdc++.h>
#define ll long long int
#define getFaster ios_base::sync_with_stdio(false),cin.tie(NULL),cout.tie(NULL)
#define rep(i,init,n) for(int i=init;i<n;i++)
#define repl(i,init,n) for(ll i=init;i<n;i++)
#define rev(i,n,init) for(int i=n;i>=init;i--)
#define MAXN 600005
using namespace std;
const double PI = atan(1.0)*4;
const ll MOD=1e9+7;
int main()
{
getFaster;
int n;
cin>>n;
int deg[n+1]={0};
rep(i,1,n)
{
int x,y;
cin>>x>>y;
deg[x]++;
deg[y]++;
}
ll cnt=0;
rep(i,1,n+1)
{
if(deg[i]==1) cnt++;
}
ll x=(cnt*(cnt-1))/2;
ll y=n-1;
if(x>=y)
cout<<"YES"<<endl;
else
cout<<"NO"<<endl;
return 0;
}
|
1189_D1. Add on a Tree
|
Note that this is the first problem of the two similar problems. You can hack this problem only if you solve both problems.
You are given a tree with n nodes. In the beginning, 0 is written on all edges. In one operation, you can choose any 2 distinct leaves u, v and any real number x and add x to values written on all edges on the simple path between u and v.
For example, on the picture below you can see the result of applying two operations to the graph: adding 2 on the path from 7 to 6, and then adding -0.5 on the path from 4 to 5.
<image>
Is it true that for any configuration of real numbers written on edges, we can achieve it with a finite number of operations?
Leaf is a node of a tree of degree 1. Simple path is a path that doesn't contain any node twice.
Input
The first line contains a single integer n (2 ≤ n ≤ 10^5) — the number of nodes.
Each of the next n-1 lines contains two integers u and v (1 ≤ u, v ≤ n, u ≠ v), meaning that there is an edge between nodes u and v. It is guaranteed that these edges form a tree.
Output
If there is a configuration of real numbers written on edges of the tree that we can't achieve by performing the operations, output "NO".
Otherwise, output "YES".
You can print each letter in any case (upper or lower).
Examples
Input
2
1 2
Output
YES
Input
3
1 2
2 3
Output
NO
Input
5
1 2
1 3
1 4
2 5
Output
NO
Input
6
1 2
1 3
1 4
2 5
2 6
Output
YES
Note
In the first example, we can add any real x to the value written on the only edge (1, 2).
<image>
In the second example, one of configurations that we can't reach is 0 written on (1, 2) and 1 written on (2, 3).
<image>
Below you can see graphs from examples 3, 4:
<image> <image>
|
{
"input": [
"2\n1 2\n",
"3\n1 2\n2 3\n",
"5\n1 2\n1 3\n1 4\n2 5\n",
"6\n1 2\n1 3\n1 4\n2 5\n2 6\n"
],
"output": [
"YES",
"NO",
"NO",
"YES"
]
}
|
{
"input": [
"50\n16 4\n17 9\n31 19\n22 10\n8 1\n40 30\n3 31\n20 29\n47 27\n22 25\n32 34\n12 15\n40 32\n10 33\n47 12\n6 24\n46 41\n14 23\n12 35\n31 42\n46 28\n31 20\n46 37\n1 39\n29 49\n37 47\n40 6\n42 36\n47 2\n24 46\n2 13\n8 45\n41 3\n32 17\n4 7\n47 26\n28 8\n41 50\n34 44\n33 21\n25 5\n16 40\n3 14\n8 18\n28 11\n32 22\n2 38\n3 48\n44 43\n",
"10\n8 1\n1 2\n8 9\n8 5\n1 3\n1 10\n1 6\n1 7\n8 4\n",
"5\n5 1\n5 4\n4 3\n1 2\n",
"7\n1 2\n2 3\n1 4\n1 5\n3 6\n3 7\n",
"3\n1 3\n2 3\n",
"60\n26 6\n59 30\n31 12\n31 3\n38 23\n59 29\n53 9\n38 56\n53 54\n29 21\n17 55\n59 38\n26 16\n24 59\n24 25\n17 35\n24 41\n30 15\n31 27\n8 44\n26 5\n26 48\n8 32\n53 17\n3 34\n3 51\n30 28\n47 10\n53 60\n36 42\n24 53\n59 22\n53 40\n26 52\n36 4\n59 8\n29 37\n36 20\n17 47\n53 18\n3 50\n30 2\n17 7\n8 58\n59 1\n31 11\n24 26\n24 43\n53 57\n59 45\n47 13\n26 46\n17 33\n30 31\n26 39\n26 19\n24 36\n8 49\n38 14\n",
"7\n1 2\n2 3\n3 4\n3 5\n1 6\n1 7\n",
"20\n19 16\n19 18\n20 7\n9 4\n6 17\n14 2\n9 15\n2 13\n5 11\n19 12\n12 20\n16 9\n11 8\n19 5\n3 1\n19 14\n5 3\n12 10\n19 6\n",
"7\n1 2\n1 3\n2 4\n2 5\n3 6\n3 7\n",
"10\n9 5\n7 1\n9 10\n7 2\n5 4\n9 6\n2 9\n10 8\n1 3\n",
"4\n2 4\n2 3\n2 1\n",
"4\n1 4\n3 2\n1 3\n",
"3\n1 2\n1 3\n",
"5\n1 2\n1 5\n1 3\n1 4\n",
"20\n14 9\n12 13\n10 15\n2 1\n20 19\n16 6\n16 3\n17 14\n3 5\n2 11\n3 10\n15 8\n14 2\n6 4\n3 20\n5 18\n1 7\n1 16\n4 12\n",
"20\n7 5\n14 13\n17 6\n3 8\n16 12\n18 9\n3 18\n14 1\n17 3\n15 2\n17 4\n9 11\n2 7\n15 17\n3 20\n16 10\n17 14\n2 16\n1 19\n",
"8\n1 2\n2 3\n3 4\n1 7\n1 8\n4 5\n4 6\n",
"5\n5 1\n5 2\n5 3\n5 4\n",
"50\n49 6\n43 7\n1 27\n19 35\n15 37\n16 12\n19 21\n16 28\n49 9\n48 39\n13 1\n2 48\n9 50\n44 3\n41 32\n48 31\n49 33\n6 11\n13 20\n49 22\n13 41\n48 29\n13 46\n15 47\n34 2\n49 13\n48 14\n34 24\n16 36\n13 40\n49 34\n49 17\n43 25\n11 23\n10 15\n19 26\n34 44\n16 42\n19 18\n46 8\n29 38\n1 45\n12 43\n13 16\n46 30\n15 5\n49 10\n11 19\n32 4\n",
"20\n13 1\n18 2\n3 7\n18 5\n20 16\n3 12\n18 9\n3 10\n18 11\n13 6\n3 18\n20 15\n20 17\n3 13\n3 4\n13 14\n3 20\n18 8\n3 19\n",
"10\n8 2\n5 6\n1 8\n2 9\n1 4\n8 10\n10 5\n2 7\n2 3\n"
],
"output": [
"NO",
"YES",
"NO",
"NO",
"NO",
"YES",
"NO",
"NO",
"NO",
"NO",
"YES",
"NO",
"NO",
"YES",
"NO",
"NO",
"NO",
"YES",
"NO",
"YES",
"NO"
]
}
|
IN-CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
using namespace std;
vector<int> conn[200100];
bool oka = true;
void dfs(int u, int par) {
if (par != -1 && conn[u].size() == 1)
;
else {
if (conn[u].size() == 1) return;
if (conn[u].size() < 3 || oka == false) {
oka = false;
return;
}
}
for (int v : conn[u]) {
if (v == par) continue;
dfs(v, u);
}
}
int main() {
int n;
cin >> n;
int u, v;
for (int i = 1; i <= n - 1; i++) {
cin >> u >> v;
conn[u].push_back(v);
conn[v].push_back(u);
}
dfs(1, -1);
if (oka)
cout << "YES" << endl;
else
cout << "NO" << endl;
return 0;
}
|
1189_D1. Add on a Tree
|
Note that this is the first problem of the two similar problems. You can hack this problem only if you solve both problems.
You are given a tree with n nodes. In the beginning, 0 is written on all edges. In one operation, you can choose any 2 distinct leaves u, v and any real number x and add x to values written on all edges on the simple path between u and v.
For example, on the picture below you can see the result of applying two operations to the graph: adding 2 on the path from 7 to 6, and then adding -0.5 on the path from 4 to 5.
<image>
Is it true that for any configuration of real numbers written on edges, we can achieve it with a finite number of operations?
Leaf is a node of a tree of degree 1. Simple path is a path that doesn't contain any node twice.
Input
The first line contains a single integer n (2 ≤ n ≤ 10^5) — the number of nodes.
Each of the next n-1 lines contains two integers u and v (1 ≤ u, v ≤ n, u ≠ v), meaning that there is an edge between nodes u and v. It is guaranteed that these edges form a tree.
Output
If there is a configuration of real numbers written on edges of the tree that we can't achieve by performing the operations, output "NO".
Otherwise, output "YES".
You can print each letter in any case (upper or lower).
Examples
Input
2
1 2
Output
YES
Input
3
1 2
2 3
Output
NO
Input
5
1 2
1 3
1 4
2 5
Output
NO
Input
6
1 2
1 3
1 4
2 5
2 6
Output
YES
Note
In the first example, we can add any real x to the value written on the only edge (1, 2).
<image>
In the second example, one of configurations that we can't reach is 0 written on (1, 2) and 1 written on (2, 3).
<image>
Below you can see graphs from examples 3, 4:
<image> <image>
|
{
"input": [
"2\n1 2\n",
"3\n1 2\n2 3\n",
"5\n1 2\n1 3\n1 4\n2 5\n",
"6\n1 2\n1 3\n1 4\n2 5\n2 6\n"
],
"output": [
"YES",
"NO",
"NO",
"YES"
]
}
|
{
"input": [
"50\n16 4\n17 9\n31 19\n22 10\n8 1\n40 30\n3 31\n20 29\n47 27\n22 25\n32 34\n12 15\n40 32\n10 33\n47 12\n6 24\n46 41\n14 23\n12 35\n31 42\n46 28\n31 20\n46 37\n1 39\n29 49\n37 47\n40 6\n42 36\n47 2\n24 46\n2 13\n8 45\n41 3\n32 17\n4 7\n47 26\n28 8\n41 50\n34 44\n33 21\n25 5\n16 40\n3 14\n8 18\n28 11\n32 22\n2 38\n3 48\n44 43\n",
"10\n8 1\n1 2\n8 9\n8 5\n1 3\n1 10\n1 6\n1 7\n8 4\n",
"5\n5 1\n5 4\n4 3\n1 2\n",
"7\n1 2\n2 3\n1 4\n1 5\n3 6\n3 7\n",
"3\n1 3\n2 3\n",
"60\n26 6\n59 30\n31 12\n31 3\n38 23\n59 29\n53 9\n38 56\n53 54\n29 21\n17 55\n59 38\n26 16\n24 59\n24 25\n17 35\n24 41\n30 15\n31 27\n8 44\n26 5\n26 48\n8 32\n53 17\n3 34\n3 51\n30 28\n47 10\n53 60\n36 42\n24 53\n59 22\n53 40\n26 52\n36 4\n59 8\n29 37\n36 20\n17 47\n53 18\n3 50\n30 2\n17 7\n8 58\n59 1\n31 11\n24 26\n24 43\n53 57\n59 45\n47 13\n26 46\n17 33\n30 31\n26 39\n26 19\n24 36\n8 49\n38 14\n",
"7\n1 2\n2 3\n3 4\n3 5\n1 6\n1 7\n",
"20\n19 16\n19 18\n20 7\n9 4\n6 17\n14 2\n9 15\n2 13\n5 11\n19 12\n12 20\n16 9\n11 8\n19 5\n3 1\n19 14\n5 3\n12 10\n19 6\n",
"7\n1 2\n1 3\n2 4\n2 5\n3 6\n3 7\n",
"10\n9 5\n7 1\n9 10\n7 2\n5 4\n9 6\n2 9\n10 8\n1 3\n",
"4\n2 4\n2 3\n2 1\n",
"4\n1 4\n3 2\n1 3\n",
"3\n1 2\n1 3\n",
"5\n1 2\n1 5\n1 3\n1 4\n",
"20\n14 9\n12 13\n10 15\n2 1\n20 19\n16 6\n16 3\n17 14\n3 5\n2 11\n3 10\n15 8\n14 2\n6 4\n3 20\n5 18\n1 7\n1 16\n4 12\n",
"20\n7 5\n14 13\n17 6\n3 8\n16 12\n18 9\n3 18\n14 1\n17 3\n15 2\n17 4\n9 11\n2 7\n15 17\n3 20\n16 10\n17 14\n2 16\n1 19\n",
"8\n1 2\n2 3\n3 4\n1 7\n1 8\n4 5\n4 6\n",
"5\n5 1\n5 2\n5 3\n5 4\n",
"50\n49 6\n43 7\n1 27\n19 35\n15 37\n16 12\n19 21\n16 28\n49 9\n48 39\n13 1\n2 48\n9 50\n44 3\n41 32\n48 31\n49 33\n6 11\n13 20\n49 22\n13 41\n48 29\n13 46\n15 47\n34 2\n49 13\n48 14\n34 24\n16 36\n13 40\n49 34\n49 17\n43 25\n11 23\n10 15\n19 26\n34 44\n16 42\n19 18\n46 8\n29 38\n1 45\n12 43\n13 16\n46 30\n15 5\n49 10\n11 19\n32 4\n",
"20\n13 1\n18 2\n3 7\n18 5\n20 16\n3 12\n18 9\n3 10\n18 11\n13 6\n3 18\n20 15\n20 17\n3 13\n3 4\n13 14\n3 20\n18 8\n3 19\n",
"10\n8 2\n5 6\n1 8\n2 9\n1 4\n8 10\n10 5\n2 7\n2 3\n"
],
"output": [
"NO",
"YES",
"NO",
"NO",
"NO",
"YES",
"NO",
"NO",
"NO",
"NO",
"YES",
"NO",
"NO",
"YES",
"NO",
"NO",
"NO",
"YES",
"NO",
"YES",
"NO"
]
}
|
IN-CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
const long long MOD = 1e9 + 7;
int n, x, y, sons[100010], all;
vector<int> g[100010];
int dfs(int pos, int no) {
for (int i = 0; i < g[pos].size(); i++)
if (g[pos][i] != no) sons[pos] += dfs(g[pos][i], pos);
return sons[pos] + (g[pos].size() == 1 ? 1 : 0);
}
void dfs2(int pos, int no) {
for (int i = 0; i < g[pos].size(); i++) {
if (g[pos][i] == no) continue;
if (sons[g[pos][i]] == 1 || all - sons[g[pos][i]] -
(g[pos].size() == 1 ? 1 : 0) -
(g[g[pos][i]].size() == 1 ? 1 : 0) ==
1) {
puts("NO");
exit(0);
}
dfs2(g[pos][i], pos);
}
}
int main() {
cin >> n;
for (int i = 0; i < n - 1; i++) {
scanf("%d%d", &x, &y);
g[x].push_back(y);
g[y].push_back(x);
}
dfs(1, 0);
all = sons[1] + (g[1].size() == 1 ? 1 : 0);
dfs2(1, 0);
puts("YES");
return 0;
}
|
1189_D1. Add on a Tree
|
Note that this is the first problem of the two similar problems. You can hack this problem only if you solve both problems.
You are given a tree with n nodes. In the beginning, 0 is written on all edges. In one operation, you can choose any 2 distinct leaves u, v and any real number x and add x to values written on all edges on the simple path between u and v.
For example, on the picture below you can see the result of applying two operations to the graph: adding 2 on the path from 7 to 6, and then adding -0.5 on the path from 4 to 5.
<image>
Is it true that for any configuration of real numbers written on edges, we can achieve it with a finite number of operations?
Leaf is a node of a tree of degree 1. Simple path is a path that doesn't contain any node twice.
Input
The first line contains a single integer n (2 ≤ n ≤ 10^5) — the number of nodes.
Each of the next n-1 lines contains two integers u and v (1 ≤ u, v ≤ n, u ≠ v), meaning that there is an edge between nodes u and v. It is guaranteed that these edges form a tree.
Output
If there is a configuration of real numbers written on edges of the tree that we can't achieve by performing the operations, output "NO".
Otherwise, output "YES".
You can print each letter in any case (upper or lower).
Examples
Input
2
1 2
Output
YES
Input
3
1 2
2 3
Output
NO
Input
5
1 2
1 3
1 4
2 5
Output
NO
Input
6
1 2
1 3
1 4
2 5
2 6
Output
YES
Note
In the first example, we can add any real x to the value written on the only edge (1, 2).
<image>
In the second example, one of configurations that we can't reach is 0 written on (1, 2) and 1 written on (2, 3).
<image>
Below you can see graphs from examples 3, 4:
<image> <image>
|
{
"input": [
"2\n1 2\n",
"3\n1 2\n2 3\n",
"5\n1 2\n1 3\n1 4\n2 5\n",
"6\n1 2\n1 3\n1 4\n2 5\n2 6\n"
],
"output": [
"YES",
"NO",
"NO",
"YES"
]
}
|
{
"input": [
"50\n16 4\n17 9\n31 19\n22 10\n8 1\n40 30\n3 31\n20 29\n47 27\n22 25\n32 34\n12 15\n40 32\n10 33\n47 12\n6 24\n46 41\n14 23\n12 35\n31 42\n46 28\n31 20\n46 37\n1 39\n29 49\n37 47\n40 6\n42 36\n47 2\n24 46\n2 13\n8 45\n41 3\n32 17\n4 7\n47 26\n28 8\n41 50\n34 44\n33 21\n25 5\n16 40\n3 14\n8 18\n28 11\n32 22\n2 38\n3 48\n44 43\n",
"10\n8 1\n1 2\n8 9\n8 5\n1 3\n1 10\n1 6\n1 7\n8 4\n",
"5\n5 1\n5 4\n4 3\n1 2\n",
"7\n1 2\n2 3\n1 4\n1 5\n3 6\n3 7\n",
"3\n1 3\n2 3\n",
"60\n26 6\n59 30\n31 12\n31 3\n38 23\n59 29\n53 9\n38 56\n53 54\n29 21\n17 55\n59 38\n26 16\n24 59\n24 25\n17 35\n24 41\n30 15\n31 27\n8 44\n26 5\n26 48\n8 32\n53 17\n3 34\n3 51\n30 28\n47 10\n53 60\n36 42\n24 53\n59 22\n53 40\n26 52\n36 4\n59 8\n29 37\n36 20\n17 47\n53 18\n3 50\n30 2\n17 7\n8 58\n59 1\n31 11\n24 26\n24 43\n53 57\n59 45\n47 13\n26 46\n17 33\n30 31\n26 39\n26 19\n24 36\n8 49\n38 14\n",
"7\n1 2\n2 3\n3 4\n3 5\n1 6\n1 7\n",
"20\n19 16\n19 18\n20 7\n9 4\n6 17\n14 2\n9 15\n2 13\n5 11\n19 12\n12 20\n16 9\n11 8\n19 5\n3 1\n19 14\n5 3\n12 10\n19 6\n",
"7\n1 2\n1 3\n2 4\n2 5\n3 6\n3 7\n",
"10\n9 5\n7 1\n9 10\n7 2\n5 4\n9 6\n2 9\n10 8\n1 3\n",
"4\n2 4\n2 3\n2 1\n",
"4\n1 4\n3 2\n1 3\n",
"3\n1 2\n1 3\n",
"5\n1 2\n1 5\n1 3\n1 4\n",
"20\n14 9\n12 13\n10 15\n2 1\n20 19\n16 6\n16 3\n17 14\n3 5\n2 11\n3 10\n15 8\n14 2\n6 4\n3 20\n5 18\n1 7\n1 16\n4 12\n",
"20\n7 5\n14 13\n17 6\n3 8\n16 12\n18 9\n3 18\n14 1\n17 3\n15 2\n17 4\n9 11\n2 7\n15 17\n3 20\n16 10\n17 14\n2 16\n1 19\n",
"8\n1 2\n2 3\n3 4\n1 7\n1 8\n4 5\n4 6\n",
"5\n5 1\n5 2\n5 3\n5 4\n",
"50\n49 6\n43 7\n1 27\n19 35\n15 37\n16 12\n19 21\n16 28\n49 9\n48 39\n13 1\n2 48\n9 50\n44 3\n41 32\n48 31\n49 33\n6 11\n13 20\n49 22\n13 41\n48 29\n13 46\n15 47\n34 2\n49 13\n48 14\n34 24\n16 36\n13 40\n49 34\n49 17\n43 25\n11 23\n10 15\n19 26\n34 44\n16 42\n19 18\n46 8\n29 38\n1 45\n12 43\n13 16\n46 30\n15 5\n49 10\n11 19\n32 4\n",
"20\n13 1\n18 2\n3 7\n18 5\n20 16\n3 12\n18 9\n3 10\n18 11\n13 6\n3 18\n20 15\n20 17\n3 13\n3 4\n13 14\n3 20\n18 8\n3 19\n",
"10\n8 2\n5 6\n1 8\n2 9\n1 4\n8 10\n10 5\n2 7\n2 3\n"
],
"output": [
"NO",
"YES",
"NO",
"NO",
"NO",
"YES",
"NO",
"NO",
"NO",
"NO",
"YES",
"NO",
"NO",
"YES",
"NO",
"NO",
"NO",
"YES",
"NO",
"YES",
"NO"
]
}
|
IN-CORRECT
|
python3
|
n=int(input())
tree=[[] for i in range(n)]
for i in range(n-1):
u,v=map(int,input().split())
tree[u-1].append(v-1)
tree[v-1].append(u-1)
if n==2:
print("YES")
exit()
elif n==3:
print("NO")
exit()
#nが4以上
check=[0]*n
for i in range(n):
if len(tree[i])==1:
check[tree[i][0]]+=1
for i in range(n):
if check[i]==1:
print("NO")
break
else:
print("YES")
|
1189_D1. Add on a Tree
|
Note that this is the first problem of the two similar problems. You can hack this problem only if you solve both problems.
You are given a tree with n nodes. In the beginning, 0 is written on all edges. In one operation, you can choose any 2 distinct leaves u, v and any real number x and add x to values written on all edges on the simple path between u and v.
For example, on the picture below you can see the result of applying two operations to the graph: adding 2 on the path from 7 to 6, and then adding -0.5 on the path from 4 to 5.
<image>
Is it true that for any configuration of real numbers written on edges, we can achieve it with a finite number of operations?
Leaf is a node of a tree of degree 1. Simple path is a path that doesn't contain any node twice.
Input
The first line contains a single integer n (2 ≤ n ≤ 10^5) — the number of nodes.
Each of the next n-1 lines contains two integers u and v (1 ≤ u, v ≤ n, u ≠ v), meaning that there is an edge between nodes u and v. It is guaranteed that these edges form a tree.
Output
If there is a configuration of real numbers written on edges of the tree that we can't achieve by performing the operations, output "NO".
Otherwise, output "YES".
You can print each letter in any case (upper or lower).
Examples
Input
2
1 2
Output
YES
Input
3
1 2
2 3
Output
NO
Input
5
1 2
1 3
1 4
2 5
Output
NO
Input
6
1 2
1 3
1 4
2 5
2 6
Output
YES
Note
In the first example, we can add any real x to the value written on the only edge (1, 2).
<image>
In the second example, one of configurations that we can't reach is 0 written on (1, 2) and 1 written on (2, 3).
<image>
Below you can see graphs from examples 3, 4:
<image> <image>
|
{
"input": [
"2\n1 2\n",
"3\n1 2\n2 3\n",
"5\n1 2\n1 3\n1 4\n2 5\n",
"6\n1 2\n1 3\n1 4\n2 5\n2 6\n"
],
"output": [
"YES",
"NO",
"NO",
"YES"
]
}
|
{
"input": [
"50\n16 4\n17 9\n31 19\n22 10\n8 1\n40 30\n3 31\n20 29\n47 27\n22 25\n32 34\n12 15\n40 32\n10 33\n47 12\n6 24\n46 41\n14 23\n12 35\n31 42\n46 28\n31 20\n46 37\n1 39\n29 49\n37 47\n40 6\n42 36\n47 2\n24 46\n2 13\n8 45\n41 3\n32 17\n4 7\n47 26\n28 8\n41 50\n34 44\n33 21\n25 5\n16 40\n3 14\n8 18\n28 11\n32 22\n2 38\n3 48\n44 43\n",
"10\n8 1\n1 2\n8 9\n8 5\n1 3\n1 10\n1 6\n1 7\n8 4\n",
"5\n5 1\n5 4\n4 3\n1 2\n",
"7\n1 2\n2 3\n1 4\n1 5\n3 6\n3 7\n",
"3\n1 3\n2 3\n",
"60\n26 6\n59 30\n31 12\n31 3\n38 23\n59 29\n53 9\n38 56\n53 54\n29 21\n17 55\n59 38\n26 16\n24 59\n24 25\n17 35\n24 41\n30 15\n31 27\n8 44\n26 5\n26 48\n8 32\n53 17\n3 34\n3 51\n30 28\n47 10\n53 60\n36 42\n24 53\n59 22\n53 40\n26 52\n36 4\n59 8\n29 37\n36 20\n17 47\n53 18\n3 50\n30 2\n17 7\n8 58\n59 1\n31 11\n24 26\n24 43\n53 57\n59 45\n47 13\n26 46\n17 33\n30 31\n26 39\n26 19\n24 36\n8 49\n38 14\n",
"7\n1 2\n2 3\n3 4\n3 5\n1 6\n1 7\n",
"20\n19 16\n19 18\n20 7\n9 4\n6 17\n14 2\n9 15\n2 13\n5 11\n19 12\n12 20\n16 9\n11 8\n19 5\n3 1\n19 14\n5 3\n12 10\n19 6\n",
"7\n1 2\n1 3\n2 4\n2 5\n3 6\n3 7\n",
"10\n9 5\n7 1\n9 10\n7 2\n5 4\n9 6\n2 9\n10 8\n1 3\n",
"4\n2 4\n2 3\n2 1\n",
"4\n1 4\n3 2\n1 3\n",
"3\n1 2\n1 3\n",
"5\n1 2\n1 5\n1 3\n1 4\n",
"20\n14 9\n12 13\n10 15\n2 1\n20 19\n16 6\n16 3\n17 14\n3 5\n2 11\n3 10\n15 8\n14 2\n6 4\n3 20\n5 18\n1 7\n1 16\n4 12\n",
"20\n7 5\n14 13\n17 6\n3 8\n16 12\n18 9\n3 18\n14 1\n17 3\n15 2\n17 4\n9 11\n2 7\n15 17\n3 20\n16 10\n17 14\n2 16\n1 19\n",
"8\n1 2\n2 3\n3 4\n1 7\n1 8\n4 5\n4 6\n",
"5\n5 1\n5 2\n5 3\n5 4\n",
"50\n49 6\n43 7\n1 27\n19 35\n15 37\n16 12\n19 21\n16 28\n49 9\n48 39\n13 1\n2 48\n9 50\n44 3\n41 32\n48 31\n49 33\n6 11\n13 20\n49 22\n13 41\n48 29\n13 46\n15 47\n34 2\n49 13\n48 14\n34 24\n16 36\n13 40\n49 34\n49 17\n43 25\n11 23\n10 15\n19 26\n34 44\n16 42\n19 18\n46 8\n29 38\n1 45\n12 43\n13 16\n46 30\n15 5\n49 10\n11 19\n32 4\n",
"20\n13 1\n18 2\n3 7\n18 5\n20 16\n3 12\n18 9\n3 10\n18 11\n13 6\n3 18\n20 15\n20 17\n3 13\n3 4\n13 14\n3 20\n18 8\n3 19\n",
"10\n8 2\n5 6\n1 8\n2 9\n1 4\n8 10\n10 5\n2 7\n2 3\n"
],
"output": [
"NO",
"YES",
"NO",
"NO",
"NO",
"YES",
"NO",
"NO",
"NO",
"NO",
"YES",
"NO",
"NO",
"YES",
"NO",
"NO",
"NO",
"YES",
"NO",
"YES",
"NO"
]
}
|
IN-CORRECT
|
java
|
/**
* BaZ :D
*/
import java.util.*;
import java.io.*;
import static java.lang.Math.*;
public class ACMIND
{
static FastReader scan;
static PrintWriter pw;
static long MOD = 1_000_000_007;
static long INF = 1_000_000_000_000_000_000L;
static long inf = 2_000_000_000;
public static void main(String[] args) {
new Thread(null,null,"BaZ",1<<25)
{
public void run()
{
try
{
solve();
}
catch(Exception e)
{
e.printStackTrace();
System.exit(1);
}
}
}.start();
}
static ArrayList<Integer> adj[];
static int sz[], total_leaf = 0;
static boolean possible = true;
static void solve() throws IOException
{
scan = new FastReader();
pw = new PrintWriter(System.out,true);
StringBuilder sb = new StringBuilder();
int n = ni();
adj = new ArrayList[n+1];
for(int i=1;i<=n;++i) {
adj[i] = new ArrayList<>();
}
for(int i=1;i<n;++i) {
int u = ni(), v = ni();
adj[u].add(v);
adj[v].add(u);
}
for(int i=1;i<=n;++i) {
if(adj[i].size()==1) {
++total_leaf;
}
}
sz = new int[n+1];
dfs(1,1);
if(possible) {
pl("YES");
}
else {
pl("NO");
}
pw.flush();
pw.close();
}
static void dfs(int v, int par) {
int leaf = 0;
for(int e : adj[v]) {
if(e==par) {
continue;
}
dfs(e, v);
sz[v]+=sz[e];
if(sz[e]==1) {
leaf++;
}
}
sz[v]++;
for(int e : adj[v]) {
if(e==par) {
continue;
}
if(sz[e]==1 && v!=1 && leaf==1) {
possible = false;
}
}
}
static int ni() throws IOException
{
return scan.nextInt();
}
static long nl() throws IOException
{
return scan.nextLong();
}
static double nd() throws IOException
{
return scan.nextDouble();
}
static void pl()
{
pw.println();
}
static void p(Object o)
{
pw.print(o+" ");
}
static void pl(Object o)
{
pw.println(o);
}
static void psb(StringBuilder sb)
{
pw.print(sb);
}
static void pa(String arrayName, Object arr[])
{
pl(arrayName+" : ");
for(Object o : arr)
p(o);
pl();
}
static void pa(String arrayName, int arr[])
{
pl(arrayName+" : ");
for(int o : arr)
p(o);
pl();
}
static void pa(String arrayName, long arr[])
{
pl(arrayName+" : ");
for(long o : arr)
p(o);
pl();
}
static void pa(String arrayName, double arr[])
{
pl(arrayName+" : ");
for(double o : arr)
p(o);
pl();
}
static void pa(String arrayName, char arr[])
{
pl(arrayName+" : ");
for(char o : arr)
p(o);
pl();
}
static void pa(String listName, List list)
{
pl(listName+" : ");
for(Object o : list)
p(o);
pl();
}
static void pa(String arrayName, Object[][] arr) {
pl(arrayName+" : ");
for(int i=0;i<arr.length;++i) {
for(Object o : arr[i])
p(o);
pl();
}
}
static void pa(String arrayName, int[][] arr) {
pl(arrayName+" : ");
for(int i=0;i<arr.length;++i) {
for(int o : arr[i])
p(o);
pl();
}
}
static void pa(String arrayName, long[][] arr) {
pl(arrayName+" : ");
for(int i=0;i<arr.length;++i) {
for(long o : arr[i])
p(o);
pl();
}
}
static void pa(String arrayName, char[][] arr) {
pl(arrayName+" : ");
for(int i=0;i<arr.length;++i) {
for(char o : arr[i])
p(o);
pl();
}
}
static void pa(String arrayName, double[][] arr) {
pl(arrayName+" : ");
for(int i=0;i<arr.length;++i) {
for(double o : arr[i])
p(o);
pl();
}
}
static class FastReader {
final private int BUFFER_SIZE = 1 << 16;
private DataInputStream din;
private byte[] buffer;
private int bufferPointer, bytesRead;
public FastReader() {
din = new DataInputStream(System.in);
buffer = new byte[BUFFER_SIZE];
bufferPointer = bytesRead = 0;
}
public FastReader(String file_name) throws IOException {
din = new DataInputStream(new FileInputStream(file_name));
buffer = new byte[BUFFER_SIZE];
bufferPointer = bytesRead = 0;
}
public String readLine() throws IOException {
byte[] buf = new byte[1000000];
int cnt = 0, c;
while ((c = read()) != -1) {
if (c == '\n') break;
buf[cnt++] = (byte) c;
}
return new String(buf, 0, cnt);
}
public int nextInt() throws IOException {
int ret = 0;
byte c = read();
while (c <= ' ') c = read();
boolean neg = (c == '-');
if (neg) c = read();
do {
ret = ret * 10 + c - '0';
} while ((c = read()) >= '0' && c <= '9');
if (neg) return -ret;
return ret;
}
public long nextLong() throws IOException {
long ret = 0;
byte c = read();
while (c <= ' ') c = read();
boolean neg = (c == '-');
if (neg) c = read();
do {
ret = ret * 10 + c - '0';
} while ((c = read()) >= '0' && c <= '9');
if (neg) return -ret;
return ret;
}
public double nextDouble() throws IOException {
double ret = 0, div = 1;
byte c = read();
while (c <= ' ') c = read();
boolean neg = (c == '-');
if (neg) c = read();
do {
ret = ret * 10 + c - '0';
} while ((c = read()) >= '0' && c <= '9');
if (c == '.') while ((c = read()) >= '0' && c <= '9') ret += (c - '0') / (div *= 10);
if (neg) return -ret;
return ret;
}
private void fillBuffer() throws IOException {
bytesRead = din.read(buffer, bufferPointer = 0, BUFFER_SIZE);
if (bytesRead == -1) buffer[0] = -1;
}
private byte read() throws IOException {
if (bufferPointer == bytesRead) fillBuffer();
return buffer[bufferPointer++];
}
public void close() throws IOException {
if (din == null) return;
din.close();
}
}
}
|
1189_D1. Add on a Tree
|
Note that this is the first problem of the two similar problems. You can hack this problem only if you solve both problems.
You are given a tree with n nodes. In the beginning, 0 is written on all edges. In one operation, you can choose any 2 distinct leaves u, v and any real number x and add x to values written on all edges on the simple path between u and v.
For example, on the picture below you can see the result of applying two operations to the graph: adding 2 on the path from 7 to 6, and then adding -0.5 on the path from 4 to 5.
<image>
Is it true that for any configuration of real numbers written on edges, we can achieve it with a finite number of operations?
Leaf is a node of a tree of degree 1. Simple path is a path that doesn't contain any node twice.
Input
The first line contains a single integer n (2 ≤ n ≤ 10^5) — the number of nodes.
Each of the next n-1 lines contains two integers u and v (1 ≤ u, v ≤ n, u ≠ v), meaning that there is an edge between nodes u and v. It is guaranteed that these edges form a tree.
Output
If there is a configuration of real numbers written on edges of the tree that we can't achieve by performing the operations, output "NO".
Otherwise, output "YES".
You can print each letter in any case (upper or lower).
Examples
Input
2
1 2
Output
YES
Input
3
1 2
2 3
Output
NO
Input
5
1 2
1 3
1 4
2 5
Output
NO
Input
6
1 2
1 3
1 4
2 5
2 6
Output
YES
Note
In the first example, we can add any real x to the value written on the only edge (1, 2).
<image>
In the second example, one of configurations that we can't reach is 0 written on (1, 2) and 1 written on (2, 3).
<image>
Below you can see graphs from examples 3, 4:
<image> <image>
|
{
"input": [
"2\n1 2\n",
"3\n1 2\n2 3\n",
"5\n1 2\n1 3\n1 4\n2 5\n",
"6\n1 2\n1 3\n1 4\n2 5\n2 6\n"
],
"output": [
"YES",
"NO",
"NO",
"YES"
]
}
|
{
"input": [
"50\n16 4\n17 9\n31 19\n22 10\n8 1\n40 30\n3 31\n20 29\n47 27\n22 25\n32 34\n12 15\n40 32\n10 33\n47 12\n6 24\n46 41\n14 23\n12 35\n31 42\n46 28\n31 20\n46 37\n1 39\n29 49\n37 47\n40 6\n42 36\n47 2\n24 46\n2 13\n8 45\n41 3\n32 17\n4 7\n47 26\n28 8\n41 50\n34 44\n33 21\n25 5\n16 40\n3 14\n8 18\n28 11\n32 22\n2 38\n3 48\n44 43\n",
"10\n8 1\n1 2\n8 9\n8 5\n1 3\n1 10\n1 6\n1 7\n8 4\n",
"5\n5 1\n5 4\n4 3\n1 2\n",
"7\n1 2\n2 3\n1 4\n1 5\n3 6\n3 7\n",
"3\n1 3\n2 3\n",
"60\n26 6\n59 30\n31 12\n31 3\n38 23\n59 29\n53 9\n38 56\n53 54\n29 21\n17 55\n59 38\n26 16\n24 59\n24 25\n17 35\n24 41\n30 15\n31 27\n8 44\n26 5\n26 48\n8 32\n53 17\n3 34\n3 51\n30 28\n47 10\n53 60\n36 42\n24 53\n59 22\n53 40\n26 52\n36 4\n59 8\n29 37\n36 20\n17 47\n53 18\n3 50\n30 2\n17 7\n8 58\n59 1\n31 11\n24 26\n24 43\n53 57\n59 45\n47 13\n26 46\n17 33\n30 31\n26 39\n26 19\n24 36\n8 49\n38 14\n",
"7\n1 2\n2 3\n3 4\n3 5\n1 6\n1 7\n",
"20\n19 16\n19 18\n20 7\n9 4\n6 17\n14 2\n9 15\n2 13\n5 11\n19 12\n12 20\n16 9\n11 8\n19 5\n3 1\n19 14\n5 3\n12 10\n19 6\n",
"7\n1 2\n1 3\n2 4\n2 5\n3 6\n3 7\n",
"10\n9 5\n7 1\n9 10\n7 2\n5 4\n9 6\n2 9\n10 8\n1 3\n",
"4\n2 4\n2 3\n2 1\n",
"4\n1 4\n3 2\n1 3\n",
"3\n1 2\n1 3\n",
"5\n1 2\n1 5\n1 3\n1 4\n",
"20\n14 9\n12 13\n10 15\n2 1\n20 19\n16 6\n16 3\n17 14\n3 5\n2 11\n3 10\n15 8\n14 2\n6 4\n3 20\n5 18\n1 7\n1 16\n4 12\n",
"20\n7 5\n14 13\n17 6\n3 8\n16 12\n18 9\n3 18\n14 1\n17 3\n15 2\n17 4\n9 11\n2 7\n15 17\n3 20\n16 10\n17 14\n2 16\n1 19\n",
"8\n1 2\n2 3\n3 4\n1 7\n1 8\n4 5\n4 6\n",
"5\n5 1\n5 2\n5 3\n5 4\n",
"50\n49 6\n43 7\n1 27\n19 35\n15 37\n16 12\n19 21\n16 28\n49 9\n48 39\n13 1\n2 48\n9 50\n44 3\n41 32\n48 31\n49 33\n6 11\n13 20\n49 22\n13 41\n48 29\n13 46\n15 47\n34 2\n49 13\n48 14\n34 24\n16 36\n13 40\n49 34\n49 17\n43 25\n11 23\n10 15\n19 26\n34 44\n16 42\n19 18\n46 8\n29 38\n1 45\n12 43\n13 16\n46 30\n15 5\n49 10\n11 19\n32 4\n",
"20\n13 1\n18 2\n3 7\n18 5\n20 16\n3 12\n18 9\n3 10\n18 11\n13 6\n3 18\n20 15\n20 17\n3 13\n3 4\n13 14\n3 20\n18 8\n3 19\n",
"10\n8 2\n5 6\n1 8\n2 9\n1 4\n8 10\n10 5\n2 7\n2 3\n"
],
"output": [
"NO",
"YES",
"NO",
"NO",
"NO",
"YES",
"NO",
"NO",
"NO",
"NO",
"YES",
"NO",
"NO",
"YES",
"NO",
"NO",
"NO",
"YES",
"NO",
"YES",
"NO"
]
}
|
IN-CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
long long in[100005];
int main() {
ios_base::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
long long n, x, y;
cin >> n;
for (long long i = 0; i < n - 1; i++) {
cin >> x >> y;
in[x]++;
in[y]++;
}
bool first = 1;
for (long long i = 1; i < n + 1; i++) {
if (in[i] % 2 == 0) {
first = 0;
break;
}
}
if (first) {
cout << "YES"
<< "\n";
} else {
cout << "NO"
<< "\n";
}
return 0;
}
|
1189_D1. Add on a Tree
|
Note that this is the first problem of the two similar problems. You can hack this problem only if you solve both problems.
You are given a tree with n nodes. In the beginning, 0 is written on all edges. In one operation, you can choose any 2 distinct leaves u, v and any real number x and add x to values written on all edges on the simple path between u and v.
For example, on the picture below you can see the result of applying two operations to the graph: adding 2 on the path from 7 to 6, and then adding -0.5 on the path from 4 to 5.
<image>
Is it true that for any configuration of real numbers written on edges, we can achieve it with a finite number of operations?
Leaf is a node of a tree of degree 1. Simple path is a path that doesn't contain any node twice.
Input
The first line contains a single integer n (2 ≤ n ≤ 10^5) — the number of nodes.
Each of the next n-1 lines contains two integers u and v (1 ≤ u, v ≤ n, u ≠ v), meaning that there is an edge between nodes u and v. It is guaranteed that these edges form a tree.
Output
If there is a configuration of real numbers written on edges of the tree that we can't achieve by performing the operations, output "NO".
Otherwise, output "YES".
You can print each letter in any case (upper or lower).
Examples
Input
2
1 2
Output
YES
Input
3
1 2
2 3
Output
NO
Input
5
1 2
1 3
1 4
2 5
Output
NO
Input
6
1 2
1 3
1 4
2 5
2 6
Output
YES
Note
In the first example, we can add any real x to the value written on the only edge (1, 2).
<image>
In the second example, one of configurations that we can't reach is 0 written on (1, 2) and 1 written on (2, 3).
<image>
Below you can see graphs from examples 3, 4:
<image> <image>
|
{
"input": [
"2\n1 2\n",
"3\n1 2\n2 3\n",
"5\n1 2\n1 3\n1 4\n2 5\n",
"6\n1 2\n1 3\n1 4\n2 5\n2 6\n"
],
"output": [
"YES",
"NO",
"NO",
"YES"
]
}
|
{
"input": [
"50\n16 4\n17 9\n31 19\n22 10\n8 1\n40 30\n3 31\n20 29\n47 27\n22 25\n32 34\n12 15\n40 32\n10 33\n47 12\n6 24\n46 41\n14 23\n12 35\n31 42\n46 28\n31 20\n46 37\n1 39\n29 49\n37 47\n40 6\n42 36\n47 2\n24 46\n2 13\n8 45\n41 3\n32 17\n4 7\n47 26\n28 8\n41 50\n34 44\n33 21\n25 5\n16 40\n3 14\n8 18\n28 11\n32 22\n2 38\n3 48\n44 43\n",
"10\n8 1\n1 2\n8 9\n8 5\n1 3\n1 10\n1 6\n1 7\n8 4\n",
"5\n5 1\n5 4\n4 3\n1 2\n",
"7\n1 2\n2 3\n1 4\n1 5\n3 6\n3 7\n",
"3\n1 3\n2 3\n",
"60\n26 6\n59 30\n31 12\n31 3\n38 23\n59 29\n53 9\n38 56\n53 54\n29 21\n17 55\n59 38\n26 16\n24 59\n24 25\n17 35\n24 41\n30 15\n31 27\n8 44\n26 5\n26 48\n8 32\n53 17\n3 34\n3 51\n30 28\n47 10\n53 60\n36 42\n24 53\n59 22\n53 40\n26 52\n36 4\n59 8\n29 37\n36 20\n17 47\n53 18\n3 50\n30 2\n17 7\n8 58\n59 1\n31 11\n24 26\n24 43\n53 57\n59 45\n47 13\n26 46\n17 33\n30 31\n26 39\n26 19\n24 36\n8 49\n38 14\n",
"7\n1 2\n2 3\n3 4\n3 5\n1 6\n1 7\n",
"20\n19 16\n19 18\n20 7\n9 4\n6 17\n14 2\n9 15\n2 13\n5 11\n19 12\n12 20\n16 9\n11 8\n19 5\n3 1\n19 14\n5 3\n12 10\n19 6\n",
"7\n1 2\n1 3\n2 4\n2 5\n3 6\n3 7\n",
"10\n9 5\n7 1\n9 10\n7 2\n5 4\n9 6\n2 9\n10 8\n1 3\n",
"4\n2 4\n2 3\n2 1\n",
"4\n1 4\n3 2\n1 3\n",
"3\n1 2\n1 3\n",
"5\n1 2\n1 5\n1 3\n1 4\n",
"20\n14 9\n12 13\n10 15\n2 1\n20 19\n16 6\n16 3\n17 14\n3 5\n2 11\n3 10\n15 8\n14 2\n6 4\n3 20\n5 18\n1 7\n1 16\n4 12\n",
"20\n7 5\n14 13\n17 6\n3 8\n16 12\n18 9\n3 18\n14 1\n17 3\n15 2\n17 4\n9 11\n2 7\n15 17\n3 20\n16 10\n17 14\n2 16\n1 19\n",
"8\n1 2\n2 3\n3 4\n1 7\n1 8\n4 5\n4 6\n",
"5\n5 1\n5 2\n5 3\n5 4\n",
"50\n49 6\n43 7\n1 27\n19 35\n15 37\n16 12\n19 21\n16 28\n49 9\n48 39\n13 1\n2 48\n9 50\n44 3\n41 32\n48 31\n49 33\n6 11\n13 20\n49 22\n13 41\n48 29\n13 46\n15 47\n34 2\n49 13\n48 14\n34 24\n16 36\n13 40\n49 34\n49 17\n43 25\n11 23\n10 15\n19 26\n34 44\n16 42\n19 18\n46 8\n29 38\n1 45\n12 43\n13 16\n46 30\n15 5\n49 10\n11 19\n32 4\n",
"20\n13 1\n18 2\n3 7\n18 5\n20 16\n3 12\n18 9\n3 10\n18 11\n13 6\n3 18\n20 15\n20 17\n3 13\n3 4\n13 14\n3 20\n18 8\n3 19\n",
"10\n8 2\n5 6\n1 8\n2 9\n1 4\n8 10\n10 5\n2 7\n2 3\n"
],
"output": [
"NO",
"YES",
"NO",
"NO",
"NO",
"YES",
"NO",
"NO",
"NO",
"NO",
"YES",
"NO",
"NO",
"YES",
"NO",
"NO",
"NO",
"YES",
"NO",
"YES",
"NO"
]
}
|
IN-CORRECT
|
java
|
import java.io.*;
import java.util.*;
public class d2prac implements Runnable
{
private boolean console=false;
boolean ans=true;
ArrayList<Integer> adj[];
public void solve()
{
int i; int n=in.ni(); adj=new ArrayList [n];
for(i=0;i<n;i++)
adj[i]=new ArrayList();
for(i=0;i<n-1;i++)
{
int u=in.ni()-1; int v=in.ni()-1;
adj[u].add(v); adj[v].add(u);
}
if(n==3)
{
out.println("NO");
return;
}
dfs(0,-1);
if(ans)
out.println("YES");
else
out.println("NO");
}
public void dfs(int v,int p)
{
boolean not=false;
if(adj[v].size()>1)
not=true;
int count=0;
for(int node:adj[v])
{
if(adj[node].size()==1)
count++;
if(node==p)
continue;
else
dfs(node,v);
}
if(not&&count<2)
ans=false;
}
@Override
public void run() {
try { init(); }
catch (FileNotFoundException e) { e.printStackTrace(); }
int t= 1;
while (t-->0) {
solve();
out.flush(); }
}
private FastInput in; private PrintWriter out;
public static void main(String[] args) throws Exception { new d2prac().run(); }
private void init() throws FileNotFoundException {
InputStream inputStream = System.in; OutputStream outputStream = System.out;
try { if (!console && System.getProperty("user.name").equals("sachan")) {
outputStream = new FileOutputStream("/home/sachan/Desktop/output.txt");
inputStream = new FileInputStream("/home/sachan/Desktop/input.txt"); }
} catch (Exception ignored) { }
out = new PrintWriter(outputStream); in = new FastInput(inputStream);
}
static class FastInput { InputStream obj;
public FastInput(InputStream obj) { this.obj = obj; }
byte inbuffer[] = new byte[1024]; int lenbuffer = 0, ptrbuffer = 0;
int readByte() { if (lenbuffer == -1) throw new InputMismatchException();
if (ptrbuffer >= lenbuffer) { ptrbuffer = 0;
try { lenbuffer = obj.read(inbuffer); }
catch (IOException e) { throw new InputMismatchException(); } }
if (lenbuffer <= 0) return -1;return inbuffer[ptrbuffer++]; }
String ns() { int b = skip();StringBuilder sb = new StringBuilder();
while (!(isSpaceChar(b))) { sb.appendCodePoint(b);b = readByte(); }return sb.toString();}
int ni() { int num = 0, b;boolean minus = false;
while ((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-')) ;
if (b == '-') { minus = true;b = readByte(); }
while (true) { if (b >= '0' && b <= '9') { num = num * 10 + (b - '0'); } else {
return minus ? -num : num; }b = readByte(); }}
long nl() { long num = 0;int b;boolean minus = false;
while ((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-')) ;
if (b == '-') { minus = true;b = readByte(); }
while (true) { if (b >= '0' && b <= '9') { num = num * 10L + (b - '0'); } else {
return minus ? -num : num; }b = readByte(); } }
boolean isSpaceChar(int c) { return (!(c >= 33 && c <= 126)); }
int skip() { int b;while ((b = readByte()) != -1 && isSpaceChar(b)) ;return b; }
float nf() {return Float.parseFloat(ns());}
double nd() {return Double.parseDouble(ns());}
char nc() {return (char) skip();}
}
}
|
1189_D1. Add on a Tree
|
Note that this is the first problem of the two similar problems. You can hack this problem only if you solve both problems.
You are given a tree with n nodes. In the beginning, 0 is written on all edges. In one operation, you can choose any 2 distinct leaves u, v and any real number x and add x to values written on all edges on the simple path between u and v.
For example, on the picture below you can see the result of applying two operations to the graph: adding 2 on the path from 7 to 6, and then adding -0.5 on the path from 4 to 5.
<image>
Is it true that for any configuration of real numbers written on edges, we can achieve it with a finite number of operations?
Leaf is a node of a tree of degree 1. Simple path is a path that doesn't contain any node twice.
Input
The first line contains a single integer n (2 ≤ n ≤ 10^5) — the number of nodes.
Each of the next n-1 lines contains two integers u and v (1 ≤ u, v ≤ n, u ≠ v), meaning that there is an edge between nodes u and v. It is guaranteed that these edges form a tree.
Output
If there is a configuration of real numbers written on edges of the tree that we can't achieve by performing the operations, output "NO".
Otherwise, output "YES".
You can print each letter in any case (upper or lower).
Examples
Input
2
1 2
Output
YES
Input
3
1 2
2 3
Output
NO
Input
5
1 2
1 3
1 4
2 5
Output
NO
Input
6
1 2
1 3
1 4
2 5
2 6
Output
YES
Note
In the first example, we can add any real x to the value written on the only edge (1, 2).
<image>
In the second example, one of configurations that we can't reach is 0 written on (1, 2) and 1 written on (2, 3).
<image>
Below you can see graphs from examples 3, 4:
<image> <image>
|
{
"input": [
"2\n1 2\n",
"3\n1 2\n2 3\n",
"5\n1 2\n1 3\n1 4\n2 5\n",
"6\n1 2\n1 3\n1 4\n2 5\n2 6\n"
],
"output": [
"YES",
"NO",
"NO",
"YES"
]
}
|
{
"input": [
"50\n16 4\n17 9\n31 19\n22 10\n8 1\n40 30\n3 31\n20 29\n47 27\n22 25\n32 34\n12 15\n40 32\n10 33\n47 12\n6 24\n46 41\n14 23\n12 35\n31 42\n46 28\n31 20\n46 37\n1 39\n29 49\n37 47\n40 6\n42 36\n47 2\n24 46\n2 13\n8 45\n41 3\n32 17\n4 7\n47 26\n28 8\n41 50\n34 44\n33 21\n25 5\n16 40\n3 14\n8 18\n28 11\n32 22\n2 38\n3 48\n44 43\n",
"10\n8 1\n1 2\n8 9\n8 5\n1 3\n1 10\n1 6\n1 7\n8 4\n",
"5\n5 1\n5 4\n4 3\n1 2\n",
"7\n1 2\n2 3\n1 4\n1 5\n3 6\n3 7\n",
"3\n1 3\n2 3\n",
"60\n26 6\n59 30\n31 12\n31 3\n38 23\n59 29\n53 9\n38 56\n53 54\n29 21\n17 55\n59 38\n26 16\n24 59\n24 25\n17 35\n24 41\n30 15\n31 27\n8 44\n26 5\n26 48\n8 32\n53 17\n3 34\n3 51\n30 28\n47 10\n53 60\n36 42\n24 53\n59 22\n53 40\n26 52\n36 4\n59 8\n29 37\n36 20\n17 47\n53 18\n3 50\n30 2\n17 7\n8 58\n59 1\n31 11\n24 26\n24 43\n53 57\n59 45\n47 13\n26 46\n17 33\n30 31\n26 39\n26 19\n24 36\n8 49\n38 14\n",
"7\n1 2\n2 3\n3 4\n3 5\n1 6\n1 7\n",
"20\n19 16\n19 18\n20 7\n9 4\n6 17\n14 2\n9 15\n2 13\n5 11\n19 12\n12 20\n16 9\n11 8\n19 5\n3 1\n19 14\n5 3\n12 10\n19 6\n",
"7\n1 2\n1 3\n2 4\n2 5\n3 6\n3 7\n",
"10\n9 5\n7 1\n9 10\n7 2\n5 4\n9 6\n2 9\n10 8\n1 3\n",
"4\n2 4\n2 3\n2 1\n",
"4\n1 4\n3 2\n1 3\n",
"3\n1 2\n1 3\n",
"5\n1 2\n1 5\n1 3\n1 4\n",
"20\n14 9\n12 13\n10 15\n2 1\n20 19\n16 6\n16 3\n17 14\n3 5\n2 11\n3 10\n15 8\n14 2\n6 4\n3 20\n5 18\n1 7\n1 16\n4 12\n",
"20\n7 5\n14 13\n17 6\n3 8\n16 12\n18 9\n3 18\n14 1\n17 3\n15 2\n17 4\n9 11\n2 7\n15 17\n3 20\n16 10\n17 14\n2 16\n1 19\n",
"8\n1 2\n2 3\n3 4\n1 7\n1 8\n4 5\n4 6\n",
"5\n5 1\n5 2\n5 3\n5 4\n",
"50\n49 6\n43 7\n1 27\n19 35\n15 37\n16 12\n19 21\n16 28\n49 9\n48 39\n13 1\n2 48\n9 50\n44 3\n41 32\n48 31\n49 33\n6 11\n13 20\n49 22\n13 41\n48 29\n13 46\n15 47\n34 2\n49 13\n48 14\n34 24\n16 36\n13 40\n49 34\n49 17\n43 25\n11 23\n10 15\n19 26\n34 44\n16 42\n19 18\n46 8\n29 38\n1 45\n12 43\n13 16\n46 30\n15 5\n49 10\n11 19\n32 4\n",
"20\n13 1\n18 2\n3 7\n18 5\n20 16\n3 12\n18 9\n3 10\n18 11\n13 6\n3 18\n20 15\n20 17\n3 13\n3 4\n13 14\n3 20\n18 8\n3 19\n",
"10\n8 2\n5 6\n1 8\n2 9\n1 4\n8 10\n10 5\n2 7\n2 3\n"
],
"output": [
"NO",
"YES",
"NO",
"NO",
"NO",
"YES",
"NO",
"NO",
"NO",
"NO",
"YES",
"NO",
"NO",
"YES",
"NO",
"NO",
"NO",
"YES",
"NO",
"YES",
"NO"
]
}
|
IN-CORRECT
|
java
|
import java.io.*;
import java.text.*;
import java.util.*;
import java.math.*;
public class template {
public static void main(String[] args) throws Exception {
new template().run();
}
LinkedList<Integer>[] adj;
int[] par, sz;
public void run() throws Exception {
FastScanner f = new FastScanner();
PrintWriter out = new PrintWriter(System.out);
int n = f.nextInt();
adj = new LinkedList[n];
for(int i = 0; i < n; i++) adj[i] = new LinkedList<>();
for(int i = 0; i < n-1; i++) {
int a = f.nextInt()-1, b = f.nextInt()-1;
adj[a].add(b);
adj[b].add(a);
}
par = new int[n];
sz = new int[n];
dfs(0, -1);
boolean works = true;
for(int i = 1; i < n; i++)
works &= sz[i] != 1;
out.println(works ? "YES" : "NO");
///
out.flush();
}
public void dfs(int i, int p) {
if(p != -1) sz[p]++;
par[i] = p;
for(int j : adj[i])
if(p != j) dfs(j, i);
}
///
static class FastScanner {
public BufferedReader reader;
public StringTokenizer tokenizer;
public FastScanner() {
reader = new BufferedReader(new InputStreamReader(System.in), 32768);
tokenizer = null;
}
public String next() {
while (tokenizer == null || !tokenizer.hasMoreTokens()) {
try {
tokenizer = new StringTokenizer(reader.readLine());
} catch (IOException e) {
throw new RuntimeException(e);
}
}
return tokenizer.nextToken();
}
public int nextInt() {
return Integer.parseInt(next());
}
public long nextLong() {
return Long.parseLong(next());
}
public double nextDouble() {
return Double.parseDouble(next());
}
public String nextLine() {
try {
return reader.readLine();
} catch(IOException e) {
throw new RuntimeException(e);
}
}
}
}
|
1189_D1. Add on a Tree
|
Note that this is the first problem of the two similar problems. You can hack this problem only if you solve both problems.
You are given a tree with n nodes. In the beginning, 0 is written on all edges. In one operation, you can choose any 2 distinct leaves u, v and any real number x and add x to values written on all edges on the simple path between u and v.
For example, on the picture below you can see the result of applying two operations to the graph: adding 2 on the path from 7 to 6, and then adding -0.5 on the path from 4 to 5.
<image>
Is it true that for any configuration of real numbers written on edges, we can achieve it with a finite number of operations?
Leaf is a node of a tree of degree 1. Simple path is a path that doesn't contain any node twice.
Input
The first line contains a single integer n (2 ≤ n ≤ 10^5) — the number of nodes.
Each of the next n-1 lines contains two integers u and v (1 ≤ u, v ≤ n, u ≠ v), meaning that there is an edge between nodes u and v. It is guaranteed that these edges form a tree.
Output
If there is a configuration of real numbers written on edges of the tree that we can't achieve by performing the operations, output "NO".
Otherwise, output "YES".
You can print each letter in any case (upper or lower).
Examples
Input
2
1 2
Output
YES
Input
3
1 2
2 3
Output
NO
Input
5
1 2
1 3
1 4
2 5
Output
NO
Input
6
1 2
1 3
1 4
2 5
2 6
Output
YES
Note
In the first example, we can add any real x to the value written on the only edge (1, 2).
<image>
In the second example, one of configurations that we can't reach is 0 written on (1, 2) and 1 written on (2, 3).
<image>
Below you can see graphs from examples 3, 4:
<image> <image>
|
{
"input": [
"2\n1 2\n",
"3\n1 2\n2 3\n",
"5\n1 2\n1 3\n1 4\n2 5\n",
"6\n1 2\n1 3\n1 4\n2 5\n2 6\n"
],
"output": [
"YES",
"NO",
"NO",
"YES"
]
}
|
{
"input": [
"50\n16 4\n17 9\n31 19\n22 10\n8 1\n40 30\n3 31\n20 29\n47 27\n22 25\n32 34\n12 15\n40 32\n10 33\n47 12\n6 24\n46 41\n14 23\n12 35\n31 42\n46 28\n31 20\n46 37\n1 39\n29 49\n37 47\n40 6\n42 36\n47 2\n24 46\n2 13\n8 45\n41 3\n32 17\n4 7\n47 26\n28 8\n41 50\n34 44\n33 21\n25 5\n16 40\n3 14\n8 18\n28 11\n32 22\n2 38\n3 48\n44 43\n",
"10\n8 1\n1 2\n8 9\n8 5\n1 3\n1 10\n1 6\n1 7\n8 4\n",
"5\n5 1\n5 4\n4 3\n1 2\n",
"7\n1 2\n2 3\n1 4\n1 5\n3 6\n3 7\n",
"3\n1 3\n2 3\n",
"60\n26 6\n59 30\n31 12\n31 3\n38 23\n59 29\n53 9\n38 56\n53 54\n29 21\n17 55\n59 38\n26 16\n24 59\n24 25\n17 35\n24 41\n30 15\n31 27\n8 44\n26 5\n26 48\n8 32\n53 17\n3 34\n3 51\n30 28\n47 10\n53 60\n36 42\n24 53\n59 22\n53 40\n26 52\n36 4\n59 8\n29 37\n36 20\n17 47\n53 18\n3 50\n30 2\n17 7\n8 58\n59 1\n31 11\n24 26\n24 43\n53 57\n59 45\n47 13\n26 46\n17 33\n30 31\n26 39\n26 19\n24 36\n8 49\n38 14\n",
"7\n1 2\n2 3\n3 4\n3 5\n1 6\n1 7\n",
"20\n19 16\n19 18\n20 7\n9 4\n6 17\n14 2\n9 15\n2 13\n5 11\n19 12\n12 20\n16 9\n11 8\n19 5\n3 1\n19 14\n5 3\n12 10\n19 6\n",
"7\n1 2\n1 3\n2 4\n2 5\n3 6\n3 7\n",
"10\n9 5\n7 1\n9 10\n7 2\n5 4\n9 6\n2 9\n10 8\n1 3\n",
"4\n2 4\n2 3\n2 1\n",
"4\n1 4\n3 2\n1 3\n",
"3\n1 2\n1 3\n",
"5\n1 2\n1 5\n1 3\n1 4\n",
"20\n14 9\n12 13\n10 15\n2 1\n20 19\n16 6\n16 3\n17 14\n3 5\n2 11\n3 10\n15 8\n14 2\n6 4\n3 20\n5 18\n1 7\n1 16\n4 12\n",
"20\n7 5\n14 13\n17 6\n3 8\n16 12\n18 9\n3 18\n14 1\n17 3\n15 2\n17 4\n9 11\n2 7\n15 17\n3 20\n16 10\n17 14\n2 16\n1 19\n",
"8\n1 2\n2 3\n3 4\n1 7\n1 8\n4 5\n4 6\n",
"5\n5 1\n5 2\n5 3\n5 4\n",
"50\n49 6\n43 7\n1 27\n19 35\n15 37\n16 12\n19 21\n16 28\n49 9\n48 39\n13 1\n2 48\n9 50\n44 3\n41 32\n48 31\n49 33\n6 11\n13 20\n49 22\n13 41\n48 29\n13 46\n15 47\n34 2\n49 13\n48 14\n34 24\n16 36\n13 40\n49 34\n49 17\n43 25\n11 23\n10 15\n19 26\n34 44\n16 42\n19 18\n46 8\n29 38\n1 45\n12 43\n13 16\n46 30\n15 5\n49 10\n11 19\n32 4\n",
"20\n13 1\n18 2\n3 7\n18 5\n20 16\n3 12\n18 9\n3 10\n18 11\n13 6\n3 18\n20 15\n20 17\n3 13\n3 4\n13 14\n3 20\n18 8\n3 19\n",
"10\n8 2\n5 6\n1 8\n2 9\n1 4\n8 10\n10 5\n2 7\n2 3\n"
],
"output": [
"NO",
"YES",
"NO",
"NO",
"NO",
"YES",
"NO",
"NO",
"NO",
"NO",
"YES",
"NO",
"NO",
"YES",
"NO",
"NO",
"NO",
"YES",
"NO",
"YES",
"NO"
]
}
|
IN-CORRECT
|
java
|
import java.io.*;
import java.util.*;
public class d2prac implements Runnable
{
private boolean console=false;
boolean ans=true;
ArrayList<Integer> adj[];
public void solve()
{
int i; int n=in.ni(); adj=new ArrayList [n];
for(i=0;i<n;i++)
adj[i]=new ArrayList();
for(i=0;i<n-1;i++)
{
int u=in.ni()-1; int v=in.ni()-1;
adj[u].add(v); adj[v].add(u);
}
if(n==3)
{
out.println("NO");
return;
}
dfs(0,-1);
if(ans)
out.println("YES");
else
out.println("NO");
}
public void dfs(int v,int p)
{
boolean not=false;
if(adj[v].size()>1)
not=true;
int count=0;
for(int node:adj[v])
{
if(adj[node].size()==1)
count++;
if(node==p)
continue;
else
dfs(node,v);
}
if(not&&count<2)
{
int c=0;
for(int node:adj[v])
{
if(adj[node].size()==1)
continue;
c++;
}
if(c<2)
ans=false;
}
}
@Override
public void run() {
try { init(); }
catch (FileNotFoundException e) { e.printStackTrace(); }
int t= 1;
while (t-->0) {
solve();
out.flush(); }
}
private FastInput in; private PrintWriter out;
public static void main(String[] args) throws Exception { new d2prac().run(); }
private void init() throws FileNotFoundException {
InputStream inputStream = System.in; OutputStream outputStream = System.out;
try { if (!console && System.getProperty("user.name").equals("sachan")) {
outputStream = new FileOutputStream("/home/sachan/Desktop/output.txt");
inputStream = new FileInputStream("/home/sachan/Desktop/input.txt"); }
} catch (Exception ignored) { }
out = new PrintWriter(outputStream); in = new FastInput(inputStream);
}
static class FastInput { InputStream obj;
public FastInput(InputStream obj) { this.obj = obj; }
byte inbuffer[] = new byte[1024]; int lenbuffer = 0, ptrbuffer = 0;
int readByte() { if (lenbuffer == -1) throw new InputMismatchException();
if (ptrbuffer >= lenbuffer) { ptrbuffer = 0;
try { lenbuffer = obj.read(inbuffer); }
catch (IOException e) { throw new InputMismatchException(); } }
if (lenbuffer <= 0) return -1;return inbuffer[ptrbuffer++]; }
String ns() { int b = skip();StringBuilder sb = new StringBuilder();
while (!(isSpaceChar(b))) { sb.appendCodePoint(b);b = readByte(); }return sb.toString();}
int ni() { int num = 0, b;boolean minus = false;
while ((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-')) ;
if (b == '-') { minus = true;b = readByte(); }
while (true) { if (b >= '0' && b <= '9') { num = num * 10 + (b - '0'); } else {
return minus ? -num : num; }b = readByte(); }}
long nl() { long num = 0;int b;boolean minus = false;
while ((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-')) ;
if (b == '-') { minus = true;b = readByte(); }
while (true) { if (b >= '0' && b <= '9') { num = num * 10L + (b - '0'); } else {
return minus ? -num : num; }b = readByte(); } }
boolean isSpaceChar(int c) { return (!(c >= 33 && c <= 126)); }
int skip() { int b;while ((b = readByte()) != -1 && isSpaceChar(b)) ;return b; }
float nf() {return Float.parseFloat(ns());}
double nd() {return Double.parseDouble(ns());}
char nc() {return (char) skip();}
}
}
|
1189_D1. Add on a Tree
|
Note that this is the first problem of the two similar problems. You can hack this problem only if you solve both problems.
You are given a tree with n nodes. In the beginning, 0 is written on all edges. In one operation, you can choose any 2 distinct leaves u, v and any real number x and add x to values written on all edges on the simple path between u and v.
For example, on the picture below you can see the result of applying two operations to the graph: adding 2 on the path from 7 to 6, and then adding -0.5 on the path from 4 to 5.
<image>
Is it true that for any configuration of real numbers written on edges, we can achieve it with a finite number of operations?
Leaf is a node of a tree of degree 1. Simple path is a path that doesn't contain any node twice.
Input
The first line contains a single integer n (2 ≤ n ≤ 10^5) — the number of nodes.
Each of the next n-1 lines contains two integers u and v (1 ≤ u, v ≤ n, u ≠ v), meaning that there is an edge between nodes u and v. It is guaranteed that these edges form a tree.
Output
If there is a configuration of real numbers written on edges of the tree that we can't achieve by performing the operations, output "NO".
Otherwise, output "YES".
You can print each letter in any case (upper or lower).
Examples
Input
2
1 2
Output
YES
Input
3
1 2
2 3
Output
NO
Input
5
1 2
1 3
1 4
2 5
Output
NO
Input
6
1 2
1 3
1 4
2 5
2 6
Output
YES
Note
In the first example, we can add any real x to the value written on the only edge (1, 2).
<image>
In the second example, one of configurations that we can't reach is 0 written on (1, 2) and 1 written on (2, 3).
<image>
Below you can see graphs from examples 3, 4:
<image> <image>
|
{
"input": [
"2\n1 2\n",
"3\n1 2\n2 3\n",
"5\n1 2\n1 3\n1 4\n2 5\n",
"6\n1 2\n1 3\n1 4\n2 5\n2 6\n"
],
"output": [
"YES",
"NO",
"NO",
"YES"
]
}
|
{
"input": [
"50\n16 4\n17 9\n31 19\n22 10\n8 1\n40 30\n3 31\n20 29\n47 27\n22 25\n32 34\n12 15\n40 32\n10 33\n47 12\n6 24\n46 41\n14 23\n12 35\n31 42\n46 28\n31 20\n46 37\n1 39\n29 49\n37 47\n40 6\n42 36\n47 2\n24 46\n2 13\n8 45\n41 3\n32 17\n4 7\n47 26\n28 8\n41 50\n34 44\n33 21\n25 5\n16 40\n3 14\n8 18\n28 11\n32 22\n2 38\n3 48\n44 43\n",
"10\n8 1\n1 2\n8 9\n8 5\n1 3\n1 10\n1 6\n1 7\n8 4\n",
"5\n5 1\n5 4\n4 3\n1 2\n",
"7\n1 2\n2 3\n1 4\n1 5\n3 6\n3 7\n",
"3\n1 3\n2 3\n",
"60\n26 6\n59 30\n31 12\n31 3\n38 23\n59 29\n53 9\n38 56\n53 54\n29 21\n17 55\n59 38\n26 16\n24 59\n24 25\n17 35\n24 41\n30 15\n31 27\n8 44\n26 5\n26 48\n8 32\n53 17\n3 34\n3 51\n30 28\n47 10\n53 60\n36 42\n24 53\n59 22\n53 40\n26 52\n36 4\n59 8\n29 37\n36 20\n17 47\n53 18\n3 50\n30 2\n17 7\n8 58\n59 1\n31 11\n24 26\n24 43\n53 57\n59 45\n47 13\n26 46\n17 33\n30 31\n26 39\n26 19\n24 36\n8 49\n38 14\n",
"7\n1 2\n2 3\n3 4\n3 5\n1 6\n1 7\n",
"20\n19 16\n19 18\n20 7\n9 4\n6 17\n14 2\n9 15\n2 13\n5 11\n19 12\n12 20\n16 9\n11 8\n19 5\n3 1\n19 14\n5 3\n12 10\n19 6\n",
"7\n1 2\n1 3\n2 4\n2 5\n3 6\n3 7\n",
"10\n9 5\n7 1\n9 10\n7 2\n5 4\n9 6\n2 9\n10 8\n1 3\n",
"4\n2 4\n2 3\n2 1\n",
"4\n1 4\n3 2\n1 3\n",
"3\n1 2\n1 3\n",
"5\n1 2\n1 5\n1 3\n1 4\n",
"20\n14 9\n12 13\n10 15\n2 1\n20 19\n16 6\n16 3\n17 14\n3 5\n2 11\n3 10\n15 8\n14 2\n6 4\n3 20\n5 18\n1 7\n1 16\n4 12\n",
"20\n7 5\n14 13\n17 6\n3 8\n16 12\n18 9\n3 18\n14 1\n17 3\n15 2\n17 4\n9 11\n2 7\n15 17\n3 20\n16 10\n17 14\n2 16\n1 19\n",
"8\n1 2\n2 3\n3 4\n1 7\n1 8\n4 5\n4 6\n",
"5\n5 1\n5 2\n5 3\n5 4\n",
"50\n49 6\n43 7\n1 27\n19 35\n15 37\n16 12\n19 21\n16 28\n49 9\n48 39\n13 1\n2 48\n9 50\n44 3\n41 32\n48 31\n49 33\n6 11\n13 20\n49 22\n13 41\n48 29\n13 46\n15 47\n34 2\n49 13\n48 14\n34 24\n16 36\n13 40\n49 34\n49 17\n43 25\n11 23\n10 15\n19 26\n34 44\n16 42\n19 18\n46 8\n29 38\n1 45\n12 43\n13 16\n46 30\n15 5\n49 10\n11 19\n32 4\n",
"20\n13 1\n18 2\n3 7\n18 5\n20 16\n3 12\n18 9\n3 10\n18 11\n13 6\n3 18\n20 15\n20 17\n3 13\n3 4\n13 14\n3 20\n18 8\n3 19\n",
"10\n8 2\n5 6\n1 8\n2 9\n1 4\n8 10\n10 5\n2 7\n2 3\n"
],
"output": [
"NO",
"YES",
"NO",
"NO",
"NO",
"YES",
"NO",
"NO",
"NO",
"NO",
"YES",
"NO",
"NO",
"YES",
"NO",
"NO",
"NO",
"YES",
"NO",
"YES",
"NO"
]
}
|
IN-CORRECT
|
cpp
|
#include <bits/stdc++.h>
#pragma GCC optimize("Ofast")
using namespace std;
long int mod = 1e9 + 7;
vector<long int> v[100500];
bool vis[100500];
long int n, m, t, c, re;
void bfs(long int s) {
queue<long int> q;
long int ans = 0;
q.push(s);
vis[s] = 1;
while (!q.empty()) {
ans = 0;
long int u = q.front();
q.pop();
for (auto i : v[u]) {
if (!vis[i]) {
vis[i] = 1;
q.push(i);
ans++;
}
}
if (!ans) re++;
}
}
int main() {
ios::sync_with_stdio(0);
cin.tie(0);
cin >> n;
for (long int i = 1; i < n; i++) {
long int x, y;
cin >> x >> y;
v[x].push_back(y);
v[y].push_back(x);
}
for (long int i = 1; i < n + 1; i++)
if (!vis[i]) bfs(i);
if (n == 2) {
cout << "YES";
return 0;
}
if (re % 2)
cout << "NO";
else
cout << "YES";
return 0;
}
|
1189_D1. Add on a Tree
|
Note that this is the first problem of the two similar problems. You can hack this problem only if you solve both problems.
You are given a tree with n nodes. In the beginning, 0 is written on all edges. In one operation, you can choose any 2 distinct leaves u, v and any real number x and add x to values written on all edges on the simple path between u and v.
For example, on the picture below you can see the result of applying two operations to the graph: adding 2 on the path from 7 to 6, and then adding -0.5 on the path from 4 to 5.
<image>
Is it true that for any configuration of real numbers written on edges, we can achieve it with a finite number of operations?
Leaf is a node of a tree of degree 1. Simple path is a path that doesn't contain any node twice.
Input
The first line contains a single integer n (2 ≤ n ≤ 10^5) — the number of nodes.
Each of the next n-1 lines contains two integers u and v (1 ≤ u, v ≤ n, u ≠ v), meaning that there is an edge between nodes u and v. It is guaranteed that these edges form a tree.
Output
If there is a configuration of real numbers written on edges of the tree that we can't achieve by performing the operations, output "NO".
Otherwise, output "YES".
You can print each letter in any case (upper or lower).
Examples
Input
2
1 2
Output
YES
Input
3
1 2
2 3
Output
NO
Input
5
1 2
1 3
1 4
2 5
Output
NO
Input
6
1 2
1 3
1 4
2 5
2 6
Output
YES
Note
In the first example, we can add any real x to the value written on the only edge (1, 2).
<image>
In the second example, one of configurations that we can't reach is 0 written on (1, 2) and 1 written on (2, 3).
<image>
Below you can see graphs from examples 3, 4:
<image> <image>
|
{
"input": [
"2\n1 2\n",
"3\n1 2\n2 3\n",
"5\n1 2\n1 3\n1 4\n2 5\n",
"6\n1 2\n1 3\n1 4\n2 5\n2 6\n"
],
"output": [
"YES",
"NO",
"NO",
"YES"
]
}
|
{
"input": [
"50\n16 4\n17 9\n31 19\n22 10\n8 1\n40 30\n3 31\n20 29\n47 27\n22 25\n32 34\n12 15\n40 32\n10 33\n47 12\n6 24\n46 41\n14 23\n12 35\n31 42\n46 28\n31 20\n46 37\n1 39\n29 49\n37 47\n40 6\n42 36\n47 2\n24 46\n2 13\n8 45\n41 3\n32 17\n4 7\n47 26\n28 8\n41 50\n34 44\n33 21\n25 5\n16 40\n3 14\n8 18\n28 11\n32 22\n2 38\n3 48\n44 43\n",
"10\n8 1\n1 2\n8 9\n8 5\n1 3\n1 10\n1 6\n1 7\n8 4\n",
"5\n5 1\n5 4\n4 3\n1 2\n",
"7\n1 2\n2 3\n1 4\n1 5\n3 6\n3 7\n",
"3\n1 3\n2 3\n",
"60\n26 6\n59 30\n31 12\n31 3\n38 23\n59 29\n53 9\n38 56\n53 54\n29 21\n17 55\n59 38\n26 16\n24 59\n24 25\n17 35\n24 41\n30 15\n31 27\n8 44\n26 5\n26 48\n8 32\n53 17\n3 34\n3 51\n30 28\n47 10\n53 60\n36 42\n24 53\n59 22\n53 40\n26 52\n36 4\n59 8\n29 37\n36 20\n17 47\n53 18\n3 50\n30 2\n17 7\n8 58\n59 1\n31 11\n24 26\n24 43\n53 57\n59 45\n47 13\n26 46\n17 33\n30 31\n26 39\n26 19\n24 36\n8 49\n38 14\n",
"7\n1 2\n2 3\n3 4\n3 5\n1 6\n1 7\n",
"20\n19 16\n19 18\n20 7\n9 4\n6 17\n14 2\n9 15\n2 13\n5 11\n19 12\n12 20\n16 9\n11 8\n19 5\n3 1\n19 14\n5 3\n12 10\n19 6\n",
"7\n1 2\n1 3\n2 4\n2 5\n3 6\n3 7\n",
"10\n9 5\n7 1\n9 10\n7 2\n5 4\n9 6\n2 9\n10 8\n1 3\n",
"4\n2 4\n2 3\n2 1\n",
"4\n1 4\n3 2\n1 3\n",
"3\n1 2\n1 3\n",
"5\n1 2\n1 5\n1 3\n1 4\n",
"20\n14 9\n12 13\n10 15\n2 1\n20 19\n16 6\n16 3\n17 14\n3 5\n2 11\n3 10\n15 8\n14 2\n6 4\n3 20\n5 18\n1 7\n1 16\n4 12\n",
"20\n7 5\n14 13\n17 6\n3 8\n16 12\n18 9\n3 18\n14 1\n17 3\n15 2\n17 4\n9 11\n2 7\n15 17\n3 20\n16 10\n17 14\n2 16\n1 19\n",
"8\n1 2\n2 3\n3 4\n1 7\n1 8\n4 5\n4 6\n",
"5\n5 1\n5 2\n5 3\n5 4\n",
"50\n49 6\n43 7\n1 27\n19 35\n15 37\n16 12\n19 21\n16 28\n49 9\n48 39\n13 1\n2 48\n9 50\n44 3\n41 32\n48 31\n49 33\n6 11\n13 20\n49 22\n13 41\n48 29\n13 46\n15 47\n34 2\n49 13\n48 14\n34 24\n16 36\n13 40\n49 34\n49 17\n43 25\n11 23\n10 15\n19 26\n34 44\n16 42\n19 18\n46 8\n29 38\n1 45\n12 43\n13 16\n46 30\n15 5\n49 10\n11 19\n32 4\n",
"20\n13 1\n18 2\n3 7\n18 5\n20 16\n3 12\n18 9\n3 10\n18 11\n13 6\n3 18\n20 15\n20 17\n3 13\n3 4\n13 14\n3 20\n18 8\n3 19\n",
"10\n8 2\n5 6\n1 8\n2 9\n1 4\n8 10\n10 5\n2 7\n2 3\n"
],
"output": [
"NO",
"YES",
"NO",
"NO",
"NO",
"YES",
"NO",
"NO",
"NO",
"NO",
"YES",
"NO",
"NO",
"YES",
"NO",
"NO",
"NO",
"YES",
"NO",
"YES",
"NO"
]
}
|
IN-CORRECT
|
python3
|
n = int(input())
tr = [0] * (n + 1)
for i in range(n - 1):
u, v = map(int, input().split())
tr[u] += 1
tr[v] += 1
for i in range(n - 1):
if tr[i] == 2:
print('No')
break
else:
print('Yes')
|
1189_D1. Add on a Tree
|
Note that this is the first problem of the two similar problems. You can hack this problem only if you solve both problems.
You are given a tree with n nodes. In the beginning, 0 is written on all edges. In one operation, you can choose any 2 distinct leaves u, v and any real number x and add x to values written on all edges on the simple path between u and v.
For example, on the picture below you can see the result of applying two operations to the graph: adding 2 on the path from 7 to 6, and then adding -0.5 on the path from 4 to 5.
<image>
Is it true that for any configuration of real numbers written on edges, we can achieve it with a finite number of operations?
Leaf is a node of a tree of degree 1. Simple path is a path that doesn't contain any node twice.
Input
The first line contains a single integer n (2 ≤ n ≤ 10^5) — the number of nodes.
Each of the next n-1 lines contains two integers u and v (1 ≤ u, v ≤ n, u ≠ v), meaning that there is an edge between nodes u and v. It is guaranteed that these edges form a tree.
Output
If there is a configuration of real numbers written on edges of the tree that we can't achieve by performing the operations, output "NO".
Otherwise, output "YES".
You can print each letter in any case (upper or lower).
Examples
Input
2
1 2
Output
YES
Input
3
1 2
2 3
Output
NO
Input
5
1 2
1 3
1 4
2 5
Output
NO
Input
6
1 2
1 3
1 4
2 5
2 6
Output
YES
Note
In the first example, we can add any real x to the value written on the only edge (1, 2).
<image>
In the second example, one of configurations that we can't reach is 0 written on (1, 2) and 1 written on (2, 3).
<image>
Below you can see graphs from examples 3, 4:
<image> <image>
|
{
"input": [
"2\n1 2\n",
"3\n1 2\n2 3\n",
"5\n1 2\n1 3\n1 4\n2 5\n",
"6\n1 2\n1 3\n1 4\n2 5\n2 6\n"
],
"output": [
"YES",
"NO",
"NO",
"YES"
]
}
|
{
"input": [
"50\n16 4\n17 9\n31 19\n22 10\n8 1\n40 30\n3 31\n20 29\n47 27\n22 25\n32 34\n12 15\n40 32\n10 33\n47 12\n6 24\n46 41\n14 23\n12 35\n31 42\n46 28\n31 20\n46 37\n1 39\n29 49\n37 47\n40 6\n42 36\n47 2\n24 46\n2 13\n8 45\n41 3\n32 17\n4 7\n47 26\n28 8\n41 50\n34 44\n33 21\n25 5\n16 40\n3 14\n8 18\n28 11\n32 22\n2 38\n3 48\n44 43\n",
"10\n8 1\n1 2\n8 9\n8 5\n1 3\n1 10\n1 6\n1 7\n8 4\n",
"5\n5 1\n5 4\n4 3\n1 2\n",
"7\n1 2\n2 3\n1 4\n1 5\n3 6\n3 7\n",
"3\n1 3\n2 3\n",
"60\n26 6\n59 30\n31 12\n31 3\n38 23\n59 29\n53 9\n38 56\n53 54\n29 21\n17 55\n59 38\n26 16\n24 59\n24 25\n17 35\n24 41\n30 15\n31 27\n8 44\n26 5\n26 48\n8 32\n53 17\n3 34\n3 51\n30 28\n47 10\n53 60\n36 42\n24 53\n59 22\n53 40\n26 52\n36 4\n59 8\n29 37\n36 20\n17 47\n53 18\n3 50\n30 2\n17 7\n8 58\n59 1\n31 11\n24 26\n24 43\n53 57\n59 45\n47 13\n26 46\n17 33\n30 31\n26 39\n26 19\n24 36\n8 49\n38 14\n",
"7\n1 2\n2 3\n3 4\n3 5\n1 6\n1 7\n",
"20\n19 16\n19 18\n20 7\n9 4\n6 17\n14 2\n9 15\n2 13\n5 11\n19 12\n12 20\n16 9\n11 8\n19 5\n3 1\n19 14\n5 3\n12 10\n19 6\n",
"7\n1 2\n1 3\n2 4\n2 5\n3 6\n3 7\n",
"10\n9 5\n7 1\n9 10\n7 2\n5 4\n9 6\n2 9\n10 8\n1 3\n",
"4\n2 4\n2 3\n2 1\n",
"4\n1 4\n3 2\n1 3\n",
"3\n1 2\n1 3\n",
"5\n1 2\n1 5\n1 3\n1 4\n",
"20\n14 9\n12 13\n10 15\n2 1\n20 19\n16 6\n16 3\n17 14\n3 5\n2 11\n3 10\n15 8\n14 2\n6 4\n3 20\n5 18\n1 7\n1 16\n4 12\n",
"20\n7 5\n14 13\n17 6\n3 8\n16 12\n18 9\n3 18\n14 1\n17 3\n15 2\n17 4\n9 11\n2 7\n15 17\n3 20\n16 10\n17 14\n2 16\n1 19\n",
"8\n1 2\n2 3\n3 4\n1 7\n1 8\n4 5\n4 6\n",
"5\n5 1\n5 2\n5 3\n5 4\n",
"50\n49 6\n43 7\n1 27\n19 35\n15 37\n16 12\n19 21\n16 28\n49 9\n48 39\n13 1\n2 48\n9 50\n44 3\n41 32\n48 31\n49 33\n6 11\n13 20\n49 22\n13 41\n48 29\n13 46\n15 47\n34 2\n49 13\n48 14\n34 24\n16 36\n13 40\n49 34\n49 17\n43 25\n11 23\n10 15\n19 26\n34 44\n16 42\n19 18\n46 8\n29 38\n1 45\n12 43\n13 16\n46 30\n15 5\n49 10\n11 19\n32 4\n",
"20\n13 1\n18 2\n3 7\n18 5\n20 16\n3 12\n18 9\n3 10\n18 11\n13 6\n3 18\n20 15\n20 17\n3 13\n3 4\n13 14\n3 20\n18 8\n3 19\n",
"10\n8 2\n5 6\n1 8\n2 9\n1 4\n8 10\n10 5\n2 7\n2 3\n"
],
"output": [
"NO",
"YES",
"NO",
"NO",
"NO",
"YES",
"NO",
"NO",
"NO",
"NO",
"YES",
"NO",
"NO",
"YES",
"NO",
"NO",
"NO",
"YES",
"NO",
"YES",
"NO"
]
}
|
IN-CORRECT
|
java
|
//Author: Patel Rag
//Java version "1.8.0_211"
import java.util.*;
import java.io.*;
public class Main
{
static Reader fr;
static Print log;
static class Reader
{
private byte[] buffer = new byte[1024];
private int index;
private InputStream in;
private int total;
public Reader()
{
in = System.in;
}
public Reader(InputStream in)
{
this.in = in;
}
private int scan() throws IOException
{
if(index >= total)
{
index = 0;
total = in.read(buffer);
if(total <= 0)
{
return -1;
}
}
return buffer[index++];
}
public final int nextInt() throws IOException
{
return (int)nextLong();
}
public final long nextLong() throws IOException
{
long res = 0;
int n = scan();
while(isWhiteSpace(n))
{
n = scan();
}
int neg = 1;
if(n == '-')
{
neg = -1;
n = scan();
}
while(!isWhiteSpace(n))
{
if(n >= '0' && n <= '9')
{
res *= 10;
res += (n - '0');
n = scan();
}
else
{
throw new InputMismatchException();
}
}
return neg*res;
}
public final double nextDouble() throws IOException
{
double doub = 0;
int n = scan();
while(isWhiteSpace(n))
{
n=scan();
}
int neg = 1;
if(n == '-')
{
neg = -1;
n = scan();
}
while(!isWhiteSpace(n) && n != '.')
{
if(n >= '0' && n <= '9')
{
doub *= 10;
doub += n-'0';
n = scan();
}
else
{
throw new InputMismatchException();
}
}
if(n == '.')
{
n = scan();
double temp = 1;
while(!isWhiteSpace(n))
{
if(n >= '0' && n <= '9')
{
temp/=10;
doub += (n-'0')*temp;
n = scan();
}
else
{
throw new InputMismatchException();
}
}
}
return neg*doub;
}
public final String next() throws IOException
{
StringBuilder sb = new StringBuilder();
int n = scan();
while(isWhiteSpace(n))
{
n = scan();
}
while(!isWhiteSpace(n))
{
sb.append((char)n);
n = scan();
}
return sb.toString();
}
public final String nextLine() throws IOException
{
StringBuilder sb = new StringBuilder();
int n = scan();
while(isWhiteSpace(n))
{
n = scan();
}
while(n != '\n' && n != '\r' && n != -1)
{
sb.append((char)n);
n = scan();
}
return sb.toString();
}
private boolean isWhiteSpace(int n)
{
if(n == ' ' || n == '\n'|| n == '\r' || n == '\t'|| n == -1)
return true;
return false;
}
}
static class FastReader
{
BufferedReader br;
StringTokenizer st;
public FastReader()
{
br = new BufferedReader(new InputStreamReader(System.in));
}
String next()
{
while (st == null || !st.hasMoreElements())
{
try
{
st = new StringTokenizer(br.readLine());
}
catch (IOException e)
{
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt()
{
return Integer.parseInt(next());
}
long nextLong()
{
return Long.parseLong(next());
}
double nextDouble()
{
return Double.parseDouble(next());
}
String nextLine()
{
String str = "";
try
{
str = br.readLine();
}
catch (IOException e)
{
e.printStackTrace();
}
return str;
}
void close() throws IOException
{
br.close();
}
}
static class Print
{
private final BufferedWriter bw;
public Print()
{
this.bw = new BufferedWriter(new OutputStreamWriter(System.out));
}
public void print(Object object)throws IOException
{
bw.append(""+object);
}
public void println(Object object)throws IOException
{
print(object);
bw.append("\n");
}
public void close()throws IOException
{
bw.close();
}
}
static long modExp(long x, long n, long mod) //Modular exponentiation
{
long result = 1;
while(n > 0)
{
if(n % 2 == 1)
result = (result%mod * x%mod)%mod;
x = (x%mod * x%mod)%mod;
n=n/2;
}
return result;
}
static long gcd(long a, long b)
{
if(a==0) return b;
return gcd(b%a,a);
}
static class TrieNode
{
TrieNode[] children;
boolean isLeaf;
ArrayList<String> word;
public TrieNode()
{
children = new TrieNode[26];
for(int i = 0; i < 26; i++) children[i] = null;
isLeaf = false;
word = new ArrayList<>();
}
}
static TrieNode root;
static void insert(String word)
{
TrieNode pCrawl = root;
for(int lev = 0; lev < word.length(); lev++)
{
if(word.charAt(lev) < 'A' || word.charAt(lev) > 'Z') continue;
int alphabet = word.charAt(lev) - 'A';
if(pCrawl.children[alphabet] == null)
{
pCrawl.children[alphabet] = new TrieNode();
}
pCrawl = pCrawl.children[alphabet];
}
pCrawl.isLeaf = true;
(pCrawl.word).add(word);
}
static void printAll(TrieNode root) throws IOException
{
if(root.isLeaf)
{
for(String str : root.word)
{
log.print(str + " ");
}
}
for(int i = 0; i < 26; i++)
{
if(root.children[i] != null)
printAll(root.children[i]);
}
}
static boolean search(String pattern) throws IOException
{
int index;
TrieNode pCrawl = root;
for(int i = 0; i < pattern.length(); i++)
{
index = pattern.charAt(i) - 'A';
if(pCrawl.children[index] == null)
return false;
pCrawl = pCrawl.children[index];
}
printAll(pCrawl);
return true;
}
public static void main(String[] args)
throws IOException
{
fr = new Reader();
log = new Print();
int n = fr.nextInt();
int[] deg = new int[n];
for(int i = 0; i < n-1; i++)
{
int u = fr.nextInt() - 1;
int v = fr.nextInt() - 1;
deg[u]++;
deg[v]++;
}
for(int i = 0; i < n-1; i++)
{
if(deg[i] == 2)
{
log.println("NO");
log.close();
return;
}
}
log.println("YES");
log.close();
}
}
class Pair<U, V> // Pair class
{
public final U first; // first field of a Pair
public final V second; // second field of a Pair
private Pair(U first, V second)
{
this.first = first;
this.second = second;
}
@Override
public boolean equals(Object o)
{
if (this == o) return true;
if (o == null || getClass() != o.getClass()) return false;
Pair<?, ?> pair = (Pair<?, ?>) o;
if (!first.equals(pair.first)) return false;
return second.equals(pair.second);
}
@Override
public int hashCode()
{
return 31 * first.hashCode() + second.hashCode();
}
public static <U, V> Pair <U, V> of(U a, V b)
{
return new Pair<>(a, b);
}
}
class myComp implements Comparator<Pair>
{
public int compare(Pair a,Pair b)
{
if((Integer)a.second <= (Integer)b.second) return 1;
return -1;
}
}
class BIT //Binary Indexed Tree
{
public long[] m_array;
public BIT(long[] dat)
{
m_array = new long[dat.length + 1];
Arrays.fill(m_array,0);
for(int i = 0; i < dat.length; i++)
{
m_array[i + 1] = dat[i];
}
for(int i = 1; i < m_array.length; i++)
{
int j = i + (i & -i);
if(j < m_array.length)
{
m_array[j] = m_array[j] + m_array[i];
}
}
}
public final long prefix_query(int i)
{
long result = 0;
for(++i; i > 0; i = i - (i & -i))
{
result = result + m_array[i];
}
return result;
}
public final long range_query(int fro, int to)
{
if(fro == 0)
{
return prefix_query(to);
}
else
{
return (prefix_query(to) - prefix_query(fro - 1));
}
}
public void update(int i, long add)
{
for(++i; i < m_array.length; i = i + (i & -i))
{
m_array[i] = m_array[i] + add;
}
}
}
|
1189_D1. Add on a Tree
|
Note that this is the first problem of the two similar problems. You can hack this problem only if you solve both problems.
You are given a tree with n nodes. In the beginning, 0 is written on all edges. In one operation, you can choose any 2 distinct leaves u, v and any real number x and add x to values written on all edges on the simple path between u and v.
For example, on the picture below you can see the result of applying two operations to the graph: adding 2 on the path from 7 to 6, and then adding -0.5 on the path from 4 to 5.
<image>
Is it true that for any configuration of real numbers written on edges, we can achieve it with a finite number of operations?
Leaf is a node of a tree of degree 1. Simple path is a path that doesn't contain any node twice.
Input
The first line contains a single integer n (2 ≤ n ≤ 10^5) — the number of nodes.
Each of the next n-1 lines contains two integers u and v (1 ≤ u, v ≤ n, u ≠ v), meaning that there is an edge between nodes u and v. It is guaranteed that these edges form a tree.
Output
If there is a configuration of real numbers written on edges of the tree that we can't achieve by performing the operations, output "NO".
Otherwise, output "YES".
You can print each letter in any case (upper or lower).
Examples
Input
2
1 2
Output
YES
Input
3
1 2
2 3
Output
NO
Input
5
1 2
1 3
1 4
2 5
Output
NO
Input
6
1 2
1 3
1 4
2 5
2 6
Output
YES
Note
In the first example, we can add any real x to the value written on the only edge (1, 2).
<image>
In the second example, one of configurations that we can't reach is 0 written on (1, 2) and 1 written on (2, 3).
<image>
Below you can see graphs from examples 3, 4:
<image> <image>
|
{
"input": [
"2\n1 2\n",
"3\n1 2\n2 3\n",
"5\n1 2\n1 3\n1 4\n2 5\n",
"6\n1 2\n1 3\n1 4\n2 5\n2 6\n"
],
"output": [
"YES",
"NO",
"NO",
"YES"
]
}
|
{
"input": [
"50\n16 4\n17 9\n31 19\n22 10\n8 1\n40 30\n3 31\n20 29\n47 27\n22 25\n32 34\n12 15\n40 32\n10 33\n47 12\n6 24\n46 41\n14 23\n12 35\n31 42\n46 28\n31 20\n46 37\n1 39\n29 49\n37 47\n40 6\n42 36\n47 2\n24 46\n2 13\n8 45\n41 3\n32 17\n4 7\n47 26\n28 8\n41 50\n34 44\n33 21\n25 5\n16 40\n3 14\n8 18\n28 11\n32 22\n2 38\n3 48\n44 43\n",
"10\n8 1\n1 2\n8 9\n8 5\n1 3\n1 10\n1 6\n1 7\n8 4\n",
"5\n5 1\n5 4\n4 3\n1 2\n",
"7\n1 2\n2 3\n1 4\n1 5\n3 6\n3 7\n",
"3\n1 3\n2 3\n",
"60\n26 6\n59 30\n31 12\n31 3\n38 23\n59 29\n53 9\n38 56\n53 54\n29 21\n17 55\n59 38\n26 16\n24 59\n24 25\n17 35\n24 41\n30 15\n31 27\n8 44\n26 5\n26 48\n8 32\n53 17\n3 34\n3 51\n30 28\n47 10\n53 60\n36 42\n24 53\n59 22\n53 40\n26 52\n36 4\n59 8\n29 37\n36 20\n17 47\n53 18\n3 50\n30 2\n17 7\n8 58\n59 1\n31 11\n24 26\n24 43\n53 57\n59 45\n47 13\n26 46\n17 33\n30 31\n26 39\n26 19\n24 36\n8 49\n38 14\n",
"7\n1 2\n2 3\n3 4\n3 5\n1 6\n1 7\n",
"20\n19 16\n19 18\n20 7\n9 4\n6 17\n14 2\n9 15\n2 13\n5 11\n19 12\n12 20\n16 9\n11 8\n19 5\n3 1\n19 14\n5 3\n12 10\n19 6\n",
"7\n1 2\n1 3\n2 4\n2 5\n3 6\n3 7\n",
"10\n9 5\n7 1\n9 10\n7 2\n5 4\n9 6\n2 9\n10 8\n1 3\n",
"4\n2 4\n2 3\n2 1\n",
"4\n1 4\n3 2\n1 3\n",
"3\n1 2\n1 3\n",
"5\n1 2\n1 5\n1 3\n1 4\n",
"20\n14 9\n12 13\n10 15\n2 1\n20 19\n16 6\n16 3\n17 14\n3 5\n2 11\n3 10\n15 8\n14 2\n6 4\n3 20\n5 18\n1 7\n1 16\n4 12\n",
"20\n7 5\n14 13\n17 6\n3 8\n16 12\n18 9\n3 18\n14 1\n17 3\n15 2\n17 4\n9 11\n2 7\n15 17\n3 20\n16 10\n17 14\n2 16\n1 19\n",
"8\n1 2\n2 3\n3 4\n1 7\n1 8\n4 5\n4 6\n",
"5\n5 1\n5 2\n5 3\n5 4\n",
"50\n49 6\n43 7\n1 27\n19 35\n15 37\n16 12\n19 21\n16 28\n49 9\n48 39\n13 1\n2 48\n9 50\n44 3\n41 32\n48 31\n49 33\n6 11\n13 20\n49 22\n13 41\n48 29\n13 46\n15 47\n34 2\n49 13\n48 14\n34 24\n16 36\n13 40\n49 34\n49 17\n43 25\n11 23\n10 15\n19 26\n34 44\n16 42\n19 18\n46 8\n29 38\n1 45\n12 43\n13 16\n46 30\n15 5\n49 10\n11 19\n32 4\n",
"20\n13 1\n18 2\n3 7\n18 5\n20 16\n3 12\n18 9\n3 10\n18 11\n13 6\n3 18\n20 15\n20 17\n3 13\n3 4\n13 14\n3 20\n18 8\n3 19\n",
"10\n8 2\n5 6\n1 8\n2 9\n1 4\n8 10\n10 5\n2 7\n2 3\n"
],
"output": [
"NO",
"YES",
"NO",
"NO",
"NO",
"YES",
"NO",
"NO",
"NO",
"NO",
"YES",
"NO",
"NO",
"YES",
"NO",
"NO",
"NO",
"YES",
"NO",
"YES",
"NO"
]
}
|
IN-CORRECT
|
python3
|
n = int(input())
deg = [0]*n
for i in range(n-1):
u, v = map(int, input().split())
deg[u-1] += 1
deg[v-1] += 1
cnt = sum(1 for i in deg if i == 1)
if cnt*(cnt-1)//2 >= n - 1:
print("YES")
else:
print("NO")
|
1189_D1. Add on a Tree
|
Note that this is the first problem of the two similar problems. You can hack this problem only if you solve both problems.
You are given a tree with n nodes. In the beginning, 0 is written on all edges. In one operation, you can choose any 2 distinct leaves u, v and any real number x and add x to values written on all edges on the simple path between u and v.
For example, on the picture below you can see the result of applying two operations to the graph: adding 2 on the path from 7 to 6, and then adding -0.5 on the path from 4 to 5.
<image>
Is it true that for any configuration of real numbers written on edges, we can achieve it with a finite number of operations?
Leaf is a node of a tree of degree 1. Simple path is a path that doesn't contain any node twice.
Input
The first line contains a single integer n (2 ≤ n ≤ 10^5) — the number of nodes.
Each of the next n-1 lines contains two integers u and v (1 ≤ u, v ≤ n, u ≠ v), meaning that there is an edge between nodes u and v. It is guaranteed that these edges form a tree.
Output
If there is a configuration of real numbers written on edges of the tree that we can't achieve by performing the operations, output "NO".
Otherwise, output "YES".
You can print each letter in any case (upper or lower).
Examples
Input
2
1 2
Output
YES
Input
3
1 2
2 3
Output
NO
Input
5
1 2
1 3
1 4
2 5
Output
NO
Input
6
1 2
1 3
1 4
2 5
2 6
Output
YES
Note
In the first example, we can add any real x to the value written on the only edge (1, 2).
<image>
In the second example, one of configurations that we can't reach is 0 written on (1, 2) and 1 written on (2, 3).
<image>
Below you can see graphs from examples 3, 4:
<image> <image>
|
{
"input": [
"2\n1 2\n",
"3\n1 2\n2 3\n",
"5\n1 2\n1 3\n1 4\n2 5\n",
"6\n1 2\n1 3\n1 4\n2 5\n2 6\n"
],
"output": [
"YES",
"NO",
"NO",
"YES"
]
}
|
{
"input": [
"50\n16 4\n17 9\n31 19\n22 10\n8 1\n40 30\n3 31\n20 29\n47 27\n22 25\n32 34\n12 15\n40 32\n10 33\n47 12\n6 24\n46 41\n14 23\n12 35\n31 42\n46 28\n31 20\n46 37\n1 39\n29 49\n37 47\n40 6\n42 36\n47 2\n24 46\n2 13\n8 45\n41 3\n32 17\n4 7\n47 26\n28 8\n41 50\n34 44\n33 21\n25 5\n16 40\n3 14\n8 18\n28 11\n32 22\n2 38\n3 48\n44 43\n",
"10\n8 1\n1 2\n8 9\n8 5\n1 3\n1 10\n1 6\n1 7\n8 4\n",
"5\n5 1\n5 4\n4 3\n1 2\n",
"7\n1 2\n2 3\n1 4\n1 5\n3 6\n3 7\n",
"3\n1 3\n2 3\n",
"60\n26 6\n59 30\n31 12\n31 3\n38 23\n59 29\n53 9\n38 56\n53 54\n29 21\n17 55\n59 38\n26 16\n24 59\n24 25\n17 35\n24 41\n30 15\n31 27\n8 44\n26 5\n26 48\n8 32\n53 17\n3 34\n3 51\n30 28\n47 10\n53 60\n36 42\n24 53\n59 22\n53 40\n26 52\n36 4\n59 8\n29 37\n36 20\n17 47\n53 18\n3 50\n30 2\n17 7\n8 58\n59 1\n31 11\n24 26\n24 43\n53 57\n59 45\n47 13\n26 46\n17 33\n30 31\n26 39\n26 19\n24 36\n8 49\n38 14\n",
"7\n1 2\n2 3\n3 4\n3 5\n1 6\n1 7\n",
"20\n19 16\n19 18\n20 7\n9 4\n6 17\n14 2\n9 15\n2 13\n5 11\n19 12\n12 20\n16 9\n11 8\n19 5\n3 1\n19 14\n5 3\n12 10\n19 6\n",
"7\n1 2\n1 3\n2 4\n2 5\n3 6\n3 7\n",
"10\n9 5\n7 1\n9 10\n7 2\n5 4\n9 6\n2 9\n10 8\n1 3\n",
"4\n2 4\n2 3\n2 1\n",
"4\n1 4\n3 2\n1 3\n",
"3\n1 2\n1 3\n",
"5\n1 2\n1 5\n1 3\n1 4\n",
"20\n14 9\n12 13\n10 15\n2 1\n20 19\n16 6\n16 3\n17 14\n3 5\n2 11\n3 10\n15 8\n14 2\n6 4\n3 20\n5 18\n1 7\n1 16\n4 12\n",
"20\n7 5\n14 13\n17 6\n3 8\n16 12\n18 9\n3 18\n14 1\n17 3\n15 2\n17 4\n9 11\n2 7\n15 17\n3 20\n16 10\n17 14\n2 16\n1 19\n",
"8\n1 2\n2 3\n3 4\n1 7\n1 8\n4 5\n4 6\n",
"5\n5 1\n5 2\n5 3\n5 4\n",
"50\n49 6\n43 7\n1 27\n19 35\n15 37\n16 12\n19 21\n16 28\n49 9\n48 39\n13 1\n2 48\n9 50\n44 3\n41 32\n48 31\n49 33\n6 11\n13 20\n49 22\n13 41\n48 29\n13 46\n15 47\n34 2\n49 13\n48 14\n34 24\n16 36\n13 40\n49 34\n49 17\n43 25\n11 23\n10 15\n19 26\n34 44\n16 42\n19 18\n46 8\n29 38\n1 45\n12 43\n13 16\n46 30\n15 5\n49 10\n11 19\n32 4\n",
"20\n13 1\n18 2\n3 7\n18 5\n20 16\n3 12\n18 9\n3 10\n18 11\n13 6\n3 18\n20 15\n20 17\n3 13\n3 4\n13 14\n3 20\n18 8\n3 19\n",
"10\n8 2\n5 6\n1 8\n2 9\n1 4\n8 10\n10 5\n2 7\n2 3\n"
],
"output": [
"NO",
"YES",
"NO",
"NO",
"NO",
"YES",
"NO",
"NO",
"NO",
"NO",
"YES",
"NO",
"NO",
"YES",
"NO",
"NO",
"NO",
"YES",
"NO",
"YES",
"NO"
]
}
|
IN-CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
const long long mod = 1000000007;
long long powmod(long long a, long long b) {
long long res = 1;
a %= mod;
assert(b >= 0);
for (; b; b >>= 1) {
if (b & 1) res = res * a % mod;
a = a * a % mod;
}
return res;
}
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
const int maxn = 1e5 + 100;
int n, x;
int a[maxn];
int main(int argc, char const *argv[]) {
ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
cin >> n;
for (int i = 0; i < n; i++) {
cin >> x;
a[x]++;
cin >> x;
a[x]++;
}
for (int i = 1; i < n + 1; i++) {
if (a[i] == 2) {
cout << "No" << endl;
return 0;
}
}
cout << "YES" << endl;
return 0;
}
|
1189_D1. Add on a Tree
|
Note that this is the first problem of the two similar problems. You can hack this problem only if you solve both problems.
You are given a tree with n nodes. In the beginning, 0 is written on all edges. In one operation, you can choose any 2 distinct leaves u, v and any real number x and add x to values written on all edges on the simple path between u and v.
For example, on the picture below you can see the result of applying two operations to the graph: adding 2 on the path from 7 to 6, and then adding -0.5 on the path from 4 to 5.
<image>
Is it true that for any configuration of real numbers written on edges, we can achieve it with a finite number of operations?
Leaf is a node of a tree of degree 1. Simple path is a path that doesn't contain any node twice.
Input
The first line contains a single integer n (2 ≤ n ≤ 10^5) — the number of nodes.
Each of the next n-1 lines contains two integers u and v (1 ≤ u, v ≤ n, u ≠ v), meaning that there is an edge between nodes u and v. It is guaranteed that these edges form a tree.
Output
If there is a configuration of real numbers written on edges of the tree that we can't achieve by performing the operations, output "NO".
Otherwise, output "YES".
You can print each letter in any case (upper or lower).
Examples
Input
2
1 2
Output
YES
Input
3
1 2
2 3
Output
NO
Input
5
1 2
1 3
1 4
2 5
Output
NO
Input
6
1 2
1 3
1 4
2 5
2 6
Output
YES
Note
In the first example, we can add any real x to the value written on the only edge (1, 2).
<image>
In the second example, one of configurations that we can't reach is 0 written on (1, 2) and 1 written on (2, 3).
<image>
Below you can see graphs from examples 3, 4:
<image> <image>
|
{
"input": [
"2\n1 2\n",
"3\n1 2\n2 3\n",
"5\n1 2\n1 3\n1 4\n2 5\n",
"6\n1 2\n1 3\n1 4\n2 5\n2 6\n"
],
"output": [
"YES",
"NO",
"NO",
"YES"
]
}
|
{
"input": [
"50\n16 4\n17 9\n31 19\n22 10\n8 1\n40 30\n3 31\n20 29\n47 27\n22 25\n32 34\n12 15\n40 32\n10 33\n47 12\n6 24\n46 41\n14 23\n12 35\n31 42\n46 28\n31 20\n46 37\n1 39\n29 49\n37 47\n40 6\n42 36\n47 2\n24 46\n2 13\n8 45\n41 3\n32 17\n4 7\n47 26\n28 8\n41 50\n34 44\n33 21\n25 5\n16 40\n3 14\n8 18\n28 11\n32 22\n2 38\n3 48\n44 43\n",
"10\n8 1\n1 2\n8 9\n8 5\n1 3\n1 10\n1 6\n1 7\n8 4\n",
"5\n5 1\n5 4\n4 3\n1 2\n",
"7\n1 2\n2 3\n1 4\n1 5\n3 6\n3 7\n",
"3\n1 3\n2 3\n",
"60\n26 6\n59 30\n31 12\n31 3\n38 23\n59 29\n53 9\n38 56\n53 54\n29 21\n17 55\n59 38\n26 16\n24 59\n24 25\n17 35\n24 41\n30 15\n31 27\n8 44\n26 5\n26 48\n8 32\n53 17\n3 34\n3 51\n30 28\n47 10\n53 60\n36 42\n24 53\n59 22\n53 40\n26 52\n36 4\n59 8\n29 37\n36 20\n17 47\n53 18\n3 50\n30 2\n17 7\n8 58\n59 1\n31 11\n24 26\n24 43\n53 57\n59 45\n47 13\n26 46\n17 33\n30 31\n26 39\n26 19\n24 36\n8 49\n38 14\n",
"7\n1 2\n2 3\n3 4\n3 5\n1 6\n1 7\n",
"20\n19 16\n19 18\n20 7\n9 4\n6 17\n14 2\n9 15\n2 13\n5 11\n19 12\n12 20\n16 9\n11 8\n19 5\n3 1\n19 14\n5 3\n12 10\n19 6\n",
"7\n1 2\n1 3\n2 4\n2 5\n3 6\n3 7\n",
"10\n9 5\n7 1\n9 10\n7 2\n5 4\n9 6\n2 9\n10 8\n1 3\n",
"4\n2 4\n2 3\n2 1\n",
"4\n1 4\n3 2\n1 3\n",
"3\n1 2\n1 3\n",
"5\n1 2\n1 5\n1 3\n1 4\n",
"20\n14 9\n12 13\n10 15\n2 1\n20 19\n16 6\n16 3\n17 14\n3 5\n2 11\n3 10\n15 8\n14 2\n6 4\n3 20\n5 18\n1 7\n1 16\n4 12\n",
"20\n7 5\n14 13\n17 6\n3 8\n16 12\n18 9\n3 18\n14 1\n17 3\n15 2\n17 4\n9 11\n2 7\n15 17\n3 20\n16 10\n17 14\n2 16\n1 19\n",
"8\n1 2\n2 3\n3 4\n1 7\n1 8\n4 5\n4 6\n",
"5\n5 1\n5 2\n5 3\n5 4\n",
"50\n49 6\n43 7\n1 27\n19 35\n15 37\n16 12\n19 21\n16 28\n49 9\n48 39\n13 1\n2 48\n9 50\n44 3\n41 32\n48 31\n49 33\n6 11\n13 20\n49 22\n13 41\n48 29\n13 46\n15 47\n34 2\n49 13\n48 14\n34 24\n16 36\n13 40\n49 34\n49 17\n43 25\n11 23\n10 15\n19 26\n34 44\n16 42\n19 18\n46 8\n29 38\n1 45\n12 43\n13 16\n46 30\n15 5\n49 10\n11 19\n32 4\n",
"20\n13 1\n18 2\n3 7\n18 5\n20 16\n3 12\n18 9\n3 10\n18 11\n13 6\n3 18\n20 15\n20 17\n3 13\n3 4\n13 14\n3 20\n18 8\n3 19\n",
"10\n8 2\n5 6\n1 8\n2 9\n1 4\n8 10\n10 5\n2 7\n2 3\n"
],
"output": [
"NO",
"YES",
"NO",
"NO",
"NO",
"YES",
"NO",
"NO",
"NO",
"NO",
"YES",
"NO",
"NO",
"YES",
"NO",
"NO",
"NO",
"YES",
"NO",
"YES",
"NO"
]
}
|
IN-CORRECT
|
python3
|
#! /usr/bin/env python3
import sys
from math import factorial
class Node:
def __init__(self, num):
self.num = num
self.binds = []
# for dijkstra
self.marker = False
self.val = None
def add_bind(self, oth):
self.binds.append(oth)
def __repr__(self):
return '<{}: {}{}>'.format(
self.num,
[i.num for i in self.binds],
', \tval: {}'.format(self.val) if self.val != None else ''
)
class Graph:
def __init__(self, size):
self.size = size
self.nodes = [None] + [Node(num) for num in range(1, size+1)]
def read_input(self):
for _ in range(1, self.size):
i, j = (int(x) for x in sys.stdin.readline().split())
self.nodes[i].add_bind(self.nodes[j])
self.nodes[j].add_bind(self.nodes[i])
def __repr__(self):
return '\n'.join(str(node) for node in self.nodes[1:])
def pairs(n):
return factorial(n) // ( factorial(n-2) * 2 )
N = int(sys.stdin.readline())
g = Graph(N)
g.read_input()
#print(g)
ends = [node for node in g.nodes[1:] if len(node.binds) == 1]
#print(pairs(len(ends)), , len(ends))
print('YES' if pairs(len(ends)) >= N-1 else 'NO')
|
1189_D1. Add on a Tree
|
Note that this is the first problem of the two similar problems. You can hack this problem only if you solve both problems.
You are given a tree with n nodes. In the beginning, 0 is written on all edges. In one operation, you can choose any 2 distinct leaves u, v and any real number x and add x to values written on all edges on the simple path between u and v.
For example, on the picture below you can see the result of applying two operations to the graph: adding 2 on the path from 7 to 6, and then adding -0.5 on the path from 4 to 5.
<image>
Is it true that for any configuration of real numbers written on edges, we can achieve it with a finite number of operations?
Leaf is a node of a tree of degree 1. Simple path is a path that doesn't contain any node twice.
Input
The first line contains a single integer n (2 ≤ n ≤ 10^5) — the number of nodes.
Each of the next n-1 lines contains two integers u and v (1 ≤ u, v ≤ n, u ≠ v), meaning that there is an edge between nodes u and v. It is guaranteed that these edges form a tree.
Output
If there is a configuration of real numbers written on edges of the tree that we can't achieve by performing the operations, output "NO".
Otherwise, output "YES".
You can print each letter in any case (upper or lower).
Examples
Input
2
1 2
Output
YES
Input
3
1 2
2 3
Output
NO
Input
5
1 2
1 3
1 4
2 5
Output
NO
Input
6
1 2
1 3
1 4
2 5
2 6
Output
YES
Note
In the first example, we can add any real x to the value written on the only edge (1, 2).
<image>
In the second example, one of configurations that we can't reach is 0 written on (1, 2) and 1 written on (2, 3).
<image>
Below you can see graphs from examples 3, 4:
<image> <image>
|
{
"input": [
"2\n1 2\n",
"3\n1 2\n2 3\n",
"5\n1 2\n1 3\n1 4\n2 5\n",
"6\n1 2\n1 3\n1 4\n2 5\n2 6\n"
],
"output": [
"YES",
"NO",
"NO",
"YES"
]
}
|
{
"input": [
"50\n16 4\n17 9\n31 19\n22 10\n8 1\n40 30\n3 31\n20 29\n47 27\n22 25\n32 34\n12 15\n40 32\n10 33\n47 12\n6 24\n46 41\n14 23\n12 35\n31 42\n46 28\n31 20\n46 37\n1 39\n29 49\n37 47\n40 6\n42 36\n47 2\n24 46\n2 13\n8 45\n41 3\n32 17\n4 7\n47 26\n28 8\n41 50\n34 44\n33 21\n25 5\n16 40\n3 14\n8 18\n28 11\n32 22\n2 38\n3 48\n44 43\n",
"10\n8 1\n1 2\n8 9\n8 5\n1 3\n1 10\n1 6\n1 7\n8 4\n",
"5\n5 1\n5 4\n4 3\n1 2\n",
"7\n1 2\n2 3\n1 4\n1 5\n3 6\n3 7\n",
"3\n1 3\n2 3\n",
"60\n26 6\n59 30\n31 12\n31 3\n38 23\n59 29\n53 9\n38 56\n53 54\n29 21\n17 55\n59 38\n26 16\n24 59\n24 25\n17 35\n24 41\n30 15\n31 27\n8 44\n26 5\n26 48\n8 32\n53 17\n3 34\n3 51\n30 28\n47 10\n53 60\n36 42\n24 53\n59 22\n53 40\n26 52\n36 4\n59 8\n29 37\n36 20\n17 47\n53 18\n3 50\n30 2\n17 7\n8 58\n59 1\n31 11\n24 26\n24 43\n53 57\n59 45\n47 13\n26 46\n17 33\n30 31\n26 39\n26 19\n24 36\n8 49\n38 14\n",
"7\n1 2\n2 3\n3 4\n3 5\n1 6\n1 7\n",
"20\n19 16\n19 18\n20 7\n9 4\n6 17\n14 2\n9 15\n2 13\n5 11\n19 12\n12 20\n16 9\n11 8\n19 5\n3 1\n19 14\n5 3\n12 10\n19 6\n",
"7\n1 2\n1 3\n2 4\n2 5\n3 6\n3 7\n",
"10\n9 5\n7 1\n9 10\n7 2\n5 4\n9 6\n2 9\n10 8\n1 3\n",
"4\n2 4\n2 3\n2 1\n",
"4\n1 4\n3 2\n1 3\n",
"3\n1 2\n1 3\n",
"5\n1 2\n1 5\n1 3\n1 4\n",
"20\n14 9\n12 13\n10 15\n2 1\n20 19\n16 6\n16 3\n17 14\n3 5\n2 11\n3 10\n15 8\n14 2\n6 4\n3 20\n5 18\n1 7\n1 16\n4 12\n",
"20\n7 5\n14 13\n17 6\n3 8\n16 12\n18 9\n3 18\n14 1\n17 3\n15 2\n17 4\n9 11\n2 7\n15 17\n3 20\n16 10\n17 14\n2 16\n1 19\n",
"8\n1 2\n2 3\n3 4\n1 7\n1 8\n4 5\n4 6\n",
"5\n5 1\n5 2\n5 3\n5 4\n",
"50\n49 6\n43 7\n1 27\n19 35\n15 37\n16 12\n19 21\n16 28\n49 9\n48 39\n13 1\n2 48\n9 50\n44 3\n41 32\n48 31\n49 33\n6 11\n13 20\n49 22\n13 41\n48 29\n13 46\n15 47\n34 2\n49 13\n48 14\n34 24\n16 36\n13 40\n49 34\n49 17\n43 25\n11 23\n10 15\n19 26\n34 44\n16 42\n19 18\n46 8\n29 38\n1 45\n12 43\n13 16\n46 30\n15 5\n49 10\n11 19\n32 4\n",
"20\n13 1\n18 2\n3 7\n18 5\n20 16\n3 12\n18 9\n3 10\n18 11\n13 6\n3 18\n20 15\n20 17\n3 13\n3 4\n13 14\n3 20\n18 8\n3 19\n",
"10\n8 2\n5 6\n1 8\n2 9\n1 4\n8 10\n10 5\n2 7\n2 3\n"
],
"output": [
"NO",
"YES",
"NO",
"NO",
"NO",
"YES",
"NO",
"NO",
"NO",
"NO",
"YES",
"NO",
"NO",
"YES",
"NO",
"NO",
"NO",
"YES",
"NO",
"YES",
"NO"
]
}
|
IN-CORRECT
|
java
|
import java.util.ArrayList;
import java.util.LinkedList;
import java.util.List;
import java.util.Scanner;
// https://codeforces.com/contest/1189/problem/D1
// 只要有一个顶点的度为 2，那么就一定有无法满足的情况，输出 NO
public class Main {
static List<List<Integer>> node = new ArrayList<List<Integer>>();
static void addList(int from, int to) {
node.get(from).add(to);
}
public static void main(String[] args) {
Scanner scanner = new Scanner(System.in);
int n = scanner.nextInt();
for (int i = 0; i < n+1; i++) {
node.add(new LinkedList<Integer>());
}
for (int i = 0; i < n-1; i++) {
int a = scanner.nextInt();
int b = scanner.nextInt();
addList(a, b);
addList(b, a);
}
scanner.close();
for (int i = 0; i < n; i++) {
if (node.get(i).size() == 2) {
System.out.println("NO");
return;
}
}
System.out.println("YES");
}
}
|
1189_D1. Add on a Tree
|
Note that this is the first problem of the two similar problems. You can hack this problem only if you solve both problems.
You are given a tree with n nodes. In the beginning, 0 is written on all edges. In one operation, you can choose any 2 distinct leaves u, v and any real number x and add x to values written on all edges on the simple path between u and v.
For example, on the picture below you can see the result of applying two operations to the graph: adding 2 on the path from 7 to 6, and then adding -0.5 on the path from 4 to 5.
<image>
Is it true that for any configuration of real numbers written on edges, we can achieve it with a finite number of operations?
Leaf is a node of a tree of degree 1. Simple path is a path that doesn't contain any node twice.
Input
The first line contains a single integer n (2 ≤ n ≤ 10^5) — the number of nodes.
Each of the next n-1 lines contains two integers u and v (1 ≤ u, v ≤ n, u ≠ v), meaning that there is an edge between nodes u and v. It is guaranteed that these edges form a tree.
Output
If there is a configuration of real numbers written on edges of the tree that we can't achieve by performing the operations, output "NO".
Otherwise, output "YES".
You can print each letter in any case (upper or lower).
Examples
Input
2
1 2
Output
YES
Input
3
1 2
2 3
Output
NO
Input
5
1 2
1 3
1 4
2 5
Output
NO
Input
6
1 2
1 3
1 4
2 5
2 6
Output
YES
Note
In the first example, we can add any real x to the value written on the only edge (1, 2).
<image>
In the second example, one of configurations that we can't reach is 0 written on (1, 2) and 1 written on (2, 3).
<image>
Below you can see graphs from examples 3, 4:
<image> <image>
|
{
"input": [
"2\n1 2\n",
"3\n1 2\n2 3\n",
"5\n1 2\n1 3\n1 4\n2 5\n",
"6\n1 2\n1 3\n1 4\n2 5\n2 6\n"
],
"output": [
"YES",
"NO",
"NO",
"YES"
]
}
|
{
"input": [
"50\n16 4\n17 9\n31 19\n22 10\n8 1\n40 30\n3 31\n20 29\n47 27\n22 25\n32 34\n12 15\n40 32\n10 33\n47 12\n6 24\n46 41\n14 23\n12 35\n31 42\n46 28\n31 20\n46 37\n1 39\n29 49\n37 47\n40 6\n42 36\n47 2\n24 46\n2 13\n8 45\n41 3\n32 17\n4 7\n47 26\n28 8\n41 50\n34 44\n33 21\n25 5\n16 40\n3 14\n8 18\n28 11\n32 22\n2 38\n3 48\n44 43\n",
"10\n8 1\n1 2\n8 9\n8 5\n1 3\n1 10\n1 6\n1 7\n8 4\n",
"5\n5 1\n5 4\n4 3\n1 2\n",
"7\n1 2\n2 3\n1 4\n1 5\n3 6\n3 7\n",
"3\n1 3\n2 3\n",
"60\n26 6\n59 30\n31 12\n31 3\n38 23\n59 29\n53 9\n38 56\n53 54\n29 21\n17 55\n59 38\n26 16\n24 59\n24 25\n17 35\n24 41\n30 15\n31 27\n8 44\n26 5\n26 48\n8 32\n53 17\n3 34\n3 51\n30 28\n47 10\n53 60\n36 42\n24 53\n59 22\n53 40\n26 52\n36 4\n59 8\n29 37\n36 20\n17 47\n53 18\n3 50\n30 2\n17 7\n8 58\n59 1\n31 11\n24 26\n24 43\n53 57\n59 45\n47 13\n26 46\n17 33\n30 31\n26 39\n26 19\n24 36\n8 49\n38 14\n",
"7\n1 2\n2 3\n3 4\n3 5\n1 6\n1 7\n",
"20\n19 16\n19 18\n20 7\n9 4\n6 17\n14 2\n9 15\n2 13\n5 11\n19 12\n12 20\n16 9\n11 8\n19 5\n3 1\n19 14\n5 3\n12 10\n19 6\n",
"7\n1 2\n1 3\n2 4\n2 5\n3 6\n3 7\n",
"10\n9 5\n7 1\n9 10\n7 2\n5 4\n9 6\n2 9\n10 8\n1 3\n",
"4\n2 4\n2 3\n2 1\n",
"4\n1 4\n3 2\n1 3\n",
"3\n1 2\n1 3\n",
"5\n1 2\n1 5\n1 3\n1 4\n",
"20\n14 9\n12 13\n10 15\n2 1\n20 19\n16 6\n16 3\n17 14\n3 5\n2 11\n3 10\n15 8\n14 2\n6 4\n3 20\n5 18\n1 7\n1 16\n4 12\n",
"20\n7 5\n14 13\n17 6\n3 8\n16 12\n18 9\n3 18\n14 1\n17 3\n15 2\n17 4\n9 11\n2 7\n15 17\n3 20\n16 10\n17 14\n2 16\n1 19\n",
"8\n1 2\n2 3\n3 4\n1 7\n1 8\n4 5\n4 6\n",
"5\n5 1\n5 2\n5 3\n5 4\n",
"50\n49 6\n43 7\n1 27\n19 35\n15 37\n16 12\n19 21\n16 28\n49 9\n48 39\n13 1\n2 48\n9 50\n44 3\n41 32\n48 31\n49 33\n6 11\n13 20\n49 22\n13 41\n48 29\n13 46\n15 47\n34 2\n49 13\n48 14\n34 24\n16 36\n13 40\n49 34\n49 17\n43 25\n11 23\n10 15\n19 26\n34 44\n16 42\n19 18\n46 8\n29 38\n1 45\n12 43\n13 16\n46 30\n15 5\n49 10\n11 19\n32 4\n",
"20\n13 1\n18 2\n3 7\n18 5\n20 16\n3 12\n18 9\n3 10\n18 11\n13 6\n3 18\n20 15\n20 17\n3 13\n3 4\n13 14\n3 20\n18 8\n3 19\n",
"10\n8 2\n5 6\n1 8\n2 9\n1 4\n8 10\n10 5\n2 7\n2 3\n"
],
"output": [
"NO",
"YES",
"NO",
"NO",
"NO",
"YES",
"NO",
"NO",
"NO",
"NO",
"YES",
"NO",
"NO",
"YES",
"NO",
"NO",
"NO",
"YES",
"NO",
"YES",
"NO"
]
}
|
IN-CORRECT
|
python3
|
n = int(input())
tree = [[] for i in range(n)]
for i in range(n - 1):
a1, a2 = map(int, input().split())
tree[a1 - 1].append(a2 - 1)
tree[a2 - 1].append(a1 - 1)
if n == 2:
print('YES')
elif n == 3:
print('NO')
else:
tree_leafs = [False for i in range(n)]
for a in range(n):
if len(tree[a]) == 1:
tree_leafs[a] = True
no_var = False
for a0 in range(n):
if tree_leafs[a0] == False:
k = 0
for a in tree[a0]:
if tree_leafs[a] == True:
k += 1
if k == 2:
break
if k == 1:
print('NO')
no_var = True
break
if not no_var:
print('YES')
|
1189_D1. Add on a Tree
|
Note that this is the first problem of the two similar problems. You can hack this problem only if you solve both problems.
You are given a tree with n nodes. In the beginning, 0 is written on all edges. In one operation, you can choose any 2 distinct leaves u, v and any real number x and add x to values written on all edges on the simple path between u and v.
For example, on the picture below you can see the result of applying two operations to the graph: adding 2 on the path from 7 to 6, and then adding -0.5 on the path from 4 to 5.
<image>
Is it true that for any configuration of real numbers written on edges, we can achieve it with a finite number of operations?
Leaf is a node of a tree of degree 1. Simple path is a path that doesn't contain any node twice.
Input
The first line contains a single integer n (2 ≤ n ≤ 10^5) — the number of nodes.
Each of the next n-1 lines contains two integers u and v (1 ≤ u, v ≤ n, u ≠ v), meaning that there is an edge between nodes u and v. It is guaranteed that these edges form a tree.
Output
If there is a configuration of real numbers written on edges of the tree that we can't achieve by performing the operations, output "NO".
Otherwise, output "YES".
You can print each letter in any case (upper or lower).
Examples
Input
2
1 2
Output
YES
Input
3
1 2
2 3
Output
NO
Input
5
1 2
1 3
1 4
2 5
Output
NO
Input
6
1 2
1 3
1 4
2 5
2 6
Output
YES
Note
In the first example, we can add any real x to the value written on the only edge (1, 2).
<image>
In the second example, one of configurations that we can't reach is 0 written on (1, 2) and 1 written on (2, 3).
<image>
Below you can see graphs from examples 3, 4:
<image> <image>
|
{
"input": [
"2\n1 2\n",
"3\n1 2\n2 3\n",
"5\n1 2\n1 3\n1 4\n2 5\n",
"6\n1 2\n1 3\n1 4\n2 5\n2 6\n"
],
"output": [
"YES",
"NO",
"NO",
"YES"
]
}
|
{
"input": [
"50\n16 4\n17 9\n31 19\n22 10\n8 1\n40 30\n3 31\n20 29\n47 27\n22 25\n32 34\n12 15\n40 32\n10 33\n47 12\n6 24\n46 41\n14 23\n12 35\n31 42\n46 28\n31 20\n46 37\n1 39\n29 49\n37 47\n40 6\n42 36\n47 2\n24 46\n2 13\n8 45\n41 3\n32 17\n4 7\n47 26\n28 8\n41 50\n34 44\n33 21\n25 5\n16 40\n3 14\n8 18\n28 11\n32 22\n2 38\n3 48\n44 43\n",
"10\n8 1\n1 2\n8 9\n8 5\n1 3\n1 10\n1 6\n1 7\n8 4\n",
"5\n5 1\n5 4\n4 3\n1 2\n",
"7\n1 2\n2 3\n1 4\n1 5\n3 6\n3 7\n",
"3\n1 3\n2 3\n",
"60\n26 6\n59 30\n31 12\n31 3\n38 23\n59 29\n53 9\n38 56\n53 54\n29 21\n17 55\n59 38\n26 16\n24 59\n24 25\n17 35\n24 41\n30 15\n31 27\n8 44\n26 5\n26 48\n8 32\n53 17\n3 34\n3 51\n30 28\n47 10\n53 60\n36 42\n24 53\n59 22\n53 40\n26 52\n36 4\n59 8\n29 37\n36 20\n17 47\n53 18\n3 50\n30 2\n17 7\n8 58\n59 1\n31 11\n24 26\n24 43\n53 57\n59 45\n47 13\n26 46\n17 33\n30 31\n26 39\n26 19\n24 36\n8 49\n38 14\n",
"7\n1 2\n2 3\n3 4\n3 5\n1 6\n1 7\n",
"20\n19 16\n19 18\n20 7\n9 4\n6 17\n14 2\n9 15\n2 13\n5 11\n19 12\n12 20\n16 9\n11 8\n19 5\n3 1\n19 14\n5 3\n12 10\n19 6\n",
"7\n1 2\n1 3\n2 4\n2 5\n3 6\n3 7\n",
"10\n9 5\n7 1\n9 10\n7 2\n5 4\n9 6\n2 9\n10 8\n1 3\n",
"4\n2 4\n2 3\n2 1\n",
"4\n1 4\n3 2\n1 3\n",
"3\n1 2\n1 3\n",
"5\n1 2\n1 5\n1 3\n1 4\n",
"20\n14 9\n12 13\n10 15\n2 1\n20 19\n16 6\n16 3\n17 14\n3 5\n2 11\n3 10\n15 8\n14 2\n6 4\n3 20\n5 18\n1 7\n1 16\n4 12\n",
"20\n7 5\n14 13\n17 6\n3 8\n16 12\n18 9\n3 18\n14 1\n17 3\n15 2\n17 4\n9 11\n2 7\n15 17\n3 20\n16 10\n17 14\n2 16\n1 19\n",
"8\n1 2\n2 3\n3 4\n1 7\n1 8\n4 5\n4 6\n",
"5\n5 1\n5 2\n5 3\n5 4\n",
"50\n49 6\n43 7\n1 27\n19 35\n15 37\n16 12\n19 21\n16 28\n49 9\n48 39\n13 1\n2 48\n9 50\n44 3\n41 32\n48 31\n49 33\n6 11\n13 20\n49 22\n13 41\n48 29\n13 46\n15 47\n34 2\n49 13\n48 14\n34 24\n16 36\n13 40\n49 34\n49 17\n43 25\n11 23\n10 15\n19 26\n34 44\n16 42\n19 18\n46 8\n29 38\n1 45\n12 43\n13 16\n46 30\n15 5\n49 10\n11 19\n32 4\n",
"20\n13 1\n18 2\n3 7\n18 5\n20 16\n3 12\n18 9\n3 10\n18 11\n13 6\n3 18\n20 15\n20 17\n3 13\n3 4\n13 14\n3 20\n18 8\n3 19\n",
"10\n8 2\n5 6\n1 8\n2 9\n1 4\n8 10\n10 5\n2 7\n2 3\n"
],
"output": [
"NO",
"YES",
"NO",
"NO",
"NO",
"YES",
"NO",
"NO",
"NO",
"NO",
"YES",
"NO",
"NO",
"YES",
"NO",
"NO",
"NO",
"YES",
"NO",
"YES",
"NO"
]
}
|
IN-CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
const int INF = 1e9 + 7;
const int BASE = 1e9 + 7;
const int dx[4] = {-1, 1, 0, 0};
const int dy[4] = {0, 0, -1, 1};
const int N = 1e5 + 1;
vector<int> a[N];
int pa[N], n;
int F[N];
void DFS(int p, int u) {
bool Check = true;
for (__typeof(a[u].begin()) it = a[u].begin(); it != a[u].end(); it++) {
int v = *it;
if (v == p) continue;
Check = false;
DFS(u, v);
pa[v] = u;
F[u] = F[u] + F[v];
}
F[u] += Check;
}
int main() {
ios_base::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
cin >> n;
int u, v;
for (int i = 1; i < n; i++) {
cin >> u >> v;
a[u].push_back(v);
a[v].push_back(u);
}
if (n == 2) {
cout << "YES";
return 0;
}
DFS(0, 1);
bool Check = true;
for (int v = 1; v <= n; v++) {
int u = pa[v];
if (u == 0) continue;
if (a[v].size() == 1) {
if (F[1] < 2) {
Check = false;
break;
}
} else {
if (!((F[v] > 1) && ((F[1] - F[v]) > 1))) {
Check = false;
break;
}
}
}
if (Check)
cout << "YES";
else
cout << "NO";
return 0;
}
|
1189_D1. Add on a Tree
|
Note that this is the first problem of the two similar problems. You can hack this problem only if you solve both problems.
You are given a tree with n nodes. In the beginning, 0 is written on all edges. In one operation, you can choose any 2 distinct leaves u, v and any real number x and add x to values written on all edges on the simple path between u and v.
For example, on the picture below you can see the result of applying two operations to the graph: adding 2 on the path from 7 to 6, and then adding -0.5 on the path from 4 to 5.
<image>
Is it true that for any configuration of real numbers written on edges, we can achieve it with a finite number of operations?
Leaf is a node of a tree of degree 1. Simple path is a path that doesn't contain any node twice.
Input
The first line contains a single integer n (2 ≤ n ≤ 10^5) — the number of nodes.
Each of the next n-1 lines contains two integers u and v (1 ≤ u, v ≤ n, u ≠ v), meaning that there is an edge between nodes u and v. It is guaranteed that these edges form a tree.
Output
If there is a configuration of real numbers written on edges of the tree that we can't achieve by performing the operations, output "NO".
Otherwise, output "YES".
You can print each letter in any case (upper or lower).
Examples
Input
2
1 2
Output
YES
Input
3
1 2
2 3
Output
NO
Input
5
1 2
1 3
1 4
2 5
Output
NO
Input
6
1 2
1 3
1 4
2 5
2 6
Output
YES
Note
In the first example, we can add any real x to the value written on the only edge (1, 2).
<image>
In the second example, one of configurations that we can't reach is 0 written on (1, 2) and 1 written on (2, 3).
<image>
Below you can see graphs from examples 3, 4:
<image> <image>
|
{
"input": [
"2\n1 2\n",
"3\n1 2\n2 3\n",
"5\n1 2\n1 3\n1 4\n2 5\n",
"6\n1 2\n1 3\n1 4\n2 5\n2 6\n"
],
"output": [
"YES",
"NO",
"NO",
"YES"
]
}
|
{
"input": [
"50\n16 4\n17 9\n31 19\n22 10\n8 1\n40 30\n3 31\n20 29\n47 27\n22 25\n32 34\n12 15\n40 32\n10 33\n47 12\n6 24\n46 41\n14 23\n12 35\n31 42\n46 28\n31 20\n46 37\n1 39\n29 49\n37 47\n40 6\n42 36\n47 2\n24 46\n2 13\n8 45\n41 3\n32 17\n4 7\n47 26\n28 8\n41 50\n34 44\n33 21\n25 5\n16 40\n3 14\n8 18\n28 11\n32 22\n2 38\n3 48\n44 43\n",
"10\n8 1\n1 2\n8 9\n8 5\n1 3\n1 10\n1 6\n1 7\n8 4\n",
"5\n5 1\n5 4\n4 3\n1 2\n",
"7\n1 2\n2 3\n1 4\n1 5\n3 6\n3 7\n",
"3\n1 3\n2 3\n",
"60\n26 6\n59 30\n31 12\n31 3\n38 23\n59 29\n53 9\n38 56\n53 54\n29 21\n17 55\n59 38\n26 16\n24 59\n24 25\n17 35\n24 41\n30 15\n31 27\n8 44\n26 5\n26 48\n8 32\n53 17\n3 34\n3 51\n30 28\n47 10\n53 60\n36 42\n24 53\n59 22\n53 40\n26 52\n36 4\n59 8\n29 37\n36 20\n17 47\n53 18\n3 50\n30 2\n17 7\n8 58\n59 1\n31 11\n24 26\n24 43\n53 57\n59 45\n47 13\n26 46\n17 33\n30 31\n26 39\n26 19\n24 36\n8 49\n38 14\n",
"7\n1 2\n2 3\n3 4\n3 5\n1 6\n1 7\n",
"20\n19 16\n19 18\n20 7\n9 4\n6 17\n14 2\n9 15\n2 13\n5 11\n19 12\n12 20\n16 9\n11 8\n19 5\n3 1\n19 14\n5 3\n12 10\n19 6\n",
"7\n1 2\n1 3\n2 4\n2 5\n3 6\n3 7\n",
"10\n9 5\n7 1\n9 10\n7 2\n5 4\n9 6\n2 9\n10 8\n1 3\n",
"4\n2 4\n2 3\n2 1\n",
"4\n1 4\n3 2\n1 3\n",
"3\n1 2\n1 3\n",
"5\n1 2\n1 5\n1 3\n1 4\n",
"20\n14 9\n12 13\n10 15\n2 1\n20 19\n16 6\n16 3\n17 14\n3 5\n2 11\n3 10\n15 8\n14 2\n6 4\n3 20\n5 18\n1 7\n1 16\n4 12\n",
"20\n7 5\n14 13\n17 6\n3 8\n16 12\n18 9\n3 18\n14 1\n17 3\n15 2\n17 4\n9 11\n2 7\n15 17\n3 20\n16 10\n17 14\n2 16\n1 19\n",
"8\n1 2\n2 3\n3 4\n1 7\n1 8\n4 5\n4 6\n",
"5\n5 1\n5 2\n5 3\n5 4\n",
"50\n49 6\n43 7\n1 27\n19 35\n15 37\n16 12\n19 21\n16 28\n49 9\n48 39\n13 1\n2 48\n9 50\n44 3\n41 32\n48 31\n49 33\n6 11\n13 20\n49 22\n13 41\n48 29\n13 46\n15 47\n34 2\n49 13\n48 14\n34 24\n16 36\n13 40\n49 34\n49 17\n43 25\n11 23\n10 15\n19 26\n34 44\n16 42\n19 18\n46 8\n29 38\n1 45\n12 43\n13 16\n46 30\n15 5\n49 10\n11 19\n32 4\n",
"20\n13 1\n18 2\n3 7\n18 5\n20 16\n3 12\n18 9\n3 10\n18 11\n13 6\n3 18\n20 15\n20 17\n3 13\n3 4\n13 14\n3 20\n18 8\n3 19\n",
"10\n8 2\n5 6\n1 8\n2 9\n1 4\n8 10\n10 5\n2 7\n2 3\n"
],
"output": [
"NO",
"YES",
"NO",
"NO",
"NO",
"YES",
"NO",
"NO",
"NO",
"NO",
"YES",
"NO",
"NO",
"YES",
"NO",
"NO",
"NO",
"YES",
"NO",
"YES",
"NO"
]
}
|
IN-CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
int main() {
int n;
cin >> n;
int ind[n];
memset(ind, 0, sizeof(ind));
int u, v;
for (int i = 0; i < n - 1; i++) {
cin >> u >> v;
u--;
v--;
ind[u]++;
ind[v]++;
}
bool f = 1;
for (int i = 0; i < n; i++) {
if (ind[i] % 2 == 0) {
return cout << "NO", 0;
}
}
cout << "YES";
}
|
1189_D1. Add on a Tree
|
Note that this is the first problem of the two similar problems. You can hack this problem only if you solve both problems.
You are given a tree with n nodes. In the beginning, 0 is written on all edges. In one operation, you can choose any 2 distinct leaves u, v and any real number x and add x to values written on all edges on the simple path between u and v.
For example, on the picture below you can see the result of applying two operations to the graph: adding 2 on the path from 7 to 6, and then adding -0.5 on the path from 4 to 5.
<image>
Is it true that for any configuration of real numbers written on edges, we can achieve it with a finite number of operations?
Leaf is a node of a tree of degree 1. Simple path is a path that doesn't contain any node twice.
Input
The first line contains a single integer n (2 ≤ n ≤ 10^5) — the number of nodes.
Each of the next n-1 lines contains two integers u and v (1 ≤ u, v ≤ n, u ≠ v), meaning that there is an edge between nodes u and v. It is guaranteed that these edges form a tree.
Output
If there is a configuration of real numbers written on edges of the tree that we can't achieve by performing the operations, output "NO".
Otherwise, output "YES".
You can print each letter in any case (upper or lower).
Examples
Input
2
1 2
Output
YES
Input
3
1 2
2 3
Output
NO
Input
5
1 2
1 3
1 4
2 5
Output
NO
Input
6
1 2
1 3
1 4
2 5
2 6
Output
YES
Note
In the first example, we can add any real x to the value written on the only edge (1, 2).
<image>
In the second example, one of configurations that we can't reach is 0 written on (1, 2) and 1 written on (2, 3).
<image>
Below you can see graphs from examples 3, 4:
<image> <image>
|
{
"input": [
"2\n1 2\n",
"3\n1 2\n2 3\n",
"5\n1 2\n1 3\n1 4\n2 5\n",
"6\n1 2\n1 3\n1 4\n2 5\n2 6\n"
],
"output": [
"YES",
"NO",
"NO",
"YES"
]
}
|
{
"input": [
"50\n16 4\n17 9\n31 19\n22 10\n8 1\n40 30\n3 31\n20 29\n47 27\n22 25\n32 34\n12 15\n40 32\n10 33\n47 12\n6 24\n46 41\n14 23\n12 35\n31 42\n46 28\n31 20\n46 37\n1 39\n29 49\n37 47\n40 6\n42 36\n47 2\n24 46\n2 13\n8 45\n41 3\n32 17\n4 7\n47 26\n28 8\n41 50\n34 44\n33 21\n25 5\n16 40\n3 14\n8 18\n28 11\n32 22\n2 38\n3 48\n44 43\n",
"10\n8 1\n1 2\n8 9\n8 5\n1 3\n1 10\n1 6\n1 7\n8 4\n",
"5\n5 1\n5 4\n4 3\n1 2\n",
"7\n1 2\n2 3\n1 4\n1 5\n3 6\n3 7\n",
"3\n1 3\n2 3\n",
"60\n26 6\n59 30\n31 12\n31 3\n38 23\n59 29\n53 9\n38 56\n53 54\n29 21\n17 55\n59 38\n26 16\n24 59\n24 25\n17 35\n24 41\n30 15\n31 27\n8 44\n26 5\n26 48\n8 32\n53 17\n3 34\n3 51\n30 28\n47 10\n53 60\n36 42\n24 53\n59 22\n53 40\n26 52\n36 4\n59 8\n29 37\n36 20\n17 47\n53 18\n3 50\n30 2\n17 7\n8 58\n59 1\n31 11\n24 26\n24 43\n53 57\n59 45\n47 13\n26 46\n17 33\n30 31\n26 39\n26 19\n24 36\n8 49\n38 14\n",
"7\n1 2\n2 3\n3 4\n3 5\n1 6\n1 7\n",
"20\n19 16\n19 18\n20 7\n9 4\n6 17\n14 2\n9 15\n2 13\n5 11\n19 12\n12 20\n16 9\n11 8\n19 5\n3 1\n19 14\n5 3\n12 10\n19 6\n",
"7\n1 2\n1 3\n2 4\n2 5\n3 6\n3 7\n",
"10\n9 5\n7 1\n9 10\n7 2\n5 4\n9 6\n2 9\n10 8\n1 3\n",
"4\n2 4\n2 3\n2 1\n",
"4\n1 4\n3 2\n1 3\n",
"3\n1 2\n1 3\n",
"5\n1 2\n1 5\n1 3\n1 4\n",
"20\n14 9\n12 13\n10 15\n2 1\n20 19\n16 6\n16 3\n17 14\n3 5\n2 11\n3 10\n15 8\n14 2\n6 4\n3 20\n5 18\n1 7\n1 16\n4 12\n",
"20\n7 5\n14 13\n17 6\n3 8\n16 12\n18 9\n3 18\n14 1\n17 3\n15 2\n17 4\n9 11\n2 7\n15 17\n3 20\n16 10\n17 14\n2 16\n1 19\n",
"8\n1 2\n2 3\n3 4\n1 7\n1 8\n4 5\n4 6\n",
"5\n5 1\n5 2\n5 3\n5 4\n",
"50\n49 6\n43 7\n1 27\n19 35\n15 37\n16 12\n19 21\n16 28\n49 9\n48 39\n13 1\n2 48\n9 50\n44 3\n41 32\n48 31\n49 33\n6 11\n13 20\n49 22\n13 41\n48 29\n13 46\n15 47\n34 2\n49 13\n48 14\n34 24\n16 36\n13 40\n49 34\n49 17\n43 25\n11 23\n10 15\n19 26\n34 44\n16 42\n19 18\n46 8\n29 38\n1 45\n12 43\n13 16\n46 30\n15 5\n49 10\n11 19\n32 4\n",
"20\n13 1\n18 2\n3 7\n18 5\n20 16\n3 12\n18 9\n3 10\n18 11\n13 6\n3 18\n20 15\n20 17\n3 13\n3 4\n13 14\n3 20\n18 8\n3 19\n",
"10\n8 2\n5 6\n1 8\n2 9\n1 4\n8 10\n10 5\n2 7\n2 3\n"
],
"output": [
"NO",
"YES",
"NO",
"NO",
"NO",
"YES",
"NO",
"NO",
"NO",
"NO",
"YES",
"NO",
"NO",
"YES",
"NO",
"NO",
"NO",
"YES",
"NO",
"YES",
"NO"
]
}
|
IN-CORRECT
|
python3
|
import sys
input = sys.stdin.readline
from collections import deque
class Graph(object):
"""docstring for Graph"""
def __init__(self,n,d): # Number of nodes and d is True if directed
self.n = n
self.graph = [[] for i in range(n)]
self.parent = [-1 for i in range(n)]
self.directed = d
def addEdge(self,x,y):
self.graph[x].append(y)
if not self.directed:
self.graph[y].append(x)
def bfs(self, root): # NORMAL BFS
queue = [root]
queue = deque(queue)
vis = [0]*self.n
while len(queue)!=0:
element = queue.popleft()
vis[element] = 1
count = 0
for i in self.graph[element]:
if vis[i]==0:
queue.append(i)
self.parent[i] = element
vis[i] = 1
count += 1
if count==1 and element!=0:
return False
return True
def dfs(self, root, ans): # Iterative DFS
stack=[root]
vis=[0]*self.n
stack2=[]
while len(stack)!=0: # INITIAL TRAVERSAL
element = stack.pop()
if vis[element]:
continue
vis[element] = 1
stack2.append(element)
for i in self.graph[element]:
if vis[i]==0:
self.parent[i] = element
stack.append(i)
while len(stack2)!=0: # BACKTRACING. Modify the loop according to the question
element = stack2.pop()
m = 0
for i in self.graph[element]:
if i!=self.parent[element]:
m += ans[i]
ans[element] = m
return ans
def shortestpath(self, source, dest): # Calculate Shortest Path between two nodes
self.bfs(source)
path = [dest]
while self.parent[path[-1]]!=-1:
path.append(parent[path[-1]])
return path[::-1]
def detect_cycle(self):
indeg = [0]*self.n
for i in range(self.n):
for j in self.graph[i]:
indeg[j] += 1
q = deque()
vis = 0
for i in range(self.n):
if indeg[i]==0:
q.append(i)
while len(q)!=0:
e = q.popleft()
vis += 1
for i in self.graph[e]:
indeg[i] -= 1
if indeg[i]==0:
q.append(i)
if vis!=self.n:
return True
return False
def reroot(self, root, ans):
stack = [root]
vis = [0]*n
while len(stack)!=0:
e = stack[-1]
if vis[e]:
stack.pop()
# Reverse_The_Change()
continue
vis[e] = 1
for i in graph[e]:
if not vis[e]:
stack.append(i)
if self.parent[e]==-1:
continue
# Change_The_Answers()
n = int(input())
g = Graph(n,False)
for i in range(n-1):
u,v = map(int,input().split())
g.addEdge(u-1,v-1)
for i in range(n):
if len(g.graph[i])==1:
leaf = i
break
if not g.bfs(leaf):
print ("NO")
else:
print ("YES")
|
1189_D1. Add on a Tree
|
Note that this is the first problem of the two similar problems. You can hack this problem only if you solve both problems.
You are given a tree with n nodes. In the beginning, 0 is written on all edges. In one operation, you can choose any 2 distinct leaves u, v and any real number x and add x to values written on all edges on the simple path between u and v.
For example, on the picture below you can see the result of applying two operations to the graph: adding 2 on the path from 7 to 6, and then adding -0.5 on the path from 4 to 5.
<image>
Is it true that for any configuration of real numbers written on edges, we can achieve it with a finite number of operations?
Leaf is a node of a tree of degree 1. Simple path is a path that doesn't contain any node twice.
Input
The first line contains a single integer n (2 ≤ n ≤ 10^5) — the number of nodes.
Each of the next n-1 lines contains two integers u and v (1 ≤ u, v ≤ n, u ≠ v), meaning that there is an edge between nodes u and v. It is guaranteed that these edges form a tree.
Output
If there is a configuration of real numbers written on edges of the tree that we can't achieve by performing the operations, output "NO".
Otherwise, output "YES".
You can print each letter in any case (upper or lower).
Examples
Input
2
1 2
Output
YES
Input
3
1 2
2 3
Output
NO
Input
5
1 2
1 3
1 4
2 5
Output
NO
Input
6
1 2
1 3
1 4
2 5
2 6
Output
YES
Note
In the first example, we can add any real x to the value written on the only edge (1, 2).
<image>
In the second example, one of configurations that we can't reach is 0 written on (1, 2) and 1 written on (2, 3).
<image>
Below you can see graphs from examples 3, 4:
<image> <image>
|
{
"input": [
"2\n1 2\n",
"3\n1 2\n2 3\n",
"5\n1 2\n1 3\n1 4\n2 5\n",
"6\n1 2\n1 3\n1 4\n2 5\n2 6\n"
],
"output": [
"YES",
"NO",
"NO",
"YES"
]
}
|
{
"input": [
"50\n16 4\n17 9\n31 19\n22 10\n8 1\n40 30\n3 31\n20 29\n47 27\n22 25\n32 34\n12 15\n40 32\n10 33\n47 12\n6 24\n46 41\n14 23\n12 35\n31 42\n46 28\n31 20\n46 37\n1 39\n29 49\n37 47\n40 6\n42 36\n47 2\n24 46\n2 13\n8 45\n41 3\n32 17\n4 7\n47 26\n28 8\n41 50\n34 44\n33 21\n25 5\n16 40\n3 14\n8 18\n28 11\n32 22\n2 38\n3 48\n44 43\n",
"10\n8 1\n1 2\n8 9\n8 5\n1 3\n1 10\n1 6\n1 7\n8 4\n",
"5\n5 1\n5 4\n4 3\n1 2\n",
"7\n1 2\n2 3\n1 4\n1 5\n3 6\n3 7\n",
"3\n1 3\n2 3\n",
"60\n26 6\n59 30\n31 12\n31 3\n38 23\n59 29\n53 9\n38 56\n53 54\n29 21\n17 55\n59 38\n26 16\n24 59\n24 25\n17 35\n24 41\n30 15\n31 27\n8 44\n26 5\n26 48\n8 32\n53 17\n3 34\n3 51\n30 28\n47 10\n53 60\n36 42\n24 53\n59 22\n53 40\n26 52\n36 4\n59 8\n29 37\n36 20\n17 47\n53 18\n3 50\n30 2\n17 7\n8 58\n59 1\n31 11\n24 26\n24 43\n53 57\n59 45\n47 13\n26 46\n17 33\n30 31\n26 39\n26 19\n24 36\n8 49\n38 14\n",
"7\n1 2\n2 3\n3 4\n3 5\n1 6\n1 7\n",
"20\n19 16\n19 18\n20 7\n9 4\n6 17\n14 2\n9 15\n2 13\n5 11\n19 12\n12 20\n16 9\n11 8\n19 5\n3 1\n19 14\n5 3\n12 10\n19 6\n",
"7\n1 2\n1 3\n2 4\n2 5\n3 6\n3 7\n",
"10\n9 5\n7 1\n9 10\n7 2\n5 4\n9 6\n2 9\n10 8\n1 3\n",
"4\n2 4\n2 3\n2 1\n",
"4\n1 4\n3 2\n1 3\n",
"3\n1 2\n1 3\n",
"5\n1 2\n1 5\n1 3\n1 4\n",
"20\n14 9\n12 13\n10 15\n2 1\n20 19\n16 6\n16 3\n17 14\n3 5\n2 11\n3 10\n15 8\n14 2\n6 4\n3 20\n5 18\n1 7\n1 16\n4 12\n",
"20\n7 5\n14 13\n17 6\n3 8\n16 12\n18 9\n3 18\n14 1\n17 3\n15 2\n17 4\n9 11\n2 7\n15 17\n3 20\n16 10\n17 14\n2 16\n1 19\n",
"8\n1 2\n2 3\n3 4\n1 7\n1 8\n4 5\n4 6\n",
"5\n5 1\n5 2\n5 3\n5 4\n",
"50\n49 6\n43 7\n1 27\n19 35\n15 37\n16 12\n19 21\n16 28\n49 9\n48 39\n13 1\n2 48\n9 50\n44 3\n41 32\n48 31\n49 33\n6 11\n13 20\n49 22\n13 41\n48 29\n13 46\n15 47\n34 2\n49 13\n48 14\n34 24\n16 36\n13 40\n49 34\n49 17\n43 25\n11 23\n10 15\n19 26\n34 44\n16 42\n19 18\n46 8\n29 38\n1 45\n12 43\n13 16\n46 30\n15 5\n49 10\n11 19\n32 4\n",
"20\n13 1\n18 2\n3 7\n18 5\n20 16\n3 12\n18 9\n3 10\n18 11\n13 6\n3 18\n20 15\n20 17\n3 13\n3 4\n13 14\n3 20\n18 8\n3 19\n",
"10\n8 2\n5 6\n1 8\n2 9\n1 4\n8 10\n10 5\n2 7\n2 3\n"
],
"output": [
"NO",
"YES",
"NO",
"NO",
"NO",
"YES",
"NO",
"NO",
"NO",
"NO",
"YES",
"NO",
"NO",
"YES",
"NO",
"NO",
"NO",
"YES",
"NO",
"YES",
"NO"
]
}
|
IN-CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
void err(istream_iterator<string> it) {}
template <typename S37, typename... Args>
void err(istream_iterator<string> it, S37 a, Args... args) {
cerr << *it << " = " << a << endl;
err(++it, args...);
}
const long long N = 200010, mod = 1e9 + 7, mod2 = 1e9 + 9, mod3 = 998244353,
sq = 450, base = 37, lg = 25, inf = 1e18 + 10, del = 67733;
long long n, m, x, y, w, z, X, Y, Z, t, k, ans, a[N];
vector<long long> v[N];
int main() {
ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
cin >> n;
for (int i = 0; i < n - 1; i++) {
cin >> x >> y;
v[x].push_back(y);
a[x]++;
a[y]++;
v[y].push_back(x);
}
if (n == 2) return cout << "YES", 0;
for (int i = 1; i <= n; i++) {
x = 0;
if (a[i] == 1) continue;
y++;
for (int j = 0; j < v[i].size(); j++)
if (a[v[i][j]] == 1) x++;
if (x % 2 == 1 || x == 0) return cout << "NO", 0;
}
if (y == 1) return cout << "NO", 0;
cout << "YES";
return 0;
}
|
1189_D1. Add on a Tree
|
Note that this is the first problem of the two similar problems. You can hack this problem only if you solve both problems.
You are given a tree with n nodes. In the beginning, 0 is written on all edges. In one operation, you can choose any 2 distinct leaves u, v and any real number x and add x to values written on all edges on the simple path between u and v.
For example, on the picture below you can see the result of applying two operations to the graph: adding 2 on the path from 7 to 6, and then adding -0.5 on the path from 4 to 5.
<image>
Is it true that for any configuration of real numbers written on edges, we can achieve it with a finite number of operations?
Leaf is a node of a tree of degree 1. Simple path is a path that doesn't contain any node twice.
Input
The first line contains a single integer n (2 ≤ n ≤ 10^5) — the number of nodes.
Each of the next n-1 lines contains two integers u and v (1 ≤ u, v ≤ n, u ≠ v), meaning that there is an edge between nodes u and v. It is guaranteed that these edges form a tree.
Output
If there is a configuration of real numbers written on edges of the tree that we can't achieve by performing the operations, output "NO".
Otherwise, output "YES".
You can print each letter in any case (upper or lower).
Examples
Input
2
1 2
Output
YES
Input
3
1 2
2 3
Output
NO
Input
5
1 2
1 3
1 4
2 5
Output
NO
Input
6
1 2
1 3
1 4
2 5
2 6
Output
YES
Note
In the first example, we can add any real x to the value written on the only edge (1, 2).
<image>
In the second example, one of configurations that we can't reach is 0 written on (1, 2) and 1 written on (2, 3).
<image>
Below you can see graphs from examples 3, 4:
<image> <image>
|
{
"input": [
"2\n1 2\n",
"3\n1 2\n2 3\n",
"5\n1 2\n1 3\n1 4\n2 5\n",
"6\n1 2\n1 3\n1 4\n2 5\n2 6\n"
],
"output": [
"YES",
"NO",
"NO",
"YES"
]
}
|
{
"input": [
"50\n16 4\n17 9\n31 19\n22 10\n8 1\n40 30\n3 31\n20 29\n47 27\n22 25\n32 34\n12 15\n40 32\n10 33\n47 12\n6 24\n46 41\n14 23\n12 35\n31 42\n46 28\n31 20\n46 37\n1 39\n29 49\n37 47\n40 6\n42 36\n47 2\n24 46\n2 13\n8 45\n41 3\n32 17\n4 7\n47 26\n28 8\n41 50\n34 44\n33 21\n25 5\n16 40\n3 14\n8 18\n28 11\n32 22\n2 38\n3 48\n44 43\n",
"10\n8 1\n1 2\n8 9\n8 5\n1 3\n1 10\n1 6\n1 7\n8 4\n",
"5\n5 1\n5 4\n4 3\n1 2\n",
"7\n1 2\n2 3\n1 4\n1 5\n3 6\n3 7\n",
"3\n1 3\n2 3\n",
"60\n26 6\n59 30\n31 12\n31 3\n38 23\n59 29\n53 9\n38 56\n53 54\n29 21\n17 55\n59 38\n26 16\n24 59\n24 25\n17 35\n24 41\n30 15\n31 27\n8 44\n26 5\n26 48\n8 32\n53 17\n3 34\n3 51\n30 28\n47 10\n53 60\n36 42\n24 53\n59 22\n53 40\n26 52\n36 4\n59 8\n29 37\n36 20\n17 47\n53 18\n3 50\n30 2\n17 7\n8 58\n59 1\n31 11\n24 26\n24 43\n53 57\n59 45\n47 13\n26 46\n17 33\n30 31\n26 39\n26 19\n24 36\n8 49\n38 14\n",
"7\n1 2\n2 3\n3 4\n3 5\n1 6\n1 7\n",
"20\n19 16\n19 18\n20 7\n9 4\n6 17\n14 2\n9 15\n2 13\n5 11\n19 12\n12 20\n16 9\n11 8\n19 5\n3 1\n19 14\n5 3\n12 10\n19 6\n",
"7\n1 2\n1 3\n2 4\n2 5\n3 6\n3 7\n",
"10\n9 5\n7 1\n9 10\n7 2\n5 4\n9 6\n2 9\n10 8\n1 3\n",
"4\n2 4\n2 3\n2 1\n",
"4\n1 4\n3 2\n1 3\n",
"3\n1 2\n1 3\n",
"5\n1 2\n1 5\n1 3\n1 4\n",
"20\n14 9\n12 13\n10 15\n2 1\n20 19\n16 6\n16 3\n17 14\n3 5\n2 11\n3 10\n15 8\n14 2\n6 4\n3 20\n5 18\n1 7\n1 16\n4 12\n",
"20\n7 5\n14 13\n17 6\n3 8\n16 12\n18 9\n3 18\n14 1\n17 3\n15 2\n17 4\n9 11\n2 7\n15 17\n3 20\n16 10\n17 14\n2 16\n1 19\n",
"8\n1 2\n2 3\n3 4\n1 7\n1 8\n4 5\n4 6\n",
"5\n5 1\n5 2\n5 3\n5 4\n",
"50\n49 6\n43 7\n1 27\n19 35\n15 37\n16 12\n19 21\n16 28\n49 9\n48 39\n13 1\n2 48\n9 50\n44 3\n41 32\n48 31\n49 33\n6 11\n13 20\n49 22\n13 41\n48 29\n13 46\n15 47\n34 2\n49 13\n48 14\n34 24\n16 36\n13 40\n49 34\n49 17\n43 25\n11 23\n10 15\n19 26\n34 44\n16 42\n19 18\n46 8\n29 38\n1 45\n12 43\n13 16\n46 30\n15 5\n49 10\n11 19\n32 4\n",
"20\n13 1\n18 2\n3 7\n18 5\n20 16\n3 12\n18 9\n3 10\n18 11\n13 6\n3 18\n20 15\n20 17\n3 13\n3 4\n13 14\n3 20\n18 8\n3 19\n",
"10\n8 2\n5 6\n1 8\n2 9\n1 4\n8 10\n10 5\n2 7\n2 3\n"
],
"output": [
"NO",
"YES",
"NO",
"NO",
"NO",
"YES",
"NO",
"NO",
"NO",
"NO",
"YES",
"NO",
"NO",
"YES",
"NO",
"NO",
"NO",
"YES",
"NO",
"YES",
"NO"
]
}
|
IN-CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
bool is_prime(long long n) {
for (long long i = 2; i * i <= n; ++i) {
if (n % i == 0) {
return false;
}
}
return true;
}
vector<long long> fact(long long n) {
n = abs(n);
vector<long long> ans;
for (int i = 1; i * i <= n; i++) {
if (n % i == 0) {
ans.push_back(i);
ans.push_back(n / i);
}
}
return ans;
}
inline long long getPow(long long a, long long b) {
long long res = 1ll, tp = a;
while (b) {
if (b & 1ll) {
res *= tp;
}
tp *= tp;
b >>= 1ll;
}
return res;
}
long long vec_mult(long long x1, long long y1, long long x2, long long y2,
long long x3, long long y3) {
return abs((x2 - x1) * (y3 - y1) - (y2 - y1) * (x3 - x1));
}
void ok() {
cout << "YES" << endl;
exit(0);
}
void no() {
cout << "NO" << endl;
exit(0);
}
inline long long nxt() {
long long x;
cin >> x;
return x;
}
const long long N = 3e5 + 5, inf = 8e18;
int days[13] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
int main() {
ios_base::sync_with_stdio(false);
cin.tie(0), cout.tie(0);
long long n = nxt();
vector<vector<long long>> g(n);
for (int i = 1; i < n; i++) {
long long t1 = nxt() - 1, t2 = nxt() - 1;
g[t1].push_back(t2);
g[t2].push_back(t1);
}
if (n % 2 == 0)
ok();
else
no();
return 0;
}
|
1189_D1. Add on a Tree
|
Note that this is the first problem of the two similar problems. You can hack this problem only if you solve both problems.
You are given a tree with n nodes. In the beginning, 0 is written on all edges. In one operation, you can choose any 2 distinct leaves u, v and any real number x and add x to values written on all edges on the simple path between u and v.
For example, on the picture below you can see the result of applying two operations to the graph: adding 2 on the path from 7 to 6, and then adding -0.5 on the path from 4 to 5.
<image>
Is it true that for any configuration of real numbers written on edges, we can achieve it with a finite number of operations?
Leaf is a node of a tree of degree 1. Simple path is a path that doesn't contain any node twice.
Input
The first line contains a single integer n (2 ≤ n ≤ 10^5) — the number of nodes.
Each of the next n-1 lines contains two integers u and v (1 ≤ u, v ≤ n, u ≠ v), meaning that there is an edge between nodes u and v. It is guaranteed that these edges form a tree.
Output
If there is a configuration of real numbers written on edges of the tree that we can't achieve by performing the operations, output "NO".
Otherwise, output "YES".
You can print each letter in any case (upper or lower).
Examples
Input
2
1 2
Output
YES
Input
3
1 2
2 3
Output
NO
Input
5
1 2
1 3
1 4
2 5
Output
NO
Input
6
1 2
1 3
1 4
2 5
2 6
Output
YES
Note
In the first example, we can add any real x to the value written on the only edge (1, 2).
<image>
In the second example, one of configurations that we can't reach is 0 written on (1, 2) and 1 written on (2, 3).
<image>
Below you can see graphs from examples 3, 4:
<image> <image>
|
{
"input": [
"2\n1 2\n",
"3\n1 2\n2 3\n",
"5\n1 2\n1 3\n1 4\n2 5\n",
"6\n1 2\n1 3\n1 4\n2 5\n2 6\n"
],
"output": [
"YES",
"NO",
"NO",
"YES"
]
}
|
{
"input": [
"50\n16 4\n17 9\n31 19\n22 10\n8 1\n40 30\n3 31\n20 29\n47 27\n22 25\n32 34\n12 15\n40 32\n10 33\n47 12\n6 24\n46 41\n14 23\n12 35\n31 42\n46 28\n31 20\n46 37\n1 39\n29 49\n37 47\n40 6\n42 36\n47 2\n24 46\n2 13\n8 45\n41 3\n32 17\n4 7\n47 26\n28 8\n41 50\n34 44\n33 21\n25 5\n16 40\n3 14\n8 18\n28 11\n32 22\n2 38\n3 48\n44 43\n",
"10\n8 1\n1 2\n8 9\n8 5\n1 3\n1 10\n1 6\n1 7\n8 4\n",
"5\n5 1\n5 4\n4 3\n1 2\n",
"7\n1 2\n2 3\n1 4\n1 5\n3 6\n3 7\n",
"3\n1 3\n2 3\n",
"60\n26 6\n59 30\n31 12\n31 3\n38 23\n59 29\n53 9\n38 56\n53 54\n29 21\n17 55\n59 38\n26 16\n24 59\n24 25\n17 35\n24 41\n30 15\n31 27\n8 44\n26 5\n26 48\n8 32\n53 17\n3 34\n3 51\n30 28\n47 10\n53 60\n36 42\n24 53\n59 22\n53 40\n26 52\n36 4\n59 8\n29 37\n36 20\n17 47\n53 18\n3 50\n30 2\n17 7\n8 58\n59 1\n31 11\n24 26\n24 43\n53 57\n59 45\n47 13\n26 46\n17 33\n30 31\n26 39\n26 19\n24 36\n8 49\n38 14\n",
"7\n1 2\n2 3\n3 4\n3 5\n1 6\n1 7\n",
"20\n19 16\n19 18\n20 7\n9 4\n6 17\n14 2\n9 15\n2 13\n5 11\n19 12\n12 20\n16 9\n11 8\n19 5\n3 1\n19 14\n5 3\n12 10\n19 6\n",
"7\n1 2\n1 3\n2 4\n2 5\n3 6\n3 7\n",
"10\n9 5\n7 1\n9 10\n7 2\n5 4\n9 6\n2 9\n10 8\n1 3\n",
"4\n2 4\n2 3\n2 1\n",
"4\n1 4\n3 2\n1 3\n",
"3\n1 2\n1 3\n",
"5\n1 2\n1 5\n1 3\n1 4\n",
"20\n14 9\n12 13\n10 15\n2 1\n20 19\n16 6\n16 3\n17 14\n3 5\n2 11\n3 10\n15 8\n14 2\n6 4\n3 20\n5 18\n1 7\n1 16\n4 12\n",
"20\n7 5\n14 13\n17 6\n3 8\n16 12\n18 9\n3 18\n14 1\n17 3\n15 2\n17 4\n9 11\n2 7\n15 17\n3 20\n16 10\n17 14\n2 16\n1 19\n",
"8\n1 2\n2 3\n3 4\n1 7\n1 8\n4 5\n4 6\n",
"5\n5 1\n5 2\n5 3\n5 4\n",
"50\n49 6\n43 7\n1 27\n19 35\n15 37\n16 12\n19 21\n16 28\n49 9\n48 39\n13 1\n2 48\n9 50\n44 3\n41 32\n48 31\n49 33\n6 11\n13 20\n49 22\n13 41\n48 29\n13 46\n15 47\n34 2\n49 13\n48 14\n34 24\n16 36\n13 40\n49 34\n49 17\n43 25\n11 23\n10 15\n19 26\n34 44\n16 42\n19 18\n46 8\n29 38\n1 45\n12 43\n13 16\n46 30\n15 5\n49 10\n11 19\n32 4\n",
"20\n13 1\n18 2\n3 7\n18 5\n20 16\n3 12\n18 9\n3 10\n18 11\n13 6\n3 18\n20 15\n20 17\n3 13\n3 4\n13 14\n3 20\n18 8\n3 19\n",
"10\n8 2\n5 6\n1 8\n2 9\n1 4\n8 10\n10 5\n2 7\n2 3\n"
],
"output": [
"NO",
"YES",
"NO",
"NO",
"NO",
"YES",
"NO",
"NO",
"NO",
"NO",
"YES",
"NO",
"NO",
"YES",
"NO",
"NO",
"NO",
"YES",
"NO",
"YES",
"NO"
]
}
|
IN-CORRECT
|
java
|
import java.util.*;
import java.io.*;
public class d {
public static void main(String[] args) throws IOException {
FastScanner sc = new FastScanner(System.in);
int n = sc.nextInt();
ArrayList<Integer>[] paths = new ArrayList[n];
for (int i = 0 ; i < n ; i++) {
paths[i] = new ArrayList<>();
}
for (int i = 0 ; i < n-1 ; i++) {
paths[sc.nextInt()-1].add(sc.nextInt()-1);
}
for (int i = 0 ; i < paths.length ; i++) {
for (Integer e : paths[i]) {
if (paths[e].size() == 1) {
System.out.println("NO"); return;
}
}
}
System.out.println("YES");
}
static class FastScanner {
BufferedReader br;
StringTokenizer st;
public FastScanner(InputStream i) {
br = new BufferedReader(new InputStreamReader(i));
st = new StringTokenizer("");
}
public String next() throws IOException {
if(st.hasMoreTokens())
return st.nextToken();
else
st = new StringTokenizer(br.readLine());
return next();
}
public int nextInt() throws IOException {
return Integer.parseInt(next());
}
public long nextLong() throws IOException {
return Long.parseLong(next());
}
public double nextDouble() throws IOException {
return Double.parseDouble(next());
}
}
}
|
1189_D1. Add on a Tree
|
Note that this is the first problem of the two similar problems. You can hack this problem only if you solve both problems.
You are given a tree with n nodes. In the beginning, 0 is written on all edges. In one operation, you can choose any 2 distinct leaves u, v and any real number x and add x to values written on all edges on the simple path between u and v.
For example, on the picture below you can see the result of applying two operations to the graph: adding 2 on the path from 7 to 6, and then adding -0.5 on the path from 4 to 5.
<image>
Is it true that for any configuration of real numbers written on edges, we can achieve it with a finite number of operations?
Leaf is a node of a tree of degree 1. Simple path is a path that doesn't contain any node twice.
Input
The first line contains a single integer n (2 ≤ n ≤ 10^5) — the number of nodes.
Each of the next n-1 lines contains two integers u and v (1 ≤ u, v ≤ n, u ≠ v), meaning that there is an edge between nodes u and v. It is guaranteed that these edges form a tree.
Output
If there is a configuration of real numbers written on edges of the tree that we can't achieve by performing the operations, output "NO".
Otherwise, output "YES".
You can print each letter in any case (upper or lower).
Examples
Input
2
1 2
Output
YES
Input
3
1 2
2 3
Output
NO
Input
5
1 2
1 3
1 4
2 5
Output
NO
Input
6
1 2
1 3
1 4
2 5
2 6
Output
YES
Note
In the first example, we can add any real x to the value written on the only edge (1, 2).
<image>
In the second example, one of configurations that we can't reach is 0 written on (1, 2) and 1 written on (2, 3).
<image>
Below you can see graphs from examples 3, 4:
<image> <image>
|
{
"input": [
"2\n1 2\n",
"3\n1 2\n2 3\n",
"5\n1 2\n1 3\n1 4\n2 5\n",
"6\n1 2\n1 3\n1 4\n2 5\n2 6\n"
],
"output": [
"YES",
"NO",
"NO",
"YES"
]
}
|
{
"input": [
"50\n16 4\n17 9\n31 19\n22 10\n8 1\n40 30\n3 31\n20 29\n47 27\n22 25\n32 34\n12 15\n40 32\n10 33\n47 12\n6 24\n46 41\n14 23\n12 35\n31 42\n46 28\n31 20\n46 37\n1 39\n29 49\n37 47\n40 6\n42 36\n47 2\n24 46\n2 13\n8 45\n41 3\n32 17\n4 7\n47 26\n28 8\n41 50\n34 44\n33 21\n25 5\n16 40\n3 14\n8 18\n28 11\n32 22\n2 38\n3 48\n44 43\n",
"10\n8 1\n1 2\n8 9\n8 5\n1 3\n1 10\n1 6\n1 7\n8 4\n",
"5\n5 1\n5 4\n4 3\n1 2\n",
"7\n1 2\n2 3\n1 4\n1 5\n3 6\n3 7\n",
"3\n1 3\n2 3\n",
"60\n26 6\n59 30\n31 12\n31 3\n38 23\n59 29\n53 9\n38 56\n53 54\n29 21\n17 55\n59 38\n26 16\n24 59\n24 25\n17 35\n24 41\n30 15\n31 27\n8 44\n26 5\n26 48\n8 32\n53 17\n3 34\n3 51\n30 28\n47 10\n53 60\n36 42\n24 53\n59 22\n53 40\n26 52\n36 4\n59 8\n29 37\n36 20\n17 47\n53 18\n3 50\n30 2\n17 7\n8 58\n59 1\n31 11\n24 26\n24 43\n53 57\n59 45\n47 13\n26 46\n17 33\n30 31\n26 39\n26 19\n24 36\n8 49\n38 14\n",
"7\n1 2\n2 3\n3 4\n3 5\n1 6\n1 7\n",
"20\n19 16\n19 18\n20 7\n9 4\n6 17\n14 2\n9 15\n2 13\n5 11\n19 12\n12 20\n16 9\n11 8\n19 5\n3 1\n19 14\n5 3\n12 10\n19 6\n",
"7\n1 2\n1 3\n2 4\n2 5\n3 6\n3 7\n",
"10\n9 5\n7 1\n9 10\n7 2\n5 4\n9 6\n2 9\n10 8\n1 3\n",
"4\n2 4\n2 3\n2 1\n",
"4\n1 4\n3 2\n1 3\n",
"3\n1 2\n1 3\n",
"5\n1 2\n1 5\n1 3\n1 4\n",
"20\n14 9\n12 13\n10 15\n2 1\n20 19\n16 6\n16 3\n17 14\n3 5\n2 11\n3 10\n15 8\n14 2\n6 4\n3 20\n5 18\n1 7\n1 16\n4 12\n",
"20\n7 5\n14 13\n17 6\n3 8\n16 12\n18 9\n3 18\n14 1\n17 3\n15 2\n17 4\n9 11\n2 7\n15 17\n3 20\n16 10\n17 14\n2 16\n1 19\n",
"8\n1 2\n2 3\n3 4\n1 7\n1 8\n4 5\n4 6\n",
"5\n5 1\n5 2\n5 3\n5 4\n",
"50\n49 6\n43 7\n1 27\n19 35\n15 37\n16 12\n19 21\n16 28\n49 9\n48 39\n13 1\n2 48\n9 50\n44 3\n41 32\n48 31\n49 33\n6 11\n13 20\n49 22\n13 41\n48 29\n13 46\n15 47\n34 2\n49 13\n48 14\n34 24\n16 36\n13 40\n49 34\n49 17\n43 25\n11 23\n10 15\n19 26\n34 44\n16 42\n19 18\n46 8\n29 38\n1 45\n12 43\n13 16\n46 30\n15 5\n49 10\n11 19\n32 4\n",
"20\n13 1\n18 2\n3 7\n18 5\n20 16\n3 12\n18 9\n3 10\n18 11\n13 6\n3 18\n20 15\n20 17\n3 13\n3 4\n13 14\n3 20\n18 8\n3 19\n",
"10\n8 2\n5 6\n1 8\n2 9\n1 4\n8 10\n10 5\n2 7\n2 3\n"
],
"output": [
"NO",
"YES",
"NO",
"NO",
"NO",
"YES",
"NO",
"NO",
"NO",
"NO",
"YES",
"NO",
"NO",
"YES",
"NO",
"NO",
"NO",
"YES",
"NO",
"YES",
"NO"
]
}
|
IN-CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
int power(long long int x, long long int y) {
long long int temp;
if (y == 0) return 1;
temp = power(x, y / 2);
if (y % 2 == 0)
return temp * temp;
else
return x * temp * temp;
}
bool prime_check(long long int n) {
long long int i, j;
if (n == 1) {
return false;
}
for (i = 2; i <= sqrt(n); i++) {
if (n % i == 0) {
return false;
}
}
return true;
}
long long int fact(long long int n) {
long long int prod = 1;
for (long long int i = 1; i <= n; i++) {
prod = (prod * i) % 1000000007;
}
return prod;
}
bool vowl(char c) {
return c == 'a' || c == 'e' || c == 'i' || c == 'o' || c == 'u';
}
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
cout.tie(NULL);
int n, i, u, v;
cin >> n;
int freq[100003] = {};
for (i = 0; i < n - 1; i++) {
cin >> u >> v;
u--;
v--;
freq[u]++;
freq[v]++;
}
for (i = 0; i < n; i++) {
if (freq[i] != 0 && freq[i] % 2 == 0) break;
}
if (i == n)
cout << "YES\n";
else
cout << "NO\n";
return 0;
}
|
1189_D1. Add on a Tree
|
Note that this is the first problem of the two similar problems. You can hack this problem only if you solve both problems.
You are given a tree with n nodes. In the beginning, 0 is written on all edges. In one operation, you can choose any 2 distinct leaves u, v and any real number x and add x to values written on all edges on the simple path between u and v.
For example, on the picture below you can see the result of applying two operations to the graph: adding 2 on the path from 7 to 6, and then adding -0.5 on the path from 4 to 5.
<image>
Is it true that for any configuration of real numbers written on edges, we can achieve it with a finite number of operations?
Leaf is a node of a tree of degree 1. Simple path is a path that doesn't contain any node twice.
Input
The first line contains a single integer n (2 ≤ n ≤ 10^5) — the number of nodes.
Each of the next n-1 lines contains two integers u and v (1 ≤ u, v ≤ n, u ≠ v), meaning that there is an edge between nodes u and v. It is guaranteed that these edges form a tree.
Output
If there is a configuration of real numbers written on edges of the tree that we can't achieve by performing the operations, output "NO".
Otherwise, output "YES".
You can print each letter in any case (upper or lower).
Examples
Input
2
1 2
Output
YES
Input
3
1 2
2 3
Output
NO
Input
5
1 2
1 3
1 4
2 5
Output
NO
Input
6
1 2
1 3
1 4
2 5
2 6
Output
YES
Note
In the first example, we can add any real x to the value written on the only edge (1, 2).
<image>
In the second example, one of configurations that we can't reach is 0 written on (1, 2) and 1 written on (2, 3).
<image>
Below you can see graphs from examples 3, 4:
<image> <image>
|
{
"input": [
"2\n1 2\n",
"3\n1 2\n2 3\n",
"5\n1 2\n1 3\n1 4\n2 5\n",
"6\n1 2\n1 3\n1 4\n2 5\n2 6\n"
],
"output": [
"YES",
"NO",
"NO",
"YES"
]
}
|
{
"input": [
"50\n16 4\n17 9\n31 19\n22 10\n8 1\n40 30\n3 31\n20 29\n47 27\n22 25\n32 34\n12 15\n40 32\n10 33\n47 12\n6 24\n46 41\n14 23\n12 35\n31 42\n46 28\n31 20\n46 37\n1 39\n29 49\n37 47\n40 6\n42 36\n47 2\n24 46\n2 13\n8 45\n41 3\n32 17\n4 7\n47 26\n28 8\n41 50\n34 44\n33 21\n25 5\n16 40\n3 14\n8 18\n28 11\n32 22\n2 38\n3 48\n44 43\n",
"10\n8 1\n1 2\n8 9\n8 5\n1 3\n1 10\n1 6\n1 7\n8 4\n",
"5\n5 1\n5 4\n4 3\n1 2\n",
"7\n1 2\n2 3\n1 4\n1 5\n3 6\n3 7\n",
"3\n1 3\n2 3\n",
"60\n26 6\n59 30\n31 12\n31 3\n38 23\n59 29\n53 9\n38 56\n53 54\n29 21\n17 55\n59 38\n26 16\n24 59\n24 25\n17 35\n24 41\n30 15\n31 27\n8 44\n26 5\n26 48\n8 32\n53 17\n3 34\n3 51\n30 28\n47 10\n53 60\n36 42\n24 53\n59 22\n53 40\n26 52\n36 4\n59 8\n29 37\n36 20\n17 47\n53 18\n3 50\n30 2\n17 7\n8 58\n59 1\n31 11\n24 26\n24 43\n53 57\n59 45\n47 13\n26 46\n17 33\n30 31\n26 39\n26 19\n24 36\n8 49\n38 14\n",
"7\n1 2\n2 3\n3 4\n3 5\n1 6\n1 7\n",
"20\n19 16\n19 18\n20 7\n9 4\n6 17\n14 2\n9 15\n2 13\n5 11\n19 12\n12 20\n16 9\n11 8\n19 5\n3 1\n19 14\n5 3\n12 10\n19 6\n",
"7\n1 2\n1 3\n2 4\n2 5\n3 6\n3 7\n",
"10\n9 5\n7 1\n9 10\n7 2\n5 4\n9 6\n2 9\n10 8\n1 3\n",
"4\n2 4\n2 3\n2 1\n",
"4\n1 4\n3 2\n1 3\n",
"3\n1 2\n1 3\n",
"5\n1 2\n1 5\n1 3\n1 4\n",
"20\n14 9\n12 13\n10 15\n2 1\n20 19\n16 6\n16 3\n17 14\n3 5\n2 11\n3 10\n15 8\n14 2\n6 4\n3 20\n5 18\n1 7\n1 16\n4 12\n",
"20\n7 5\n14 13\n17 6\n3 8\n16 12\n18 9\n3 18\n14 1\n17 3\n15 2\n17 4\n9 11\n2 7\n15 17\n3 20\n16 10\n17 14\n2 16\n1 19\n",
"8\n1 2\n2 3\n3 4\n1 7\n1 8\n4 5\n4 6\n",
"5\n5 1\n5 2\n5 3\n5 4\n",
"50\n49 6\n43 7\n1 27\n19 35\n15 37\n16 12\n19 21\n16 28\n49 9\n48 39\n13 1\n2 48\n9 50\n44 3\n41 32\n48 31\n49 33\n6 11\n13 20\n49 22\n13 41\n48 29\n13 46\n15 47\n34 2\n49 13\n48 14\n34 24\n16 36\n13 40\n49 34\n49 17\n43 25\n11 23\n10 15\n19 26\n34 44\n16 42\n19 18\n46 8\n29 38\n1 45\n12 43\n13 16\n46 30\n15 5\n49 10\n11 19\n32 4\n",
"20\n13 1\n18 2\n3 7\n18 5\n20 16\n3 12\n18 9\n3 10\n18 11\n13 6\n3 18\n20 15\n20 17\n3 13\n3 4\n13 14\n3 20\n18 8\n3 19\n",
"10\n8 2\n5 6\n1 8\n2 9\n1 4\n8 10\n10 5\n2 7\n2 3\n"
],
"output": [
"NO",
"YES",
"NO",
"NO",
"NO",
"YES",
"NO",
"NO",
"NO",
"NO",
"YES",
"NO",
"NO",
"YES",
"NO",
"NO",
"NO",
"YES",
"NO",
"YES",
"NO"
]
}
|
IN-CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
long long in[100005];
int main() {
ios_base::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
long long n, x, y;
cin >> n;
for (long long i = 0; i < n - 1; i++) {
cin >> x >> y;
in[x]++;
in[y]++;
}
long long cnt = 0;
for (long long i = 1; i < n + 1; i++) {
if (in[i] == 1) {
cnt++;
}
}
if (cnt % 2 == 0) {
cout << "YES"
<< "\n";
} else {
cout << "NO"
<< "\n";
}
return 0;
}
|
1189_D1. Add on a Tree
|
Note that this is the first problem of the two similar problems. You can hack this problem only if you solve both problems.
You are given a tree with n nodes. In the beginning, 0 is written on all edges. In one operation, you can choose any 2 distinct leaves u, v and any real number x and add x to values written on all edges on the simple path between u and v.
For example, on the picture below you can see the result of applying two operations to the graph: adding 2 on the path from 7 to 6, and then adding -0.5 on the path from 4 to 5.
<image>
Is it true that for any configuration of real numbers written on edges, we can achieve it with a finite number of operations?
Leaf is a node of a tree of degree 1. Simple path is a path that doesn't contain any node twice.
Input
The first line contains a single integer n (2 ≤ n ≤ 10^5) — the number of nodes.
Each of the next n-1 lines contains two integers u and v (1 ≤ u, v ≤ n, u ≠ v), meaning that there is an edge between nodes u and v. It is guaranteed that these edges form a tree.
Output
If there is a configuration of real numbers written on edges of the tree that we can't achieve by performing the operations, output "NO".
Otherwise, output "YES".
You can print each letter in any case (upper or lower).
Examples
Input
2
1 2
Output
YES
Input
3
1 2
2 3
Output
NO
Input
5
1 2
1 3
1 4
2 5
Output
NO
Input
6
1 2
1 3
1 4
2 5
2 6
Output
YES
Note
In the first example, we can add any real x to the value written on the only edge (1, 2).
<image>
In the second example, one of configurations that we can't reach is 0 written on (1, 2) and 1 written on (2, 3).
<image>
Below you can see graphs from examples 3, 4:
<image> <image>
|
{
"input": [
"2\n1 2\n",
"3\n1 2\n2 3\n",
"5\n1 2\n1 3\n1 4\n2 5\n",
"6\n1 2\n1 3\n1 4\n2 5\n2 6\n"
],
"output": [
"YES",
"NO",
"NO",
"YES"
]
}
|
{
"input": [
"50\n16 4\n17 9\n31 19\n22 10\n8 1\n40 30\n3 31\n20 29\n47 27\n22 25\n32 34\n12 15\n40 32\n10 33\n47 12\n6 24\n46 41\n14 23\n12 35\n31 42\n46 28\n31 20\n46 37\n1 39\n29 49\n37 47\n40 6\n42 36\n47 2\n24 46\n2 13\n8 45\n41 3\n32 17\n4 7\n47 26\n28 8\n41 50\n34 44\n33 21\n25 5\n16 40\n3 14\n8 18\n28 11\n32 22\n2 38\n3 48\n44 43\n",
"10\n8 1\n1 2\n8 9\n8 5\n1 3\n1 10\n1 6\n1 7\n8 4\n",
"5\n5 1\n5 4\n4 3\n1 2\n",
"7\n1 2\n2 3\n1 4\n1 5\n3 6\n3 7\n",
"3\n1 3\n2 3\n",
"60\n26 6\n59 30\n31 12\n31 3\n38 23\n59 29\n53 9\n38 56\n53 54\n29 21\n17 55\n59 38\n26 16\n24 59\n24 25\n17 35\n24 41\n30 15\n31 27\n8 44\n26 5\n26 48\n8 32\n53 17\n3 34\n3 51\n30 28\n47 10\n53 60\n36 42\n24 53\n59 22\n53 40\n26 52\n36 4\n59 8\n29 37\n36 20\n17 47\n53 18\n3 50\n30 2\n17 7\n8 58\n59 1\n31 11\n24 26\n24 43\n53 57\n59 45\n47 13\n26 46\n17 33\n30 31\n26 39\n26 19\n24 36\n8 49\n38 14\n",
"7\n1 2\n2 3\n3 4\n3 5\n1 6\n1 7\n",
"20\n19 16\n19 18\n20 7\n9 4\n6 17\n14 2\n9 15\n2 13\n5 11\n19 12\n12 20\n16 9\n11 8\n19 5\n3 1\n19 14\n5 3\n12 10\n19 6\n",
"7\n1 2\n1 3\n2 4\n2 5\n3 6\n3 7\n",
"10\n9 5\n7 1\n9 10\n7 2\n5 4\n9 6\n2 9\n10 8\n1 3\n",
"4\n2 4\n2 3\n2 1\n",
"4\n1 4\n3 2\n1 3\n",
"3\n1 2\n1 3\n",
"5\n1 2\n1 5\n1 3\n1 4\n",
"20\n14 9\n12 13\n10 15\n2 1\n20 19\n16 6\n16 3\n17 14\n3 5\n2 11\n3 10\n15 8\n14 2\n6 4\n3 20\n5 18\n1 7\n1 16\n4 12\n",
"20\n7 5\n14 13\n17 6\n3 8\n16 12\n18 9\n3 18\n14 1\n17 3\n15 2\n17 4\n9 11\n2 7\n15 17\n3 20\n16 10\n17 14\n2 16\n1 19\n",
"8\n1 2\n2 3\n3 4\n1 7\n1 8\n4 5\n4 6\n",
"5\n5 1\n5 2\n5 3\n5 4\n",
"50\n49 6\n43 7\n1 27\n19 35\n15 37\n16 12\n19 21\n16 28\n49 9\n48 39\n13 1\n2 48\n9 50\n44 3\n41 32\n48 31\n49 33\n6 11\n13 20\n49 22\n13 41\n48 29\n13 46\n15 47\n34 2\n49 13\n48 14\n34 24\n16 36\n13 40\n49 34\n49 17\n43 25\n11 23\n10 15\n19 26\n34 44\n16 42\n19 18\n46 8\n29 38\n1 45\n12 43\n13 16\n46 30\n15 5\n49 10\n11 19\n32 4\n",
"20\n13 1\n18 2\n3 7\n18 5\n20 16\n3 12\n18 9\n3 10\n18 11\n13 6\n3 18\n20 15\n20 17\n3 13\n3 4\n13 14\n3 20\n18 8\n3 19\n",
"10\n8 2\n5 6\n1 8\n2 9\n1 4\n8 10\n10 5\n2 7\n2 3\n"
],
"output": [
"NO",
"YES",
"NO",
"NO",
"NO",
"YES",
"NO",
"NO",
"NO",
"NO",
"YES",
"NO",
"NO",
"YES",
"NO",
"NO",
"NO",
"YES",
"NO",
"YES",
"NO"
]
}
|
IN-CORRECT
|
java
|
import java.io.*;
import java.util.*;
public class Main {
public static void main(String[] args) throws IOException {
InputStream inputStream = System.in;
OutputStream outputStream = System.out;
InputReader in = new InputReader(inputStream);
PrintWriter out = new PrintWriter(outputStream);
// InputReader in = new InputReader(".in");
// PrintWriter out = new PrintWriter(new FileWriter(".out"));
Task solver = new Task();
solver.solve(1, in, out);
out.close();
}
static class Task {
public void solve(int testNumber, InputReader in, PrintWriter out) {
int n = in.nextInt();
int[] a = new int[1000000];
Set<Integer> set = new HashSet<>();
for (int i = 0; i < n - 1; i++) {
int u = in.nextInt();
int v = in.nextInt();
if (a[u] < 0) {
a[u] = 2;
} else {
a[u]++;
}
if (a[v] == 0) {
a[v] = -u;
} else {
a[v]++;
}
set.add(u);
set.add(v);
}
for (int e : set) {
if (a[e] < 0 && a[-a[e]] == 2) {
out.println("NO");
return;
}
}
out.println("YES");
}
}
static class InputReader {
public BufferedReader reader;
public StringTokenizer tokenizer;
public InputReader(InputStream stream) {
reader = new BufferedReader(new InputStreamReader(stream), 32768);
tokenizer = null;
}
public InputReader(String s) {
try {
reader = new BufferedReader(new FileReader(s));
} catch (FileNotFoundException e) {
e.printStackTrace();
}
}
public String nextToken() {
while (tokenizer == null || !tokenizer.hasMoreTokens()) {
try {
tokenizer = new StringTokenizer(reader.readLine());
} catch (IOException e) {
throw new RuntimeException(e);
}
}
return tokenizer.nextToken();
}
public int nextInt() {
return Integer.parseInt(nextToken());
}
public long nextLong() {
return Long.parseLong(nextToken());
}
public double nextDouble() {
return Double.parseDouble(nextToken());
}
}
}
|
1189_D1. Add on a Tree
|
Note that this is the first problem of the two similar problems. You can hack this problem only if you solve both problems.
You are given a tree with n nodes. In the beginning, 0 is written on all edges. In one operation, you can choose any 2 distinct leaves u, v and any real number x and add x to values written on all edges on the simple path between u and v.
For example, on the picture below you can see the result of applying two operations to the graph: adding 2 on the path from 7 to 6, and then adding -0.5 on the path from 4 to 5.
<image>
Is it true that for any configuration of real numbers written on edges, we can achieve it with a finite number of operations?
Leaf is a node of a tree of degree 1. Simple path is a path that doesn't contain any node twice.
Input
The first line contains a single integer n (2 ≤ n ≤ 10^5) — the number of nodes.
Each of the next n-1 lines contains two integers u and v (1 ≤ u, v ≤ n, u ≠ v), meaning that there is an edge between nodes u and v. It is guaranteed that these edges form a tree.
Output
If there is a configuration of real numbers written on edges of the tree that we can't achieve by performing the operations, output "NO".
Otherwise, output "YES".
You can print each letter in any case (upper or lower).
Examples
Input
2
1 2
Output
YES
Input
3
1 2
2 3
Output
NO
Input
5
1 2
1 3
1 4
2 5
Output
NO
Input
6
1 2
1 3
1 4
2 5
2 6
Output
YES
Note
In the first example, we can add any real x to the value written on the only edge (1, 2).
<image>
In the second example, one of configurations that we can't reach is 0 written on (1, 2) and 1 written on (2, 3).
<image>
Below you can see graphs from examples 3, 4:
<image> <image>
|
{
"input": [
"2\n1 2\n",
"3\n1 2\n2 3\n",
"5\n1 2\n1 3\n1 4\n2 5\n",
"6\n1 2\n1 3\n1 4\n2 5\n2 6\n"
],
"output": [
"YES",
"NO",
"NO",
"YES"
]
}
|
{
"input": [
"50\n16 4\n17 9\n31 19\n22 10\n8 1\n40 30\n3 31\n20 29\n47 27\n22 25\n32 34\n12 15\n40 32\n10 33\n47 12\n6 24\n46 41\n14 23\n12 35\n31 42\n46 28\n31 20\n46 37\n1 39\n29 49\n37 47\n40 6\n42 36\n47 2\n24 46\n2 13\n8 45\n41 3\n32 17\n4 7\n47 26\n28 8\n41 50\n34 44\n33 21\n25 5\n16 40\n3 14\n8 18\n28 11\n32 22\n2 38\n3 48\n44 43\n",
"10\n8 1\n1 2\n8 9\n8 5\n1 3\n1 10\n1 6\n1 7\n8 4\n",
"5\n5 1\n5 4\n4 3\n1 2\n",
"7\n1 2\n2 3\n1 4\n1 5\n3 6\n3 7\n",
"3\n1 3\n2 3\n",
"60\n26 6\n59 30\n31 12\n31 3\n38 23\n59 29\n53 9\n38 56\n53 54\n29 21\n17 55\n59 38\n26 16\n24 59\n24 25\n17 35\n24 41\n30 15\n31 27\n8 44\n26 5\n26 48\n8 32\n53 17\n3 34\n3 51\n30 28\n47 10\n53 60\n36 42\n24 53\n59 22\n53 40\n26 52\n36 4\n59 8\n29 37\n36 20\n17 47\n53 18\n3 50\n30 2\n17 7\n8 58\n59 1\n31 11\n24 26\n24 43\n53 57\n59 45\n47 13\n26 46\n17 33\n30 31\n26 39\n26 19\n24 36\n8 49\n38 14\n",
"7\n1 2\n2 3\n3 4\n3 5\n1 6\n1 7\n",
"20\n19 16\n19 18\n20 7\n9 4\n6 17\n14 2\n9 15\n2 13\n5 11\n19 12\n12 20\n16 9\n11 8\n19 5\n3 1\n19 14\n5 3\n12 10\n19 6\n",
"7\n1 2\n1 3\n2 4\n2 5\n3 6\n3 7\n",
"10\n9 5\n7 1\n9 10\n7 2\n5 4\n9 6\n2 9\n10 8\n1 3\n",
"4\n2 4\n2 3\n2 1\n",
"4\n1 4\n3 2\n1 3\n",
"3\n1 2\n1 3\n",
"5\n1 2\n1 5\n1 3\n1 4\n",
"20\n14 9\n12 13\n10 15\n2 1\n20 19\n16 6\n16 3\n17 14\n3 5\n2 11\n3 10\n15 8\n14 2\n6 4\n3 20\n5 18\n1 7\n1 16\n4 12\n",
"20\n7 5\n14 13\n17 6\n3 8\n16 12\n18 9\n3 18\n14 1\n17 3\n15 2\n17 4\n9 11\n2 7\n15 17\n3 20\n16 10\n17 14\n2 16\n1 19\n",
"8\n1 2\n2 3\n3 4\n1 7\n1 8\n4 5\n4 6\n",
"5\n5 1\n5 2\n5 3\n5 4\n",
"50\n49 6\n43 7\n1 27\n19 35\n15 37\n16 12\n19 21\n16 28\n49 9\n48 39\n13 1\n2 48\n9 50\n44 3\n41 32\n48 31\n49 33\n6 11\n13 20\n49 22\n13 41\n48 29\n13 46\n15 47\n34 2\n49 13\n48 14\n34 24\n16 36\n13 40\n49 34\n49 17\n43 25\n11 23\n10 15\n19 26\n34 44\n16 42\n19 18\n46 8\n29 38\n1 45\n12 43\n13 16\n46 30\n15 5\n49 10\n11 19\n32 4\n",
"20\n13 1\n18 2\n3 7\n18 5\n20 16\n3 12\n18 9\n3 10\n18 11\n13 6\n3 18\n20 15\n20 17\n3 13\n3 4\n13 14\n3 20\n18 8\n3 19\n",
"10\n8 2\n5 6\n1 8\n2 9\n1 4\n8 10\n10 5\n2 7\n2 3\n"
],
"output": [
"NO",
"YES",
"NO",
"NO",
"NO",
"YES",
"NO",
"NO",
"NO",
"NO",
"YES",
"NO",
"NO",
"YES",
"NO",
"NO",
"NO",
"YES",
"NO",
"YES",
"NO"
]
}
|
IN-CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
const long long max_n = 1e6 + 20;
long long n, m, k, ans, sum;
long long a[max_n];
long long mark[max_n];
vector<long long> v, adj[max_n], jda[max_n];
void dfs(long long v) {
mark[v] = 1;
if (jda[v].size() == 1) ans = -1;
if (jda[v].size() == 2 && jda[jda[v][0]].size() == 0 &&
jda[jda[v][1]].size() == 0 && adj[v].size() == 0)
ans = -1;
for (auto i : adj[v]) {
if (!mark[i]) {
dfs(i);
}
}
}
int32_t main() {
cin >> n;
for (long long i = 1; i < n; i++) {
long long u, v;
cin >> u >> v;
u--, v--;
adj[v].push_back(u);
jda[u].push_back(v);
}
if (n == 2) {
cout << "YES";
return 0;
}
for (long long i = 0; i < n; i++) {
if (jda[i].size() == 0) {
dfs(i);
}
}
if (ans == -1) {
cout << "NO";
} else
cout << "YES";
return 0;
}
|
1189_D1. Add on a Tree
|
Note that this is the first problem of the two similar problems. You can hack this problem only if you solve both problems.
You are given a tree with n nodes. In the beginning, 0 is written on all edges. In one operation, you can choose any 2 distinct leaves u, v and any real number x and add x to values written on all edges on the simple path between u and v.
For example, on the picture below you can see the result of applying two operations to the graph: adding 2 on the path from 7 to 6, and then adding -0.5 on the path from 4 to 5.
<image>
Is it true that for any configuration of real numbers written on edges, we can achieve it with a finite number of operations?
Leaf is a node of a tree of degree 1. Simple path is a path that doesn't contain any node twice.
Input
The first line contains a single integer n (2 ≤ n ≤ 10^5) — the number of nodes.
Each of the next n-1 lines contains two integers u and v (1 ≤ u, v ≤ n, u ≠ v), meaning that there is an edge between nodes u and v. It is guaranteed that these edges form a tree.
Output
If there is a configuration of real numbers written on edges of the tree that we can't achieve by performing the operations, output "NO".
Otherwise, output "YES".
You can print each letter in any case (upper or lower).
Examples
Input
2
1 2
Output
YES
Input
3
1 2
2 3
Output
NO
Input
5
1 2
1 3
1 4
2 5
Output
NO
Input
6
1 2
1 3
1 4
2 5
2 6
Output
YES
Note
In the first example, we can add any real x to the value written on the only edge (1, 2).
<image>
In the second example, one of configurations that we can't reach is 0 written on (1, 2) and 1 written on (2, 3).
<image>
Below you can see graphs from examples 3, 4:
<image> <image>
|
{
"input": [
"2\n1 2\n",
"3\n1 2\n2 3\n",
"5\n1 2\n1 3\n1 4\n2 5\n",
"6\n1 2\n1 3\n1 4\n2 5\n2 6\n"
],
"output": [
"YES",
"NO",
"NO",
"YES"
]
}
|
{
"input": [
"50\n16 4\n17 9\n31 19\n22 10\n8 1\n40 30\n3 31\n20 29\n47 27\n22 25\n32 34\n12 15\n40 32\n10 33\n47 12\n6 24\n46 41\n14 23\n12 35\n31 42\n46 28\n31 20\n46 37\n1 39\n29 49\n37 47\n40 6\n42 36\n47 2\n24 46\n2 13\n8 45\n41 3\n32 17\n4 7\n47 26\n28 8\n41 50\n34 44\n33 21\n25 5\n16 40\n3 14\n8 18\n28 11\n32 22\n2 38\n3 48\n44 43\n",
"10\n8 1\n1 2\n8 9\n8 5\n1 3\n1 10\n1 6\n1 7\n8 4\n",
"5\n5 1\n5 4\n4 3\n1 2\n",
"7\n1 2\n2 3\n1 4\n1 5\n3 6\n3 7\n",
"3\n1 3\n2 3\n",
"60\n26 6\n59 30\n31 12\n31 3\n38 23\n59 29\n53 9\n38 56\n53 54\n29 21\n17 55\n59 38\n26 16\n24 59\n24 25\n17 35\n24 41\n30 15\n31 27\n8 44\n26 5\n26 48\n8 32\n53 17\n3 34\n3 51\n30 28\n47 10\n53 60\n36 42\n24 53\n59 22\n53 40\n26 52\n36 4\n59 8\n29 37\n36 20\n17 47\n53 18\n3 50\n30 2\n17 7\n8 58\n59 1\n31 11\n24 26\n24 43\n53 57\n59 45\n47 13\n26 46\n17 33\n30 31\n26 39\n26 19\n24 36\n8 49\n38 14\n",
"7\n1 2\n2 3\n3 4\n3 5\n1 6\n1 7\n",
"20\n19 16\n19 18\n20 7\n9 4\n6 17\n14 2\n9 15\n2 13\n5 11\n19 12\n12 20\n16 9\n11 8\n19 5\n3 1\n19 14\n5 3\n12 10\n19 6\n",
"7\n1 2\n1 3\n2 4\n2 5\n3 6\n3 7\n",
"10\n9 5\n7 1\n9 10\n7 2\n5 4\n9 6\n2 9\n10 8\n1 3\n",
"4\n2 4\n2 3\n2 1\n",
"4\n1 4\n3 2\n1 3\n",
"3\n1 2\n1 3\n",
"5\n1 2\n1 5\n1 3\n1 4\n",
"20\n14 9\n12 13\n10 15\n2 1\n20 19\n16 6\n16 3\n17 14\n3 5\n2 11\n3 10\n15 8\n14 2\n6 4\n3 20\n5 18\n1 7\n1 16\n4 12\n",
"20\n7 5\n14 13\n17 6\n3 8\n16 12\n18 9\n3 18\n14 1\n17 3\n15 2\n17 4\n9 11\n2 7\n15 17\n3 20\n16 10\n17 14\n2 16\n1 19\n",
"8\n1 2\n2 3\n3 4\n1 7\n1 8\n4 5\n4 6\n",
"5\n5 1\n5 2\n5 3\n5 4\n",
"50\n49 6\n43 7\n1 27\n19 35\n15 37\n16 12\n19 21\n16 28\n49 9\n48 39\n13 1\n2 48\n9 50\n44 3\n41 32\n48 31\n49 33\n6 11\n13 20\n49 22\n13 41\n48 29\n13 46\n15 47\n34 2\n49 13\n48 14\n34 24\n16 36\n13 40\n49 34\n49 17\n43 25\n11 23\n10 15\n19 26\n34 44\n16 42\n19 18\n46 8\n29 38\n1 45\n12 43\n13 16\n46 30\n15 5\n49 10\n11 19\n32 4\n",
"20\n13 1\n18 2\n3 7\n18 5\n20 16\n3 12\n18 9\n3 10\n18 11\n13 6\n3 18\n20 15\n20 17\n3 13\n3 4\n13 14\n3 20\n18 8\n3 19\n",
"10\n8 2\n5 6\n1 8\n2 9\n1 4\n8 10\n10 5\n2 7\n2 3\n"
],
"output": [
"NO",
"YES",
"NO",
"NO",
"NO",
"YES",
"NO",
"NO",
"NO",
"NO",
"YES",
"NO",
"NO",
"YES",
"NO",
"NO",
"NO",
"YES",
"NO",
"YES",
"NO"
]
}
|
IN-CORRECT
|
java
|
import java.util.*;
import java.io.*;
public class Main {
static final int MOD_PRIME = 1000000007;
public static void main(String[] args) {
InputStream inputStream = System.in;
OutputStream outputStream = System.out;
InputReader in = new InputReader(inputStream);
PrintWriter out = new PrintWriter(outputStream);
Task solver = new Task();
int i = 0;
int t = 1;
// t = in.nextInt();
for (; i < t; i++)
solver.solve(i, in, out);
out.close();
}
static class Task {
public void solve(int testNumber, InputReader in, PrintWriter out) {
int n = in.nextInt();
HashMap<Integer , ArrayList<Integer>> adj = new HashMap<Integer, ArrayList<Integer>>();
for(int i = 0; i < n - 1; i++) {
int u = in.nextInt();
int v = in.nextInt();
if(!adj.containsKey(u)) adj.put(u, new ArrayList<Integer>());
if(!adj.containsKey(v)) adj.put(v, new ArrayList<Integer>());
adj.get(u).add(v);
adj.get(v).add(u);
}
long leaf = 0;
for(int k : adj.keySet()) {
if(adj.get(k).size() == 1) leaf++;
}
if(n - 1 > (leaf*(leaf - 1))/2) {
out.println("NO");
}else {
out.println("YES");
}
}
}
// template code
static class InputReader {
public BufferedReader reader;
public StringTokenizer tokenizer;
public InputReader(InputStream stream) {
reader = new BufferedReader(new InputStreamReader(stream), 32768);
tokenizer = null;
}
public String next() {
while (tokenizer == null || !tokenizer.hasMoreTokens()) {
try {
tokenizer = new StringTokenizer(reader.readLine());
} catch (IOException e) {
throw new RuntimeException(e);
}
}
return tokenizer.nextToken();
}
public int nextInt() {
return Integer.parseInt(next());
}
public long nextLong() {
return Long.parseLong(next());
}
public double nextDouble() {
return Double.parseDouble(next());
}
}
static long modexp(long a, long b, long p) {
// returns a to the power b mod p by modular exponentiation
long res = 1;
long mult = a % p;
while (b > 0) {
if (b % 2 == 1) {
res = (res * mult) % p;
}
b /= 2;
mult = (mult * mult) % p;
}
return res;
}
static double log(double arg, double base) {
// returns log of a base b, contains floating point errors, dont use for exact
// calculations.
if (base < 0 || base == 1) {
throw new ArithmeticException("base cant be 1 or negative");
}
if (arg < 0) {
throw new ArithmeticException("log of negative number undefined");
}
return Math.log10(arg) / Math.log10(base);
}
static int gcd(int a, int b) {
if (b == 0) {
return a;
}
return gcd(b, a % b);
}
static long gcd(long a, long b) {
if (b == 0) {
return a;
}
return gcd(b, a % b);
}
// scope for improvement
static ArrayList<Integer> sieveOfEratosthenes(int n) {
boolean[] check = new boolean[n + 1];
ArrayList<Integer> prime = new ArrayList<Integer>();
for (long i = 2; i <= n; i++) {
if (!check[(int) i]) {
prime.add((int) i);
for (long j = i * i; j <= n; j += i) {
check[(int) j] = true;
}
}
}
return prime;
}
static int modInverse(int a, int n) {
// returns inverse of a mod n by extended euclidean algorithm
int t = 0;
int newt = 1;
int r = n;
int newr = a;
int quotient;
int tempr, tempt;
while (newr != 0) {
quotient = r / newr;
tempt = newt;
tempr = newr;
newr = r - quotient * tempr;
newt = t - quotient * tempt;
t = tempt;
r = tempr;
}
if (r > 1) {
throw new ArithmeticException("inverse of " + a + " mod " + n + " does not exist");
} else {
if (t < 0) {
t += n;
}
return t;
}
}
static long primeModInverse(long a, long n) {
// returns inverse of a mod n by mod exponentiation, use only if n is prime
return modexp(a, n - 2, n);
}
static void dfs(HashMap<Integer, ArrayList<Integer>> adj, Set<Integer> ans, Set<Integer> open,
HashMap<String, Integer> edge, boolean[] vis, int i) {
vis[i] = true;
open.add(i);
if (adj.get(i) != null) {
for (int k : adj.get(i)) {
if (!vis[k]) {
dfs(adj, ans, open, edge, vis, k);
} else if (open.contains(k)) {
ans.add(edge.get(String.valueOf(i) + " " + String.valueOf(k)));
}
}
}
open.remove(i);
}
}
|
1189_D1. Add on a Tree
|
Note that this is the first problem of the two similar problems. You can hack this problem only if you solve both problems.
You are given a tree with n nodes. In the beginning, 0 is written on all edges. In one operation, you can choose any 2 distinct leaves u, v and any real number x and add x to values written on all edges on the simple path between u and v.
For example, on the picture below you can see the result of applying two operations to the graph: adding 2 on the path from 7 to 6, and then adding -0.5 on the path from 4 to 5.
<image>
Is it true that for any configuration of real numbers written on edges, we can achieve it with a finite number of operations?
Leaf is a node of a tree of degree 1. Simple path is a path that doesn't contain any node twice.
Input
The first line contains a single integer n (2 ≤ n ≤ 10^5) — the number of nodes.
Each of the next n-1 lines contains two integers u and v (1 ≤ u, v ≤ n, u ≠ v), meaning that there is an edge between nodes u and v. It is guaranteed that these edges form a tree.
Output
If there is a configuration of real numbers written on edges of the tree that we can't achieve by performing the operations, output "NO".
Otherwise, output "YES".
You can print each letter in any case (upper or lower).
Examples
Input
2
1 2
Output
YES
Input
3
1 2
2 3
Output
NO
Input
5
1 2
1 3
1 4
2 5
Output
NO
Input
6
1 2
1 3
1 4
2 5
2 6
Output
YES
Note
In the first example, we can add any real x to the value written on the only edge (1, 2).
<image>
In the second example, one of configurations that we can't reach is 0 written on (1, 2) and 1 written on (2, 3).
<image>
Below you can see graphs from examples 3, 4:
<image> <image>
|
{
"input": [
"2\n1 2\n",
"3\n1 2\n2 3\n",
"5\n1 2\n1 3\n1 4\n2 5\n",
"6\n1 2\n1 3\n1 4\n2 5\n2 6\n"
],
"output": [
"YES",
"NO",
"NO",
"YES"
]
}
|
{
"input": [
"50\n16 4\n17 9\n31 19\n22 10\n8 1\n40 30\n3 31\n20 29\n47 27\n22 25\n32 34\n12 15\n40 32\n10 33\n47 12\n6 24\n46 41\n14 23\n12 35\n31 42\n46 28\n31 20\n46 37\n1 39\n29 49\n37 47\n40 6\n42 36\n47 2\n24 46\n2 13\n8 45\n41 3\n32 17\n4 7\n47 26\n28 8\n41 50\n34 44\n33 21\n25 5\n16 40\n3 14\n8 18\n28 11\n32 22\n2 38\n3 48\n44 43\n",
"10\n8 1\n1 2\n8 9\n8 5\n1 3\n1 10\n1 6\n1 7\n8 4\n",
"5\n5 1\n5 4\n4 3\n1 2\n",
"7\n1 2\n2 3\n1 4\n1 5\n3 6\n3 7\n",
"3\n1 3\n2 3\n",
"60\n26 6\n59 30\n31 12\n31 3\n38 23\n59 29\n53 9\n38 56\n53 54\n29 21\n17 55\n59 38\n26 16\n24 59\n24 25\n17 35\n24 41\n30 15\n31 27\n8 44\n26 5\n26 48\n8 32\n53 17\n3 34\n3 51\n30 28\n47 10\n53 60\n36 42\n24 53\n59 22\n53 40\n26 52\n36 4\n59 8\n29 37\n36 20\n17 47\n53 18\n3 50\n30 2\n17 7\n8 58\n59 1\n31 11\n24 26\n24 43\n53 57\n59 45\n47 13\n26 46\n17 33\n30 31\n26 39\n26 19\n24 36\n8 49\n38 14\n",
"7\n1 2\n2 3\n3 4\n3 5\n1 6\n1 7\n",
"20\n19 16\n19 18\n20 7\n9 4\n6 17\n14 2\n9 15\n2 13\n5 11\n19 12\n12 20\n16 9\n11 8\n19 5\n3 1\n19 14\n5 3\n12 10\n19 6\n",
"7\n1 2\n1 3\n2 4\n2 5\n3 6\n3 7\n",
"10\n9 5\n7 1\n9 10\n7 2\n5 4\n9 6\n2 9\n10 8\n1 3\n",
"4\n2 4\n2 3\n2 1\n",
"4\n1 4\n3 2\n1 3\n",
"3\n1 2\n1 3\n",
"5\n1 2\n1 5\n1 3\n1 4\n",
"20\n14 9\n12 13\n10 15\n2 1\n20 19\n16 6\n16 3\n17 14\n3 5\n2 11\n3 10\n15 8\n14 2\n6 4\n3 20\n5 18\n1 7\n1 16\n4 12\n",
"20\n7 5\n14 13\n17 6\n3 8\n16 12\n18 9\n3 18\n14 1\n17 3\n15 2\n17 4\n9 11\n2 7\n15 17\n3 20\n16 10\n17 14\n2 16\n1 19\n",
"8\n1 2\n2 3\n3 4\n1 7\n1 8\n4 5\n4 6\n",
"5\n5 1\n5 2\n5 3\n5 4\n",
"50\n49 6\n43 7\n1 27\n19 35\n15 37\n16 12\n19 21\n16 28\n49 9\n48 39\n13 1\n2 48\n9 50\n44 3\n41 32\n48 31\n49 33\n6 11\n13 20\n49 22\n13 41\n48 29\n13 46\n15 47\n34 2\n49 13\n48 14\n34 24\n16 36\n13 40\n49 34\n49 17\n43 25\n11 23\n10 15\n19 26\n34 44\n16 42\n19 18\n46 8\n29 38\n1 45\n12 43\n13 16\n46 30\n15 5\n49 10\n11 19\n32 4\n",
"20\n13 1\n18 2\n3 7\n18 5\n20 16\n3 12\n18 9\n3 10\n18 11\n13 6\n3 18\n20 15\n20 17\n3 13\n3 4\n13 14\n3 20\n18 8\n3 19\n",
"10\n8 2\n5 6\n1 8\n2 9\n1 4\n8 10\n10 5\n2 7\n2 3\n"
],
"output": [
"NO",
"YES",
"NO",
"NO",
"NO",
"YES",
"NO",
"NO",
"NO",
"NO",
"YES",
"NO",
"NO",
"YES",
"NO",
"NO",
"NO",
"YES",
"NO",
"YES",
"NO"
]
}
|
IN-CORRECT
|
java
|
//package Contest572;
//package ... ctrl+alt+l //выравнять код
import java.io.*;
import java.util.LinkedList;
import java.util.StringTokenizer;
public class mainD1 {
public static PrintWriter out = new PrintWriter(System.out);
public static FastScanner enter = new FastScanner(System.in);
public static void main(String[] args) throws IOException {
int n=enter.nextInt();
Graph g=new Graph(n);
int[] mass1=new int[n];
int[] mass2=new int[n];
for (int i = 0; i <n-1 ; i++) {
mass1[i]=enter.nextInt();
mass2[i]=enter.nextInt();
g.addEdge(mass1[i],mass2[i]);
g.addEdge(mass2[i],mass1[i]);
}
if(n==2){
out.println("YES");
out.close();
return;
}
for (int i = 0; i <n-1 ; i++) {
if(g.adj[mass1[i]].size()<=2 && g.adj[mass2[i]].size()<=2){
out.println("NO");
out.close();
return;
}
}
out.println("YES");
out.close();
}
static class FastScanner {
BufferedReader br;
StringTokenizer stok;
FastScanner(InputStream is) {
br = new BufferedReader(new InputStreamReader(is));
}
String next() throws IOException {
while (stok == null || !stok.hasMoreTokens()) {
String s = br.readLine();
if (s == null) {
return null;
}
stok = new StringTokenizer(s);
}
return stok.nextToken();
}
int nextInt() throws IOException {
return Integer.parseInt(next());
}
long nextLong() throws IOException {
return Long.parseLong(next());
}
double nextDouble() throws IOException {
return Double.parseDouble(next());
}
char nextChar() throws IOException {
return (char) (br.read());
}
String nextLine() throws IOException {
return br.readLine();
}
}
static class Graph{
private int n; // No. of vertices
private int m;
private LinkedList<Integer> adj[]; //Adjacency Lists
boolean visited[];
int[] color;
// Constructor
Graph(int v)
{
n = v;
adj = new LinkedList[v+1];
for (int i=1; i<v+1; ++i) adj[i] = new LinkedList();
visited= new boolean[n+1];
color= new int[n+1];
}
public Graph(int n, int m) {
this.n = n;
this.m = m;
adj = new LinkedList[n+1];
for (int i=1; i<n+1; ++i) adj[i] = new LinkedList();
visited= new boolean[n+1];
color= new int[n+1];
}
void addEdge(int v, int w)
{
adj[v].add(w);
}
void sout_tree(){
for (int i = 1; i <=n ; i++) {
System.out.print(i +" ");
System.out.println(adj[i]);
}
}
}
}
|
1189_D1. Add on a Tree
|
Note that this is the first problem of the two similar problems. You can hack this problem only if you solve both problems.
You are given a tree with n nodes. In the beginning, 0 is written on all edges. In one operation, you can choose any 2 distinct leaves u, v and any real number x and add x to values written on all edges on the simple path between u and v.
For example, on the picture below you can see the result of applying two operations to the graph: adding 2 on the path from 7 to 6, and then adding -0.5 on the path from 4 to 5.
<image>
Is it true that for any configuration of real numbers written on edges, we can achieve it with a finite number of operations?
Leaf is a node of a tree of degree 1. Simple path is a path that doesn't contain any node twice.
Input
The first line contains a single integer n (2 ≤ n ≤ 10^5) — the number of nodes.
Each of the next n-1 lines contains two integers u and v (1 ≤ u, v ≤ n, u ≠ v), meaning that there is an edge between nodes u and v. It is guaranteed that these edges form a tree.
Output
If there is a configuration of real numbers written on edges of the tree that we can't achieve by performing the operations, output "NO".
Otherwise, output "YES".
You can print each letter in any case (upper or lower).
Examples
Input
2
1 2
Output
YES
Input
3
1 2
2 3
Output
NO
Input
5
1 2
1 3
1 4
2 5
Output
NO
Input
6
1 2
1 3
1 4
2 5
2 6
Output
YES
Note
In the first example, we can add any real x to the value written on the only edge (1, 2).
<image>
In the second example, one of configurations that we can't reach is 0 written on (1, 2) and 1 written on (2, 3).
<image>
Below you can see graphs from examples 3, 4:
<image> <image>
|
{
"input": [
"2\n1 2\n",
"3\n1 2\n2 3\n",
"5\n1 2\n1 3\n1 4\n2 5\n",
"6\n1 2\n1 3\n1 4\n2 5\n2 6\n"
],
"output": [
"YES",
"NO",
"NO",
"YES"
]
}
|
{
"input": [
"50\n16 4\n17 9\n31 19\n22 10\n8 1\n40 30\n3 31\n20 29\n47 27\n22 25\n32 34\n12 15\n40 32\n10 33\n47 12\n6 24\n46 41\n14 23\n12 35\n31 42\n46 28\n31 20\n46 37\n1 39\n29 49\n37 47\n40 6\n42 36\n47 2\n24 46\n2 13\n8 45\n41 3\n32 17\n4 7\n47 26\n28 8\n41 50\n34 44\n33 21\n25 5\n16 40\n3 14\n8 18\n28 11\n32 22\n2 38\n3 48\n44 43\n",
"10\n8 1\n1 2\n8 9\n8 5\n1 3\n1 10\n1 6\n1 7\n8 4\n",
"5\n5 1\n5 4\n4 3\n1 2\n",
"7\n1 2\n2 3\n1 4\n1 5\n3 6\n3 7\n",
"3\n1 3\n2 3\n",
"60\n26 6\n59 30\n31 12\n31 3\n38 23\n59 29\n53 9\n38 56\n53 54\n29 21\n17 55\n59 38\n26 16\n24 59\n24 25\n17 35\n24 41\n30 15\n31 27\n8 44\n26 5\n26 48\n8 32\n53 17\n3 34\n3 51\n30 28\n47 10\n53 60\n36 42\n24 53\n59 22\n53 40\n26 52\n36 4\n59 8\n29 37\n36 20\n17 47\n53 18\n3 50\n30 2\n17 7\n8 58\n59 1\n31 11\n24 26\n24 43\n53 57\n59 45\n47 13\n26 46\n17 33\n30 31\n26 39\n26 19\n24 36\n8 49\n38 14\n",
"7\n1 2\n2 3\n3 4\n3 5\n1 6\n1 7\n",
"20\n19 16\n19 18\n20 7\n9 4\n6 17\n14 2\n9 15\n2 13\n5 11\n19 12\n12 20\n16 9\n11 8\n19 5\n3 1\n19 14\n5 3\n12 10\n19 6\n",
"7\n1 2\n1 3\n2 4\n2 5\n3 6\n3 7\n",
"10\n9 5\n7 1\n9 10\n7 2\n5 4\n9 6\n2 9\n10 8\n1 3\n",
"4\n2 4\n2 3\n2 1\n",
"4\n1 4\n3 2\n1 3\n",
"3\n1 2\n1 3\n",
"5\n1 2\n1 5\n1 3\n1 4\n",
"20\n14 9\n12 13\n10 15\n2 1\n20 19\n16 6\n16 3\n17 14\n3 5\n2 11\n3 10\n15 8\n14 2\n6 4\n3 20\n5 18\n1 7\n1 16\n4 12\n",
"20\n7 5\n14 13\n17 6\n3 8\n16 12\n18 9\n3 18\n14 1\n17 3\n15 2\n17 4\n9 11\n2 7\n15 17\n3 20\n16 10\n17 14\n2 16\n1 19\n",
"8\n1 2\n2 3\n3 4\n1 7\n1 8\n4 5\n4 6\n",
"5\n5 1\n5 2\n5 3\n5 4\n",
"50\n49 6\n43 7\n1 27\n19 35\n15 37\n16 12\n19 21\n16 28\n49 9\n48 39\n13 1\n2 48\n9 50\n44 3\n41 32\n48 31\n49 33\n6 11\n13 20\n49 22\n13 41\n48 29\n13 46\n15 47\n34 2\n49 13\n48 14\n34 24\n16 36\n13 40\n49 34\n49 17\n43 25\n11 23\n10 15\n19 26\n34 44\n16 42\n19 18\n46 8\n29 38\n1 45\n12 43\n13 16\n46 30\n15 5\n49 10\n11 19\n32 4\n",
"20\n13 1\n18 2\n3 7\n18 5\n20 16\n3 12\n18 9\n3 10\n18 11\n13 6\n3 18\n20 15\n20 17\n3 13\n3 4\n13 14\n3 20\n18 8\n3 19\n",
"10\n8 2\n5 6\n1 8\n2 9\n1 4\n8 10\n10 5\n2 7\n2 3\n"
],
"output": [
"NO",
"YES",
"NO",
"NO",
"NO",
"YES",
"NO",
"NO",
"NO",
"NO",
"YES",
"NO",
"NO",
"YES",
"NO",
"NO",
"NO",
"YES",
"NO",
"YES",
"NO"
]
}
|
IN-CORRECT
|
java
|
import java.util.*;
import java.lang.*;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.*;
public class Main {
int flag=1;
public void dfs(ArrayList<Integer>[] tree,int[] visited,int s) {
visited[s]=1;
int count=0;
for(int i:tree[s]) {
if(visited[i]==0) {
count++;
this.dfs(tree, visited, i);
}
}
if(count==1) {
flag=0;
}
}
public static void main(String[] args) throws Exception{
FastReader sc=new FastReader();
OutputStream outputStream = System.out;
PrintWriter out = new PrintWriter(outputStream);
Main mm=new Main();
int n=sc.nextInt();
ArrayList<Integer>[] tree=new ArrayList[n];
for(int i=0;i<n;i++) {
tree[i]=new ArrayList<Integer>();
}
for(int i=0;i<n-1;i++) {
int a=sc.nextInt()-1;
int b=sc.nextInt()-1;
tree[a].add(b);
tree[b].add(a);
}
if(n==2) {
System.out.println("YES");
}
else {
int s=-1;
for(int i=0;i<n;i++) {
if(tree[i].size()!=1) {
s=i;
break;
}
}
mm.dfs(tree,new int[n], s);
if(mm.flag==1) {
System.out.println("YES");
}
else {
System.out.println("NO");
}
}
}
}
class FastReader
{
BufferedReader br;
StringTokenizer st;
public FastReader()
{
br = new BufferedReader(new
InputStreamReader(System.in));
}
String next()
{
while (st == null || !st.hasMoreElements())
{
try
{
st = new StringTokenizer(br.readLine());
}
catch (IOException e)
{
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt()
{
return Integer.parseInt(next());
}
long nextLong()
{
return Long.parseLong(next());
}
double nextDouble()
{
return Double.parseDouble(next());
}
String nextLine()
{
String str = "";
try
{
str = br.readLine();
}
catch (IOException e)
{
e.printStackTrace();
}
return str;
}
}
|
1189_D1. Add on a Tree
|
Note that this is the first problem of the two similar problems. You can hack this problem only if you solve both problems.
You are given a tree with n nodes. In the beginning, 0 is written on all edges. In one operation, you can choose any 2 distinct leaves u, v and any real number x and add x to values written on all edges on the simple path between u and v.
For example, on the picture below you can see the result of applying two operations to the graph: adding 2 on the path from 7 to 6, and then adding -0.5 on the path from 4 to 5.
<image>
Is it true that for any configuration of real numbers written on edges, we can achieve it with a finite number of operations?
Leaf is a node of a tree of degree 1. Simple path is a path that doesn't contain any node twice.
Input
The first line contains a single integer n (2 ≤ n ≤ 10^5) — the number of nodes.
Each of the next n-1 lines contains two integers u and v (1 ≤ u, v ≤ n, u ≠ v), meaning that there is an edge between nodes u and v. It is guaranteed that these edges form a tree.
Output
If there is a configuration of real numbers written on edges of the tree that we can't achieve by performing the operations, output "NO".
Otherwise, output "YES".
You can print each letter in any case (upper or lower).
Examples
Input
2
1 2
Output
YES
Input
3
1 2
2 3
Output
NO
Input
5
1 2
1 3
1 4
2 5
Output
NO
Input
6
1 2
1 3
1 4
2 5
2 6
Output
YES
Note
In the first example, we can add any real x to the value written on the only edge (1, 2).
<image>
In the second example, one of configurations that we can't reach is 0 written on (1, 2) and 1 written on (2, 3).
<image>
Below you can see graphs from examples 3, 4:
<image> <image>
|
{
"input": [
"2\n1 2\n",
"3\n1 2\n2 3\n",
"5\n1 2\n1 3\n1 4\n2 5\n",
"6\n1 2\n1 3\n1 4\n2 5\n2 6\n"
],
"output": [
"YES",
"NO",
"NO",
"YES"
]
}
|
{
"input": [
"50\n16 4\n17 9\n31 19\n22 10\n8 1\n40 30\n3 31\n20 29\n47 27\n22 25\n32 34\n12 15\n40 32\n10 33\n47 12\n6 24\n46 41\n14 23\n12 35\n31 42\n46 28\n31 20\n46 37\n1 39\n29 49\n37 47\n40 6\n42 36\n47 2\n24 46\n2 13\n8 45\n41 3\n32 17\n4 7\n47 26\n28 8\n41 50\n34 44\n33 21\n25 5\n16 40\n3 14\n8 18\n28 11\n32 22\n2 38\n3 48\n44 43\n",
"10\n8 1\n1 2\n8 9\n8 5\n1 3\n1 10\n1 6\n1 7\n8 4\n",
"5\n5 1\n5 4\n4 3\n1 2\n",
"7\n1 2\n2 3\n1 4\n1 5\n3 6\n3 7\n",
"3\n1 3\n2 3\n",
"60\n26 6\n59 30\n31 12\n31 3\n38 23\n59 29\n53 9\n38 56\n53 54\n29 21\n17 55\n59 38\n26 16\n24 59\n24 25\n17 35\n24 41\n30 15\n31 27\n8 44\n26 5\n26 48\n8 32\n53 17\n3 34\n3 51\n30 28\n47 10\n53 60\n36 42\n24 53\n59 22\n53 40\n26 52\n36 4\n59 8\n29 37\n36 20\n17 47\n53 18\n3 50\n30 2\n17 7\n8 58\n59 1\n31 11\n24 26\n24 43\n53 57\n59 45\n47 13\n26 46\n17 33\n30 31\n26 39\n26 19\n24 36\n8 49\n38 14\n",
"7\n1 2\n2 3\n3 4\n3 5\n1 6\n1 7\n",
"20\n19 16\n19 18\n20 7\n9 4\n6 17\n14 2\n9 15\n2 13\n5 11\n19 12\n12 20\n16 9\n11 8\n19 5\n3 1\n19 14\n5 3\n12 10\n19 6\n",
"7\n1 2\n1 3\n2 4\n2 5\n3 6\n3 7\n",
"10\n9 5\n7 1\n9 10\n7 2\n5 4\n9 6\n2 9\n10 8\n1 3\n",
"4\n2 4\n2 3\n2 1\n",
"4\n1 4\n3 2\n1 3\n",
"3\n1 2\n1 3\n",
"5\n1 2\n1 5\n1 3\n1 4\n",
"20\n14 9\n12 13\n10 15\n2 1\n20 19\n16 6\n16 3\n17 14\n3 5\n2 11\n3 10\n15 8\n14 2\n6 4\n3 20\n5 18\n1 7\n1 16\n4 12\n",
"20\n7 5\n14 13\n17 6\n3 8\n16 12\n18 9\n3 18\n14 1\n17 3\n15 2\n17 4\n9 11\n2 7\n15 17\n3 20\n16 10\n17 14\n2 16\n1 19\n",
"8\n1 2\n2 3\n3 4\n1 7\n1 8\n4 5\n4 6\n",
"5\n5 1\n5 2\n5 3\n5 4\n",
"50\n49 6\n43 7\n1 27\n19 35\n15 37\n16 12\n19 21\n16 28\n49 9\n48 39\n13 1\n2 48\n9 50\n44 3\n41 32\n48 31\n49 33\n6 11\n13 20\n49 22\n13 41\n48 29\n13 46\n15 47\n34 2\n49 13\n48 14\n34 24\n16 36\n13 40\n49 34\n49 17\n43 25\n11 23\n10 15\n19 26\n34 44\n16 42\n19 18\n46 8\n29 38\n1 45\n12 43\n13 16\n46 30\n15 5\n49 10\n11 19\n32 4\n",
"20\n13 1\n18 2\n3 7\n18 5\n20 16\n3 12\n18 9\n3 10\n18 11\n13 6\n3 18\n20 15\n20 17\n3 13\n3 4\n13 14\n3 20\n18 8\n3 19\n",
"10\n8 2\n5 6\n1 8\n2 9\n1 4\n8 10\n10 5\n2 7\n2 3\n"
],
"output": [
"NO",
"YES",
"NO",
"NO",
"NO",
"YES",
"NO",
"NO",
"NO",
"NO",
"YES",
"NO",
"NO",
"YES",
"NO",
"NO",
"NO",
"YES",
"NO",
"YES",
"NO"
]
}
|
IN-CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
const int MAX_N = 1e5 + 5;
int n;
vector<int> G[MAX_N];
vector<int> leaf;
bool ans = true;
int main() {
scanf("%d", &n);
for (int i = 0, t1, t2; i < n - 1; i++) {
scanf("%d%d", &t1, &t2);
G[t1].emplace_back(t2);
G[t2].emplace_back(t1);
}
for (int i = 1; i <= n; i++)
if (G[i].size() == 1) leaf.emplace_back(i);
for (auto node : leaf) {
if (G[G[node].back()].size() == 2) {
ans = false;
break;
}
}
puts(ans ? "YES" : "NO");
return 0;
}
|
1189_D1. Add on a Tree
|
Note that this is the first problem of the two similar problems. You can hack this problem only if you solve both problems.
You are given a tree with n nodes. In the beginning, 0 is written on all edges. In one operation, you can choose any 2 distinct leaves u, v and any real number x and add x to values written on all edges on the simple path between u and v.
For example, on the picture below you can see the result of applying two operations to the graph: adding 2 on the path from 7 to 6, and then adding -0.5 on the path from 4 to 5.
<image>
Is it true that for any configuration of real numbers written on edges, we can achieve it with a finite number of operations?
Leaf is a node of a tree of degree 1. Simple path is a path that doesn't contain any node twice.
Input
The first line contains a single integer n (2 ≤ n ≤ 10^5) — the number of nodes.
Each of the next n-1 lines contains two integers u and v (1 ≤ u, v ≤ n, u ≠ v), meaning that there is an edge between nodes u and v. It is guaranteed that these edges form a tree.
Output
If there is a configuration of real numbers written on edges of the tree that we can't achieve by performing the operations, output "NO".
Otherwise, output "YES".
You can print each letter in any case (upper or lower).
Examples
Input
2
1 2
Output
YES
Input
3
1 2
2 3
Output
NO
Input
5
1 2
1 3
1 4
2 5
Output
NO
Input
6
1 2
1 3
1 4
2 5
2 6
Output
YES
Note
In the first example, we can add any real x to the value written on the only edge (1, 2).
<image>
In the second example, one of configurations that we can't reach is 0 written on (1, 2) and 1 written on (2, 3).
<image>
Below you can see graphs from examples 3, 4:
<image> <image>
|
{
"input": [
"2\n1 2\n",
"3\n1 2\n2 3\n",
"5\n1 2\n1 3\n1 4\n2 5\n",
"6\n1 2\n1 3\n1 4\n2 5\n2 6\n"
],
"output": [
"YES",
"NO",
"NO",
"YES"
]
}
|
{
"input": [
"50\n16 4\n17 9\n31 19\n22 10\n8 1\n40 30\n3 31\n20 29\n47 27\n22 25\n32 34\n12 15\n40 32\n10 33\n47 12\n6 24\n46 41\n14 23\n12 35\n31 42\n46 28\n31 20\n46 37\n1 39\n29 49\n37 47\n40 6\n42 36\n47 2\n24 46\n2 13\n8 45\n41 3\n32 17\n4 7\n47 26\n28 8\n41 50\n34 44\n33 21\n25 5\n16 40\n3 14\n8 18\n28 11\n32 22\n2 38\n3 48\n44 43\n",
"10\n8 1\n1 2\n8 9\n8 5\n1 3\n1 10\n1 6\n1 7\n8 4\n",
"5\n5 1\n5 4\n4 3\n1 2\n",
"7\n1 2\n2 3\n1 4\n1 5\n3 6\n3 7\n",
"3\n1 3\n2 3\n",
"60\n26 6\n59 30\n31 12\n31 3\n38 23\n59 29\n53 9\n38 56\n53 54\n29 21\n17 55\n59 38\n26 16\n24 59\n24 25\n17 35\n24 41\n30 15\n31 27\n8 44\n26 5\n26 48\n8 32\n53 17\n3 34\n3 51\n30 28\n47 10\n53 60\n36 42\n24 53\n59 22\n53 40\n26 52\n36 4\n59 8\n29 37\n36 20\n17 47\n53 18\n3 50\n30 2\n17 7\n8 58\n59 1\n31 11\n24 26\n24 43\n53 57\n59 45\n47 13\n26 46\n17 33\n30 31\n26 39\n26 19\n24 36\n8 49\n38 14\n",
"7\n1 2\n2 3\n3 4\n3 5\n1 6\n1 7\n",
"20\n19 16\n19 18\n20 7\n9 4\n6 17\n14 2\n9 15\n2 13\n5 11\n19 12\n12 20\n16 9\n11 8\n19 5\n3 1\n19 14\n5 3\n12 10\n19 6\n",
"7\n1 2\n1 3\n2 4\n2 5\n3 6\n3 7\n",
"10\n9 5\n7 1\n9 10\n7 2\n5 4\n9 6\n2 9\n10 8\n1 3\n",
"4\n2 4\n2 3\n2 1\n",
"4\n1 4\n3 2\n1 3\n",
"3\n1 2\n1 3\n",
"5\n1 2\n1 5\n1 3\n1 4\n",
"20\n14 9\n12 13\n10 15\n2 1\n20 19\n16 6\n16 3\n17 14\n3 5\n2 11\n3 10\n15 8\n14 2\n6 4\n3 20\n5 18\n1 7\n1 16\n4 12\n",
"20\n7 5\n14 13\n17 6\n3 8\n16 12\n18 9\n3 18\n14 1\n17 3\n15 2\n17 4\n9 11\n2 7\n15 17\n3 20\n16 10\n17 14\n2 16\n1 19\n",
"8\n1 2\n2 3\n3 4\n1 7\n1 8\n4 5\n4 6\n",
"5\n5 1\n5 2\n5 3\n5 4\n",
"50\n49 6\n43 7\n1 27\n19 35\n15 37\n16 12\n19 21\n16 28\n49 9\n48 39\n13 1\n2 48\n9 50\n44 3\n41 32\n48 31\n49 33\n6 11\n13 20\n49 22\n13 41\n48 29\n13 46\n15 47\n34 2\n49 13\n48 14\n34 24\n16 36\n13 40\n49 34\n49 17\n43 25\n11 23\n10 15\n19 26\n34 44\n16 42\n19 18\n46 8\n29 38\n1 45\n12 43\n13 16\n46 30\n15 5\n49 10\n11 19\n32 4\n",
"20\n13 1\n18 2\n3 7\n18 5\n20 16\n3 12\n18 9\n3 10\n18 11\n13 6\n3 18\n20 15\n20 17\n3 13\n3 4\n13 14\n3 20\n18 8\n3 19\n",
"10\n8 2\n5 6\n1 8\n2 9\n1 4\n8 10\n10 5\n2 7\n2 3\n"
],
"output": [
"NO",
"YES",
"NO",
"NO",
"NO",
"YES",
"NO",
"NO",
"NO",
"NO",
"YES",
"NO",
"NO",
"YES",
"NO",
"NO",
"NO",
"YES",
"NO",
"YES",
"NO"
]
}
|
IN-CORRECT
|
java
|
import java.util.Scanner;
public class TaskD {
public static void main(String[] args) {
Scanner s = new Scanner(System.in);
int n = s.nextInt();
int[] deg = new int[n];
for(int i=0;i<n-1;i++) {
int a = s.nextInt()-1;
int b = s.nextInt()-1;
deg[a]++;
deg[b]++;
}
for(int i=0;i<n;i++) {
if(deg[i] <= 2) {
System.out.println("NO");
return;
}
}
System.out.println("YES");
}
}
|
1189_D1. Add on a Tree
|
Note that this is the first problem of the two similar problems. You can hack this problem only if you solve both problems.
You are given a tree with n nodes. In the beginning, 0 is written on all edges. In one operation, you can choose any 2 distinct leaves u, v and any real number x and add x to values written on all edges on the simple path between u and v.
For example, on the picture below you can see the result of applying two operations to the graph: adding 2 on the path from 7 to 6, and then adding -0.5 on the path from 4 to 5.
<image>
Is it true that for any configuration of real numbers written on edges, we can achieve it with a finite number of operations?
Leaf is a node of a tree of degree 1. Simple path is a path that doesn't contain any node twice.
Input
The first line contains a single integer n (2 ≤ n ≤ 10^5) — the number of nodes.
Each of the next n-1 lines contains two integers u and v (1 ≤ u, v ≤ n, u ≠ v), meaning that there is an edge between nodes u and v. It is guaranteed that these edges form a tree.
Output
If there is a configuration of real numbers written on edges of the tree that we can't achieve by performing the operations, output "NO".
Otherwise, output "YES".
You can print each letter in any case (upper or lower).
Examples
Input
2
1 2
Output
YES
Input
3
1 2
2 3
Output
NO
Input
5
1 2
1 3
1 4
2 5
Output
NO
Input
6
1 2
1 3
1 4
2 5
2 6
Output
YES
Note
In the first example, we can add any real x to the value written on the only edge (1, 2).
<image>
In the second example, one of configurations that we can't reach is 0 written on (1, 2) and 1 written on (2, 3).
<image>
Below you can see graphs from examples 3, 4:
<image> <image>
|
{
"input": [
"2\n1 2\n",
"3\n1 2\n2 3\n",
"5\n1 2\n1 3\n1 4\n2 5\n",
"6\n1 2\n1 3\n1 4\n2 5\n2 6\n"
],
"output": [
"YES",
"NO",
"NO",
"YES"
]
}
|
{
"input": [
"50\n16 4\n17 9\n31 19\n22 10\n8 1\n40 30\n3 31\n20 29\n47 27\n22 25\n32 34\n12 15\n40 32\n10 33\n47 12\n6 24\n46 41\n14 23\n12 35\n31 42\n46 28\n31 20\n46 37\n1 39\n29 49\n37 47\n40 6\n42 36\n47 2\n24 46\n2 13\n8 45\n41 3\n32 17\n4 7\n47 26\n28 8\n41 50\n34 44\n33 21\n25 5\n16 40\n3 14\n8 18\n28 11\n32 22\n2 38\n3 48\n44 43\n",
"10\n8 1\n1 2\n8 9\n8 5\n1 3\n1 10\n1 6\n1 7\n8 4\n",
"5\n5 1\n5 4\n4 3\n1 2\n",
"7\n1 2\n2 3\n1 4\n1 5\n3 6\n3 7\n",
"3\n1 3\n2 3\n",
"60\n26 6\n59 30\n31 12\n31 3\n38 23\n59 29\n53 9\n38 56\n53 54\n29 21\n17 55\n59 38\n26 16\n24 59\n24 25\n17 35\n24 41\n30 15\n31 27\n8 44\n26 5\n26 48\n8 32\n53 17\n3 34\n3 51\n30 28\n47 10\n53 60\n36 42\n24 53\n59 22\n53 40\n26 52\n36 4\n59 8\n29 37\n36 20\n17 47\n53 18\n3 50\n30 2\n17 7\n8 58\n59 1\n31 11\n24 26\n24 43\n53 57\n59 45\n47 13\n26 46\n17 33\n30 31\n26 39\n26 19\n24 36\n8 49\n38 14\n",
"7\n1 2\n2 3\n3 4\n3 5\n1 6\n1 7\n",
"20\n19 16\n19 18\n20 7\n9 4\n6 17\n14 2\n9 15\n2 13\n5 11\n19 12\n12 20\n16 9\n11 8\n19 5\n3 1\n19 14\n5 3\n12 10\n19 6\n",
"7\n1 2\n1 3\n2 4\n2 5\n3 6\n3 7\n",
"10\n9 5\n7 1\n9 10\n7 2\n5 4\n9 6\n2 9\n10 8\n1 3\n",
"4\n2 4\n2 3\n2 1\n",
"4\n1 4\n3 2\n1 3\n",
"3\n1 2\n1 3\n",
"5\n1 2\n1 5\n1 3\n1 4\n",
"20\n14 9\n12 13\n10 15\n2 1\n20 19\n16 6\n16 3\n17 14\n3 5\n2 11\n3 10\n15 8\n14 2\n6 4\n3 20\n5 18\n1 7\n1 16\n4 12\n",
"20\n7 5\n14 13\n17 6\n3 8\n16 12\n18 9\n3 18\n14 1\n17 3\n15 2\n17 4\n9 11\n2 7\n15 17\n3 20\n16 10\n17 14\n2 16\n1 19\n",
"8\n1 2\n2 3\n3 4\n1 7\n1 8\n4 5\n4 6\n",
"5\n5 1\n5 2\n5 3\n5 4\n",
"50\n49 6\n43 7\n1 27\n19 35\n15 37\n16 12\n19 21\n16 28\n49 9\n48 39\n13 1\n2 48\n9 50\n44 3\n41 32\n48 31\n49 33\n6 11\n13 20\n49 22\n13 41\n48 29\n13 46\n15 47\n34 2\n49 13\n48 14\n34 24\n16 36\n13 40\n49 34\n49 17\n43 25\n11 23\n10 15\n19 26\n34 44\n16 42\n19 18\n46 8\n29 38\n1 45\n12 43\n13 16\n46 30\n15 5\n49 10\n11 19\n32 4\n",
"20\n13 1\n18 2\n3 7\n18 5\n20 16\n3 12\n18 9\n3 10\n18 11\n13 6\n3 18\n20 15\n20 17\n3 13\n3 4\n13 14\n3 20\n18 8\n3 19\n",
"10\n8 2\n5 6\n1 8\n2 9\n1 4\n8 10\n10 5\n2 7\n2 3\n"
],
"output": [
"NO",
"YES",
"NO",
"NO",
"NO",
"YES",
"NO",
"NO",
"NO",
"NO",
"YES",
"NO",
"NO",
"YES",
"NO",
"NO",
"NO",
"YES",
"NO",
"YES",
"NO"
]
}
|
IN-CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
void pr_init() {
ios_base::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
}
vector<vector<long long> > tree;
vector<long long> deg;
bool dfs(long long u, long long p) {
if (deg[u] == 1) return true;
if (deg[u] < 3) return false;
bool fur = true;
long long cnt = 0;
for (auto x : tree[u])
if (x != p) {
fur = fur && dfs(x, u);
if (deg[x] == 1) cnt++;
}
if (cnt < 2) return false;
return fur;
}
void solve() {
long long n;
cin >> n;
tree.assign(n + 1, vector<long long>());
deg.assign(n + 1, 0);
for (long long i = 0; i < n - 1; i++) {
long long u, v;
cin >> u >> v;
deg[u]++;
deg[v]++;
tree[u].emplace_back(v);
tree[v].emplace_back(u);
}
if (n == 2) {
cout << "YES\n";
return;
} else if (n == 3) {
cout << "NO\n";
return;
}
if (dfs(1, -1))
cout << "YES\n";
else
cout << "NO\n";
}
int32_t main() {
solve();
return 0;
}
|
1189_D1. Add on a Tree
|
Note that this is the first problem of the two similar problems. You can hack this problem only if you solve both problems.
You are given a tree with n nodes. In the beginning, 0 is written on all edges. In one operation, you can choose any 2 distinct leaves u, v and any real number x and add x to values written on all edges on the simple path between u and v.
For example, on the picture below you can see the result of applying two operations to the graph: adding 2 on the path from 7 to 6, and then adding -0.5 on the path from 4 to 5.
<image>
Is it true that for any configuration of real numbers written on edges, we can achieve it with a finite number of operations?
Leaf is a node of a tree of degree 1. Simple path is a path that doesn't contain any node twice.
Input
The first line contains a single integer n (2 ≤ n ≤ 10^5) — the number of nodes.
Each of the next n-1 lines contains two integers u and v (1 ≤ u, v ≤ n, u ≠ v), meaning that there is an edge between nodes u and v. It is guaranteed that these edges form a tree.
Output
If there is a configuration of real numbers written on edges of the tree that we can't achieve by performing the operations, output "NO".
Otherwise, output "YES".
You can print each letter in any case (upper or lower).
Examples
Input
2
1 2
Output
YES
Input
3
1 2
2 3
Output
NO
Input
5
1 2
1 3
1 4
2 5
Output
NO
Input
6
1 2
1 3
1 4
2 5
2 6
Output
YES
Note
In the first example, we can add any real x to the value written on the only edge (1, 2).
<image>
In the second example, one of configurations that we can't reach is 0 written on (1, 2) and 1 written on (2, 3).
<image>
Below you can see graphs from examples 3, 4:
<image> <image>
|
{
"input": [
"2\n1 2\n",
"3\n1 2\n2 3\n",
"5\n1 2\n1 3\n1 4\n2 5\n",
"6\n1 2\n1 3\n1 4\n2 5\n2 6\n"
],
"output": [
"YES",
"NO",
"NO",
"YES"
]
}
|
{
"input": [
"50\n16 4\n17 9\n31 19\n22 10\n8 1\n40 30\n3 31\n20 29\n47 27\n22 25\n32 34\n12 15\n40 32\n10 33\n47 12\n6 24\n46 41\n14 23\n12 35\n31 42\n46 28\n31 20\n46 37\n1 39\n29 49\n37 47\n40 6\n42 36\n47 2\n24 46\n2 13\n8 45\n41 3\n32 17\n4 7\n47 26\n28 8\n41 50\n34 44\n33 21\n25 5\n16 40\n3 14\n8 18\n28 11\n32 22\n2 38\n3 48\n44 43\n",
"10\n8 1\n1 2\n8 9\n8 5\n1 3\n1 10\n1 6\n1 7\n8 4\n",
"5\n5 1\n5 4\n4 3\n1 2\n",
"7\n1 2\n2 3\n1 4\n1 5\n3 6\n3 7\n",
"3\n1 3\n2 3\n",
"60\n26 6\n59 30\n31 12\n31 3\n38 23\n59 29\n53 9\n38 56\n53 54\n29 21\n17 55\n59 38\n26 16\n24 59\n24 25\n17 35\n24 41\n30 15\n31 27\n8 44\n26 5\n26 48\n8 32\n53 17\n3 34\n3 51\n30 28\n47 10\n53 60\n36 42\n24 53\n59 22\n53 40\n26 52\n36 4\n59 8\n29 37\n36 20\n17 47\n53 18\n3 50\n30 2\n17 7\n8 58\n59 1\n31 11\n24 26\n24 43\n53 57\n59 45\n47 13\n26 46\n17 33\n30 31\n26 39\n26 19\n24 36\n8 49\n38 14\n",
"7\n1 2\n2 3\n3 4\n3 5\n1 6\n1 7\n",
"20\n19 16\n19 18\n20 7\n9 4\n6 17\n14 2\n9 15\n2 13\n5 11\n19 12\n12 20\n16 9\n11 8\n19 5\n3 1\n19 14\n5 3\n12 10\n19 6\n",
"7\n1 2\n1 3\n2 4\n2 5\n3 6\n3 7\n",
"10\n9 5\n7 1\n9 10\n7 2\n5 4\n9 6\n2 9\n10 8\n1 3\n",
"4\n2 4\n2 3\n2 1\n",
"4\n1 4\n3 2\n1 3\n",
"3\n1 2\n1 3\n",
"5\n1 2\n1 5\n1 3\n1 4\n",
"20\n14 9\n12 13\n10 15\n2 1\n20 19\n16 6\n16 3\n17 14\n3 5\n2 11\n3 10\n15 8\n14 2\n6 4\n3 20\n5 18\n1 7\n1 16\n4 12\n",
"20\n7 5\n14 13\n17 6\n3 8\n16 12\n18 9\n3 18\n14 1\n17 3\n15 2\n17 4\n9 11\n2 7\n15 17\n3 20\n16 10\n17 14\n2 16\n1 19\n",
"8\n1 2\n2 3\n3 4\n1 7\n1 8\n4 5\n4 6\n",
"5\n5 1\n5 2\n5 3\n5 4\n",
"50\n49 6\n43 7\n1 27\n19 35\n15 37\n16 12\n19 21\n16 28\n49 9\n48 39\n13 1\n2 48\n9 50\n44 3\n41 32\n48 31\n49 33\n6 11\n13 20\n49 22\n13 41\n48 29\n13 46\n15 47\n34 2\n49 13\n48 14\n34 24\n16 36\n13 40\n49 34\n49 17\n43 25\n11 23\n10 15\n19 26\n34 44\n16 42\n19 18\n46 8\n29 38\n1 45\n12 43\n13 16\n46 30\n15 5\n49 10\n11 19\n32 4\n",
"20\n13 1\n18 2\n3 7\n18 5\n20 16\n3 12\n18 9\n3 10\n18 11\n13 6\n3 18\n20 15\n20 17\n3 13\n3 4\n13 14\n3 20\n18 8\n3 19\n",
"10\n8 2\n5 6\n1 8\n2 9\n1 4\n8 10\n10 5\n2 7\n2 3\n"
],
"output": [
"NO",
"YES",
"NO",
"NO",
"NO",
"YES",
"NO",
"NO",
"NO",
"NO",
"YES",
"NO",
"NO",
"YES",
"NO",
"NO",
"NO",
"YES",
"NO",
"YES",
"NO"
]
}
|
IN-CORRECT
|
java
|
import java.util.Scanner;
public class Code {
public static int[] arr = new int[200005], sum = new int[100005];
public static void main(String[] args) {
Scanner scanner = new Scanner(System.in);
int n = scanner.nextInt();
for(int i=0;i<=n;i++)arr[i]=0;
for(int i=1;i<n;i++) {
int a = scanner.nextInt(), b = scanner.nextInt();
arr[a]++;
arr[b]++;
}
for(int i=1;i<=n;i++) {
if(arr[i]%2==0) {
System.out.printf("NO");
return;
}
}
System.out.printf("YES");
}
}
|
1189_D1. Add on a Tree
|
Note that this is the first problem of the two similar problems. You can hack this problem only if you solve both problems.
You are given a tree with n nodes. In the beginning, 0 is written on all edges. In one operation, you can choose any 2 distinct leaves u, v and any real number x and add x to values written on all edges on the simple path between u and v.
For example, on the picture below you can see the result of applying two operations to the graph: adding 2 on the path from 7 to 6, and then adding -0.5 on the path from 4 to 5.
<image>
Is it true that for any configuration of real numbers written on edges, we can achieve it with a finite number of operations?
Leaf is a node of a tree of degree 1. Simple path is a path that doesn't contain any node twice.
Input
The first line contains a single integer n (2 ≤ n ≤ 10^5) — the number of nodes.
Each of the next n-1 lines contains two integers u and v (1 ≤ u, v ≤ n, u ≠ v), meaning that there is an edge between nodes u and v. It is guaranteed that these edges form a tree.
Output
If there is a configuration of real numbers written on edges of the tree that we can't achieve by performing the operations, output "NO".
Otherwise, output "YES".
You can print each letter in any case (upper or lower).
Examples
Input
2
1 2
Output
YES
Input
3
1 2
2 3
Output
NO
Input
5
1 2
1 3
1 4
2 5
Output
NO
Input
6
1 2
1 3
1 4
2 5
2 6
Output
YES
Note
In the first example, we can add any real x to the value written on the only edge (1, 2).
<image>
In the second example, one of configurations that we can't reach is 0 written on (1, 2) and 1 written on (2, 3).
<image>
Below you can see graphs from examples 3, 4:
<image> <image>
|
{
"input": [
"2\n1 2\n",
"3\n1 2\n2 3\n",
"5\n1 2\n1 3\n1 4\n2 5\n",
"6\n1 2\n1 3\n1 4\n2 5\n2 6\n"
],
"output": [
"YES",
"NO",
"NO",
"YES"
]
}
|
{
"input": [
"50\n16 4\n17 9\n31 19\n22 10\n8 1\n40 30\n3 31\n20 29\n47 27\n22 25\n32 34\n12 15\n40 32\n10 33\n47 12\n6 24\n46 41\n14 23\n12 35\n31 42\n46 28\n31 20\n46 37\n1 39\n29 49\n37 47\n40 6\n42 36\n47 2\n24 46\n2 13\n8 45\n41 3\n32 17\n4 7\n47 26\n28 8\n41 50\n34 44\n33 21\n25 5\n16 40\n3 14\n8 18\n28 11\n32 22\n2 38\n3 48\n44 43\n",
"10\n8 1\n1 2\n8 9\n8 5\n1 3\n1 10\n1 6\n1 7\n8 4\n",
"5\n5 1\n5 4\n4 3\n1 2\n",
"7\n1 2\n2 3\n1 4\n1 5\n3 6\n3 7\n",
"3\n1 3\n2 3\n",
"60\n26 6\n59 30\n31 12\n31 3\n38 23\n59 29\n53 9\n38 56\n53 54\n29 21\n17 55\n59 38\n26 16\n24 59\n24 25\n17 35\n24 41\n30 15\n31 27\n8 44\n26 5\n26 48\n8 32\n53 17\n3 34\n3 51\n30 28\n47 10\n53 60\n36 42\n24 53\n59 22\n53 40\n26 52\n36 4\n59 8\n29 37\n36 20\n17 47\n53 18\n3 50\n30 2\n17 7\n8 58\n59 1\n31 11\n24 26\n24 43\n53 57\n59 45\n47 13\n26 46\n17 33\n30 31\n26 39\n26 19\n24 36\n8 49\n38 14\n",
"7\n1 2\n2 3\n3 4\n3 5\n1 6\n1 7\n",
"20\n19 16\n19 18\n20 7\n9 4\n6 17\n14 2\n9 15\n2 13\n5 11\n19 12\n12 20\n16 9\n11 8\n19 5\n3 1\n19 14\n5 3\n12 10\n19 6\n",
"7\n1 2\n1 3\n2 4\n2 5\n3 6\n3 7\n",
"10\n9 5\n7 1\n9 10\n7 2\n5 4\n9 6\n2 9\n10 8\n1 3\n",
"4\n2 4\n2 3\n2 1\n",
"4\n1 4\n3 2\n1 3\n",
"3\n1 2\n1 3\n",
"5\n1 2\n1 5\n1 3\n1 4\n",
"20\n14 9\n12 13\n10 15\n2 1\n20 19\n16 6\n16 3\n17 14\n3 5\n2 11\n3 10\n15 8\n14 2\n6 4\n3 20\n5 18\n1 7\n1 16\n4 12\n",
"20\n7 5\n14 13\n17 6\n3 8\n16 12\n18 9\n3 18\n14 1\n17 3\n15 2\n17 4\n9 11\n2 7\n15 17\n3 20\n16 10\n17 14\n2 16\n1 19\n",
"8\n1 2\n2 3\n3 4\n1 7\n1 8\n4 5\n4 6\n",
"5\n5 1\n5 2\n5 3\n5 4\n",
"50\n49 6\n43 7\n1 27\n19 35\n15 37\n16 12\n19 21\n16 28\n49 9\n48 39\n13 1\n2 48\n9 50\n44 3\n41 32\n48 31\n49 33\n6 11\n13 20\n49 22\n13 41\n48 29\n13 46\n15 47\n34 2\n49 13\n48 14\n34 24\n16 36\n13 40\n49 34\n49 17\n43 25\n11 23\n10 15\n19 26\n34 44\n16 42\n19 18\n46 8\n29 38\n1 45\n12 43\n13 16\n46 30\n15 5\n49 10\n11 19\n32 4\n",
"20\n13 1\n18 2\n3 7\n18 5\n20 16\n3 12\n18 9\n3 10\n18 11\n13 6\n3 18\n20 15\n20 17\n3 13\n3 4\n13 14\n3 20\n18 8\n3 19\n",
"10\n8 2\n5 6\n1 8\n2 9\n1 4\n8 10\n10 5\n2 7\n2 3\n"
],
"output": [
"NO",
"YES",
"NO",
"NO",
"NO",
"YES",
"NO",
"NO",
"NO",
"NO",
"YES",
"NO",
"NO",
"YES",
"NO",
"NO",
"NO",
"YES",
"NO",
"YES",
"NO"
]
}
|
IN-CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
const int N = 500010;
int n, m, d[N];
inline int read() {
int sym = 0, res = 0;
char ch = 0;
while (!isdigit(ch)) sym |= (ch == '-'), ch = getchar();
while (isdigit(ch)) res = (res << 3) + (res << 1) + (ch ^ 48), ch = getchar();
return sym ? -res : res;
}
void file() {
freopen("read.in", "r", stdin);
freopen("write.out", "w", stdout);
}
int main() {
n = read();
for (int i = 1; i <= n - 1; i++) {
int x = read(), y = read();
d[x]++, d[y]++;
}
for (int i = 1; i <= n; i++) {
if (d[i] % 2 == 0) {
printf("NO");
return 0;
}
}
printf("YES");
return 0;
}
|
1189_D1. Add on a Tree
|
Note that this is the first problem of the two similar problems. You can hack this problem only if you solve both problems.
You are given a tree with n nodes. In the beginning, 0 is written on all edges. In one operation, you can choose any 2 distinct leaves u, v and any real number x and add x to values written on all edges on the simple path between u and v.
For example, on the picture below you can see the result of applying two operations to the graph: adding 2 on the path from 7 to 6, and then adding -0.5 on the path from 4 to 5.
<image>
Is it true that for any configuration of real numbers written on edges, we can achieve it with a finite number of operations?
Leaf is a node of a tree of degree 1. Simple path is a path that doesn't contain any node twice.
Input
The first line contains a single integer n (2 ≤ n ≤ 10^5) — the number of nodes.
Each of the next n-1 lines contains two integers u and v (1 ≤ u, v ≤ n, u ≠ v), meaning that there is an edge between nodes u and v. It is guaranteed that these edges form a tree.
Output
If there is a configuration of real numbers written on edges of the tree that we can't achieve by performing the operations, output "NO".
Otherwise, output "YES".
You can print each letter in any case (upper or lower).
Examples
Input
2
1 2
Output
YES
Input
3
1 2
2 3
Output
NO
Input
5
1 2
1 3
1 4
2 5
Output
NO
Input
6
1 2
1 3
1 4
2 5
2 6
Output
YES
Note
In the first example, we can add any real x to the value written on the only edge (1, 2).
<image>
In the second example, one of configurations that we can't reach is 0 written on (1, 2) and 1 written on (2, 3).
<image>
Below you can see graphs from examples 3, 4:
<image> <image>
|
{
"input": [
"2\n1 2\n",
"3\n1 2\n2 3\n",
"5\n1 2\n1 3\n1 4\n2 5\n",
"6\n1 2\n1 3\n1 4\n2 5\n2 6\n"
],
"output": [
"YES",
"NO",
"NO",
"YES"
]
}
|
{
"input": [
"50\n16 4\n17 9\n31 19\n22 10\n8 1\n40 30\n3 31\n20 29\n47 27\n22 25\n32 34\n12 15\n40 32\n10 33\n47 12\n6 24\n46 41\n14 23\n12 35\n31 42\n46 28\n31 20\n46 37\n1 39\n29 49\n37 47\n40 6\n42 36\n47 2\n24 46\n2 13\n8 45\n41 3\n32 17\n4 7\n47 26\n28 8\n41 50\n34 44\n33 21\n25 5\n16 40\n3 14\n8 18\n28 11\n32 22\n2 38\n3 48\n44 43\n",
"10\n8 1\n1 2\n8 9\n8 5\n1 3\n1 10\n1 6\n1 7\n8 4\n",
"5\n5 1\n5 4\n4 3\n1 2\n",
"7\n1 2\n2 3\n1 4\n1 5\n3 6\n3 7\n",
"3\n1 3\n2 3\n",
"60\n26 6\n59 30\n31 12\n31 3\n38 23\n59 29\n53 9\n38 56\n53 54\n29 21\n17 55\n59 38\n26 16\n24 59\n24 25\n17 35\n24 41\n30 15\n31 27\n8 44\n26 5\n26 48\n8 32\n53 17\n3 34\n3 51\n30 28\n47 10\n53 60\n36 42\n24 53\n59 22\n53 40\n26 52\n36 4\n59 8\n29 37\n36 20\n17 47\n53 18\n3 50\n30 2\n17 7\n8 58\n59 1\n31 11\n24 26\n24 43\n53 57\n59 45\n47 13\n26 46\n17 33\n30 31\n26 39\n26 19\n24 36\n8 49\n38 14\n",
"7\n1 2\n2 3\n3 4\n3 5\n1 6\n1 7\n",
"20\n19 16\n19 18\n20 7\n9 4\n6 17\n14 2\n9 15\n2 13\n5 11\n19 12\n12 20\n16 9\n11 8\n19 5\n3 1\n19 14\n5 3\n12 10\n19 6\n",
"7\n1 2\n1 3\n2 4\n2 5\n3 6\n3 7\n",
"10\n9 5\n7 1\n9 10\n7 2\n5 4\n9 6\n2 9\n10 8\n1 3\n",
"4\n2 4\n2 3\n2 1\n",
"4\n1 4\n3 2\n1 3\n",
"3\n1 2\n1 3\n",
"5\n1 2\n1 5\n1 3\n1 4\n",
"20\n14 9\n12 13\n10 15\n2 1\n20 19\n16 6\n16 3\n17 14\n3 5\n2 11\n3 10\n15 8\n14 2\n6 4\n3 20\n5 18\n1 7\n1 16\n4 12\n",
"20\n7 5\n14 13\n17 6\n3 8\n16 12\n18 9\n3 18\n14 1\n17 3\n15 2\n17 4\n9 11\n2 7\n15 17\n3 20\n16 10\n17 14\n2 16\n1 19\n",
"8\n1 2\n2 3\n3 4\n1 7\n1 8\n4 5\n4 6\n",
"5\n5 1\n5 2\n5 3\n5 4\n",
"50\n49 6\n43 7\n1 27\n19 35\n15 37\n16 12\n19 21\n16 28\n49 9\n48 39\n13 1\n2 48\n9 50\n44 3\n41 32\n48 31\n49 33\n6 11\n13 20\n49 22\n13 41\n48 29\n13 46\n15 47\n34 2\n49 13\n48 14\n34 24\n16 36\n13 40\n49 34\n49 17\n43 25\n11 23\n10 15\n19 26\n34 44\n16 42\n19 18\n46 8\n29 38\n1 45\n12 43\n13 16\n46 30\n15 5\n49 10\n11 19\n32 4\n",
"20\n13 1\n18 2\n3 7\n18 5\n20 16\n3 12\n18 9\n3 10\n18 11\n13 6\n3 18\n20 15\n20 17\n3 13\n3 4\n13 14\n3 20\n18 8\n3 19\n",
"10\n8 2\n5 6\n1 8\n2 9\n1 4\n8 10\n10 5\n2 7\n2 3\n"
],
"output": [
"NO",
"YES",
"NO",
"NO",
"NO",
"YES",
"NO",
"NO",
"NO",
"NO",
"YES",
"NO",
"NO",
"YES",
"NO",
"NO",
"NO",
"YES",
"NO",
"YES",
"NO"
]
}
|
IN-CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
const long long maxn = 1e5 + 10;
struct node {
long long to, nxt;
} e[maxn << 1];
long long n, tot, head[maxn], deg[maxn], f[maxn];
long long read() {
long long x = 0, f = 1;
char ch = getchar();
while (ch < '0' || ch > '9') {
if (ch == '-') f = -1;
ch = getchar();
}
while (ch >= '0' && ch <= '9') {
x = x * 10 + ch - '0';
ch = getchar();
}
return x * f;
}
void print(long long x) {
if (x < 0) putchar('-'), x = -x;
if (x > 9) print(x / 10);
putchar(x % 10 + '0');
}
void write(long long x) {
print(x);
puts("");
}
void add(long long u, long long v) {
e[++tot].to = v;
e[tot].nxt = head[u];
head[u] = tot;
deg[v]++;
}
void insert(long long u, long long v) {
add(u, v);
add(v, u);
}
signed main() {
n = read();
bool flag = 0;
for (long long i = 2; i <= n; i++) insert(read(), read());
for (long long i = 1; i <= n; i++)
if (deg[i] == 2) {
puts("NO");
flag = 1;
}
if (!flag) puts("YES");
return 0;
}
|
1189_D1. Add on a Tree
|
Note that this is the first problem of the two similar problems. You can hack this problem only if you solve both problems.
You are given a tree with n nodes. In the beginning, 0 is written on all edges. In one operation, you can choose any 2 distinct leaves u, v and any real number x and add x to values written on all edges on the simple path between u and v.
For example, on the picture below you can see the result of applying two operations to the graph: adding 2 on the path from 7 to 6, and then adding -0.5 on the path from 4 to 5.
<image>
Is it true that for any configuration of real numbers written on edges, we can achieve it with a finite number of operations?
Leaf is a node of a tree of degree 1. Simple path is a path that doesn't contain any node twice.
Input
The first line contains a single integer n (2 ≤ n ≤ 10^5) — the number of nodes.
Each of the next n-1 lines contains two integers u and v (1 ≤ u, v ≤ n, u ≠ v), meaning that there is an edge between nodes u and v. It is guaranteed that these edges form a tree.
Output
If there is a configuration of real numbers written on edges of the tree that we can't achieve by performing the operations, output "NO".
Otherwise, output "YES".
You can print each letter in any case (upper or lower).
Examples
Input
2
1 2
Output
YES
Input
3
1 2
2 3
Output
NO
Input
5
1 2
1 3
1 4
2 5
Output
NO
Input
6
1 2
1 3
1 4
2 5
2 6
Output
YES
Note
In the first example, we can add any real x to the value written on the only edge (1, 2).
<image>
In the second example, one of configurations that we can't reach is 0 written on (1, 2) and 1 written on (2, 3).
<image>
Below you can see graphs from examples 3, 4:
<image> <image>
|
{
"input": [
"2\n1 2\n",
"3\n1 2\n2 3\n",
"5\n1 2\n1 3\n1 4\n2 5\n",
"6\n1 2\n1 3\n1 4\n2 5\n2 6\n"
],
"output": [
"YES",
"NO",
"NO",
"YES"
]
}
|
{
"input": [
"50\n16 4\n17 9\n31 19\n22 10\n8 1\n40 30\n3 31\n20 29\n47 27\n22 25\n32 34\n12 15\n40 32\n10 33\n47 12\n6 24\n46 41\n14 23\n12 35\n31 42\n46 28\n31 20\n46 37\n1 39\n29 49\n37 47\n40 6\n42 36\n47 2\n24 46\n2 13\n8 45\n41 3\n32 17\n4 7\n47 26\n28 8\n41 50\n34 44\n33 21\n25 5\n16 40\n3 14\n8 18\n28 11\n32 22\n2 38\n3 48\n44 43\n",
"10\n8 1\n1 2\n8 9\n8 5\n1 3\n1 10\n1 6\n1 7\n8 4\n",
"5\n5 1\n5 4\n4 3\n1 2\n",
"7\n1 2\n2 3\n1 4\n1 5\n3 6\n3 7\n",
"3\n1 3\n2 3\n",
"60\n26 6\n59 30\n31 12\n31 3\n38 23\n59 29\n53 9\n38 56\n53 54\n29 21\n17 55\n59 38\n26 16\n24 59\n24 25\n17 35\n24 41\n30 15\n31 27\n8 44\n26 5\n26 48\n8 32\n53 17\n3 34\n3 51\n30 28\n47 10\n53 60\n36 42\n24 53\n59 22\n53 40\n26 52\n36 4\n59 8\n29 37\n36 20\n17 47\n53 18\n3 50\n30 2\n17 7\n8 58\n59 1\n31 11\n24 26\n24 43\n53 57\n59 45\n47 13\n26 46\n17 33\n30 31\n26 39\n26 19\n24 36\n8 49\n38 14\n",
"7\n1 2\n2 3\n3 4\n3 5\n1 6\n1 7\n",
"20\n19 16\n19 18\n20 7\n9 4\n6 17\n14 2\n9 15\n2 13\n5 11\n19 12\n12 20\n16 9\n11 8\n19 5\n3 1\n19 14\n5 3\n12 10\n19 6\n",
"7\n1 2\n1 3\n2 4\n2 5\n3 6\n3 7\n",
"10\n9 5\n7 1\n9 10\n7 2\n5 4\n9 6\n2 9\n10 8\n1 3\n",
"4\n2 4\n2 3\n2 1\n",
"4\n1 4\n3 2\n1 3\n",
"3\n1 2\n1 3\n",
"5\n1 2\n1 5\n1 3\n1 4\n",
"20\n14 9\n12 13\n10 15\n2 1\n20 19\n16 6\n16 3\n17 14\n3 5\n2 11\n3 10\n15 8\n14 2\n6 4\n3 20\n5 18\n1 7\n1 16\n4 12\n",
"20\n7 5\n14 13\n17 6\n3 8\n16 12\n18 9\n3 18\n14 1\n17 3\n15 2\n17 4\n9 11\n2 7\n15 17\n3 20\n16 10\n17 14\n2 16\n1 19\n",
"8\n1 2\n2 3\n3 4\n1 7\n1 8\n4 5\n4 6\n",
"5\n5 1\n5 2\n5 3\n5 4\n",
"50\n49 6\n43 7\n1 27\n19 35\n15 37\n16 12\n19 21\n16 28\n49 9\n48 39\n13 1\n2 48\n9 50\n44 3\n41 32\n48 31\n49 33\n6 11\n13 20\n49 22\n13 41\n48 29\n13 46\n15 47\n34 2\n49 13\n48 14\n34 24\n16 36\n13 40\n49 34\n49 17\n43 25\n11 23\n10 15\n19 26\n34 44\n16 42\n19 18\n46 8\n29 38\n1 45\n12 43\n13 16\n46 30\n15 5\n49 10\n11 19\n32 4\n",
"20\n13 1\n18 2\n3 7\n18 5\n20 16\n3 12\n18 9\n3 10\n18 11\n13 6\n3 18\n20 15\n20 17\n3 13\n3 4\n13 14\n3 20\n18 8\n3 19\n",
"10\n8 2\n5 6\n1 8\n2 9\n1 4\n8 10\n10 5\n2 7\n2 3\n"
],
"output": [
"NO",
"YES",
"NO",
"NO",
"NO",
"YES",
"NO",
"NO",
"NO",
"NO",
"YES",
"NO",
"NO",
"YES",
"NO",
"NO",
"NO",
"YES",
"NO",
"YES",
"NO"
]
}
|
IN-CORRECT
|
cpp
|
#include <bits/stdc++.h>
long long mod = 998244353;
using namespace std;
const long long N = 1e5 + 5;
vector<long long> g[N];
long long n;
void fail() {
cout << "NO\n";
exit(0);
}
long long cnt;
void dfs(long long node, long long p = 0) {
if (g[node].size() == 1) cnt++;
for (long long nx : g[node]) {
if (nx == p) continue;
dfs(nx, node);
}
}
int32_t main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
cin >> n;
for (long long i = 1, u, v; i < n; i++) {
cin >> u >> v;
g[u].push_back(v);
g[v].push_back(u);
}
dfs(1);
if (cnt >= 4)
cout << "YES";
else
cout << "NO";
return 0;
}
|
1189_D1. Add on a Tree
|
Note that this is the first problem of the two similar problems. You can hack this problem only if you solve both problems.
You are given a tree with n nodes. In the beginning, 0 is written on all edges. In one operation, you can choose any 2 distinct leaves u, v and any real number x and add x to values written on all edges on the simple path between u and v.
For example, on the picture below you can see the result of applying two operations to the graph: adding 2 on the path from 7 to 6, and then adding -0.5 on the path from 4 to 5.
<image>
Is it true that for any configuration of real numbers written on edges, we can achieve it with a finite number of operations?
Leaf is a node of a tree of degree 1. Simple path is a path that doesn't contain any node twice.
Input
The first line contains a single integer n (2 ≤ n ≤ 10^5) — the number of nodes.
Each of the next n-1 lines contains two integers u and v (1 ≤ u, v ≤ n, u ≠ v), meaning that there is an edge between nodes u and v. It is guaranteed that these edges form a tree.
Output
If there is a configuration of real numbers written on edges of the tree that we can't achieve by performing the operations, output "NO".
Otherwise, output "YES".
You can print each letter in any case (upper or lower).
Examples
Input
2
1 2
Output
YES
Input
3
1 2
2 3
Output
NO
Input
5
1 2
1 3
1 4
2 5
Output
NO
Input
6
1 2
1 3
1 4
2 5
2 6
Output
YES
Note
In the first example, we can add any real x to the value written on the only edge (1, 2).
<image>
In the second example, one of configurations that we can't reach is 0 written on (1, 2) and 1 written on (2, 3).
<image>
Below you can see graphs from examples 3, 4:
<image> <image>
|
{
"input": [
"2\n1 2\n",
"3\n1 2\n2 3\n",
"5\n1 2\n1 3\n1 4\n2 5\n",
"6\n1 2\n1 3\n1 4\n2 5\n2 6\n"
],
"output": [
"YES",
"NO",
"NO",
"YES"
]
}
|
{
"input": [
"50\n16 4\n17 9\n31 19\n22 10\n8 1\n40 30\n3 31\n20 29\n47 27\n22 25\n32 34\n12 15\n40 32\n10 33\n47 12\n6 24\n46 41\n14 23\n12 35\n31 42\n46 28\n31 20\n46 37\n1 39\n29 49\n37 47\n40 6\n42 36\n47 2\n24 46\n2 13\n8 45\n41 3\n32 17\n4 7\n47 26\n28 8\n41 50\n34 44\n33 21\n25 5\n16 40\n3 14\n8 18\n28 11\n32 22\n2 38\n3 48\n44 43\n",
"10\n8 1\n1 2\n8 9\n8 5\n1 3\n1 10\n1 6\n1 7\n8 4\n",
"5\n5 1\n5 4\n4 3\n1 2\n",
"7\n1 2\n2 3\n1 4\n1 5\n3 6\n3 7\n",
"3\n1 3\n2 3\n",
"60\n26 6\n59 30\n31 12\n31 3\n38 23\n59 29\n53 9\n38 56\n53 54\n29 21\n17 55\n59 38\n26 16\n24 59\n24 25\n17 35\n24 41\n30 15\n31 27\n8 44\n26 5\n26 48\n8 32\n53 17\n3 34\n3 51\n30 28\n47 10\n53 60\n36 42\n24 53\n59 22\n53 40\n26 52\n36 4\n59 8\n29 37\n36 20\n17 47\n53 18\n3 50\n30 2\n17 7\n8 58\n59 1\n31 11\n24 26\n24 43\n53 57\n59 45\n47 13\n26 46\n17 33\n30 31\n26 39\n26 19\n24 36\n8 49\n38 14\n",
"7\n1 2\n2 3\n3 4\n3 5\n1 6\n1 7\n",
"20\n19 16\n19 18\n20 7\n9 4\n6 17\n14 2\n9 15\n2 13\n5 11\n19 12\n12 20\n16 9\n11 8\n19 5\n3 1\n19 14\n5 3\n12 10\n19 6\n",
"7\n1 2\n1 3\n2 4\n2 5\n3 6\n3 7\n",
"10\n9 5\n7 1\n9 10\n7 2\n5 4\n9 6\n2 9\n10 8\n1 3\n",
"4\n2 4\n2 3\n2 1\n",
"4\n1 4\n3 2\n1 3\n",
"3\n1 2\n1 3\n",
"5\n1 2\n1 5\n1 3\n1 4\n",
"20\n14 9\n12 13\n10 15\n2 1\n20 19\n16 6\n16 3\n17 14\n3 5\n2 11\n3 10\n15 8\n14 2\n6 4\n3 20\n5 18\n1 7\n1 16\n4 12\n",
"20\n7 5\n14 13\n17 6\n3 8\n16 12\n18 9\n3 18\n14 1\n17 3\n15 2\n17 4\n9 11\n2 7\n15 17\n3 20\n16 10\n17 14\n2 16\n1 19\n",
"8\n1 2\n2 3\n3 4\n1 7\n1 8\n4 5\n4 6\n",
"5\n5 1\n5 2\n5 3\n5 4\n",
"50\n49 6\n43 7\n1 27\n19 35\n15 37\n16 12\n19 21\n16 28\n49 9\n48 39\n13 1\n2 48\n9 50\n44 3\n41 32\n48 31\n49 33\n6 11\n13 20\n49 22\n13 41\n48 29\n13 46\n15 47\n34 2\n49 13\n48 14\n34 24\n16 36\n13 40\n49 34\n49 17\n43 25\n11 23\n10 15\n19 26\n34 44\n16 42\n19 18\n46 8\n29 38\n1 45\n12 43\n13 16\n46 30\n15 5\n49 10\n11 19\n32 4\n",
"20\n13 1\n18 2\n3 7\n18 5\n20 16\n3 12\n18 9\n3 10\n18 11\n13 6\n3 18\n20 15\n20 17\n3 13\n3 4\n13 14\n3 20\n18 8\n3 19\n",
"10\n8 2\n5 6\n1 8\n2 9\n1 4\n8 10\n10 5\n2 7\n2 3\n"
],
"output": [
"NO",
"YES",
"NO",
"NO",
"NO",
"YES",
"NO",
"NO",
"NO",
"NO",
"YES",
"NO",
"NO",
"YES",
"NO",
"NO",
"NO",
"YES",
"NO",
"YES",
"NO"
]
}
|
IN-CORRECT
|
python3
|
# https://codeforces.com/contest/1189/problem/D1
n = int(input())
g = {}
p = {}
path = {}
flg = True
for _ in range(n-1):
u,v = map(int, input().split())
if u not in g:
g[u] = []
g[u].append(v)
if v not in g:
g[v] = []
g[v].append(u)
def solve(g, p, path):
if len(g) == 2:
return True
root = None
for u in g:
if len(g[u]) >= 2:
root = u
break
S = [root]
p[root] = 0
i = 0
while i < len(S):
cur = S[i]
for x in g[cur]:
if x == p[cur]:continue
p[x] = cur
S.append(x)
i+=1
for cur in S[::-1]:
if len(g[cur]) == 1:
path[cur] = 0
else:
cnt = 0
for x in g[cur]:
if x == p[cur]:continue
if path[x] == 1:
return False
cnt += 1
path[cur] = cnt
return True
flg = solve(g, p, path)
if flg == True:
print('YES')
else:
print('NO')
|
1189_D1. Add on a Tree
|
Note that this is the first problem of the two similar problems. You can hack this problem only if you solve both problems.
You are given a tree with n nodes. In the beginning, 0 is written on all edges. In one operation, you can choose any 2 distinct leaves u, v and any real number x and add x to values written on all edges on the simple path between u and v.
For example, on the picture below you can see the result of applying two operations to the graph: adding 2 on the path from 7 to 6, and then adding -0.5 on the path from 4 to 5.
<image>
Is it true that for any configuration of real numbers written on edges, we can achieve it with a finite number of operations?
Leaf is a node of a tree of degree 1. Simple path is a path that doesn't contain any node twice.
Input
The first line contains a single integer n (2 ≤ n ≤ 10^5) — the number of nodes.
Each of the next n-1 lines contains two integers u and v (1 ≤ u, v ≤ n, u ≠ v), meaning that there is an edge between nodes u and v. It is guaranteed that these edges form a tree.
Output
If there is a configuration of real numbers written on edges of the tree that we can't achieve by performing the operations, output "NO".
Otherwise, output "YES".
You can print each letter in any case (upper or lower).
Examples
Input
2
1 2
Output
YES
Input
3
1 2
2 3
Output
NO
Input
5
1 2
1 3
1 4
2 5
Output
NO
Input
6
1 2
1 3
1 4
2 5
2 6
Output
YES
Note
In the first example, we can add any real x to the value written on the only edge (1, 2).
<image>
In the second example, one of configurations that we can't reach is 0 written on (1, 2) and 1 written on (2, 3).
<image>
Below you can see graphs from examples 3, 4:
<image> <image>
|
{
"input": [
"2\n1 2\n",
"3\n1 2\n2 3\n",
"5\n1 2\n1 3\n1 4\n2 5\n",
"6\n1 2\n1 3\n1 4\n2 5\n2 6\n"
],
"output": [
"YES",
"NO",
"NO",
"YES"
]
}
|
{
"input": [
"50\n16 4\n17 9\n31 19\n22 10\n8 1\n40 30\n3 31\n20 29\n47 27\n22 25\n32 34\n12 15\n40 32\n10 33\n47 12\n6 24\n46 41\n14 23\n12 35\n31 42\n46 28\n31 20\n46 37\n1 39\n29 49\n37 47\n40 6\n42 36\n47 2\n24 46\n2 13\n8 45\n41 3\n32 17\n4 7\n47 26\n28 8\n41 50\n34 44\n33 21\n25 5\n16 40\n3 14\n8 18\n28 11\n32 22\n2 38\n3 48\n44 43\n",
"10\n8 1\n1 2\n8 9\n8 5\n1 3\n1 10\n1 6\n1 7\n8 4\n",
"5\n5 1\n5 4\n4 3\n1 2\n",
"7\n1 2\n2 3\n1 4\n1 5\n3 6\n3 7\n",
"3\n1 3\n2 3\n",
"60\n26 6\n59 30\n31 12\n31 3\n38 23\n59 29\n53 9\n38 56\n53 54\n29 21\n17 55\n59 38\n26 16\n24 59\n24 25\n17 35\n24 41\n30 15\n31 27\n8 44\n26 5\n26 48\n8 32\n53 17\n3 34\n3 51\n30 28\n47 10\n53 60\n36 42\n24 53\n59 22\n53 40\n26 52\n36 4\n59 8\n29 37\n36 20\n17 47\n53 18\n3 50\n30 2\n17 7\n8 58\n59 1\n31 11\n24 26\n24 43\n53 57\n59 45\n47 13\n26 46\n17 33\n30 31\n26 39\n26 19\n24 36\n8 49\n38 14\n",
"7\n1 2\n2 3\n3 4\n3 5\n1 6\n1 7\n",
"20\n19 16\n19 18\n20 7\n9 4\n6 17\n14 2\n9 15\n2 13\n5 11\n19 12\n12 20\n16 9\n11 8\n19 5\n3 1\n19 14\n5 3\n12 10\n19 6\n",
"7\n1 2\n1 3\n2 4\n2 5\n3 6\n3 7\n",
"10\n9 5\n7 1\n9 10\n7 2\n5 4\n9 6\n2 9\n10 8\n1 3\n",
"4\n2 4\n2 3\n2 1\n",
"4\n1 4\n3 2\n1 3\n",
"3\n1 2\n1 3\n",
"5\n1 2\n1 5\n1 3\n1 4\n",
"20\n14 9\n12 13\n10 15\n2 1\n20 19\n16 6\n16 3\n17 14\n3 5\n2 11\n3 10\n15 8\n14 2\n6 4\n3 20\n5 18\n1 7\n1 16\n4 12\n",
"20\n7 5\n14 13\n17 6\n3 8\n16 12\n18 9\n3 18\n14 1\n17 3\n15 2\n17 4\n9 11\n2 7\n15 17\n3 20\n16 10\n17 14\n2 16\n1 19\n",
"8\n1 2\n2 3\n3 4\n1 7\n1 8\n4 5\n4 6\n",
"5\n5 1\n5 2\n5 3\n5 4\n",
"50\n49 6\n43 7\n1 27\n19 35\n15 37\n16 12\n19 21\n16 28\n49 9\n48 39\n13 1\n2 48\n9 50\n44 3\n41 32\n48 31\n49 33\n6 11\n13 20\n49 22\n13 41\n48 29\n13 46\n15 47\n34 2\n49 13\n48 14\n34 24\n16 36\n13 40\n49 34\n49 17\n43 25\n11 23\n10 15\n19 26\n34 44\n16 42\n19 18\n46 8\n29 38\n1 45\n12 43\n13 16\n46 30\n15 5\n49 10\n11 19\n32 4\n",
"20\n13 1\n18 2\n3 7\n18 5\n20 16\n3 12\n18 9\n3 10\n18 11\n13 6\n3 18\n20 15\n20 17\n3 13\n3 4\n13 14\n3 20\n18 8\n3 19\n",
"10\n8 2\n5 6\n1 8\n2 9\n1 4\n8 10\n10 5\n2 7\n2 3\n"
],
"output": [
"NO",
"YES",
"NO",
"NO",
"NO",
"YES",
"NO",
"NO",
"NO",
"NO",
"YES",
"NO",
"NO",
"YES",
"NO",
"NO",
"NO",
"YES",
"NO",
"YES",
"NO"
]
}
|
IN-CORRECT
|
cpp
|
#include <bits/stdc++.h>
#pragma GCC optimize("O3")
using namespace std;
const int N = 3e5 + 5, M = 1e8 + 5, OO = 1000000;
int T, n, m;
int u, v;
int deg[N];
long long l;
int main() {
scanf("%d", &n);
l = n;
for (int i = 1; i < n; ++i) {
scanf("%d %d", &u, &v);
++deg[u], ++deg[v];
if (deg[u] == 2) --l;
if (deg[v] == 2) --l;
}
l = (l * (l - 1)) >> 1ll;
if (l >= n - 1) {
printf("YES\n");
} else {
printf("NO\n");
}
}
|
1189_D1. Add on a Tree
|
Note that this is the first problem of the two similar problems. You can hack this problem only if you solve both problems.
You are given a tree with n nodes. In the beginning, 0 is written on all edges. In one operation, you can choose any 2 distinct leaves u, v and any real number x and add x to values written on all edges on the simple path between u and v.
For example, on the picture below you can see the result of applying two operations to the graph: adding 2 on the path from 7 to 6, and then adding -0.5 on the path from 4 to 5.
<image>
Is it true that for any configuration of real numbers written on edges, we can achieve it with a finite number of operations?
Leaf is a node of a tree of degree 1. Simple path is a path that doesn't contain any node twice.
Input
The first line contains a single integer n (2 ≤ n ≤ 10^5) — the number of nodes.
Each of the next n-1 lines contains two integers u and v (1 ≤ u, v ≤ n, u ≠ v), meaning that there is an edge between nodes u and v. It is guaranteed that these edges form a tree.
Output
If there is a configuration of real numbers written on edges of the tree that we can't achieve by performing the operations, output "NO".
Otherwise, output "YES".
You can print each letter in any case (upper or lower).
Examples
Input
2
1 2
Output
YES
Input
3
1 2
2 3
Output
NO
Input
5
1 2
1 3
1 4
2 5
Output
NO
Input
6
1 2
1 3
1 4
2 5
2 6
Output
YES
Note
In the first example, we can add any real x to the value written on the only edge (1, 2).
<image>
In the second example, one of configurations that we can't reach is 0 written on (1, 2) and 1 written on (2, 3).
<image>
Below you can see graphs from examples 3, 4:
<image> <image>
|
{
"input": [
"2\n1 2\n",
"3\n1 2\n2 3\n",
"5\n1 2\n1 3\n1 4\n2 5\n",
"6\n1 2\n1 3\n1 4\n2 5\n2 6\n"
],
"output": [
"YES",
"NO",
"NO",
"YES"
]
}
|
{
"input": [
"50\n16 4\n17 9\n31 19\n22 10\n8 1\n40 30\n3 31\n20 29\n47 27\n22 25\n32 34\n12 15\n40 32\n10 33\n47 12\n6 24\n46 41\n14 23\n12 35\n31 42\n46 28\n31 20\n46 37\n1 39\n29 49\n37 47\n40 6\n42 36\n47 2\n24 46\n2 13\n8 45\n41 3\n32 17\n4 7\n47 26\n28 8\n41 50\n34 44\n33 21\n25 5\n16 40\n3 14\n8 18\n28 11\n32 22\n2 38\n3 48\n44 43\n",
"10\n8 1\n1 2\n8 9\n8 5\n1 3\n1 10\n1 6\n1 7\n8 4\n",
"5\n5 1\n5 4\n4 3\n1 2\n",
"7\n1 2\n2 3\n1 4\n1 5\n3 6\n3 7\n",
"3\n1 3\n2 3\n",
"60\n26 6\n59 30\n31 12\n31 3\n38 23\n59 29\n53 9\n38 56\n53 54\n29 21\n17 55\n59 38\n26 16\n24 59\n24 25\n17 35\n24 41\n30 15\n31 27\n8 44\n26 5\n26 48\n8 32\n53 17\n3 34\n3 51\n30 28\n47 10\n53 60\n36 42\n24 53\n59 22\n53 40\n26 52\n36 4\n59 8\n29 37\n36 20\n17 47\n53 18\n3 50\n30 2\n17 7\n8 58\n59 1\n31 11\n24 26\n24 43\n53 57\n59 45\n47 13\n26 46\n17 33\n30 31\n26 39\n26 19\n24 36\n8 49\n38 14\n",
"7\n1 2\n2 3\n3 4\n3 5\n1 6\n1 7\n",
"20\n19 16\n19 18\n20 7\n9 4\n6 17\n14 2\n9 15\n2 13\n5 11\n19 12\n12 20\n16 9\n11 8\n19 5\n3 1\n19 14\n5 3\n12 10\n19 6\n",
"7\n1 2\n1 3\n2 4\n2 5\n3 6\n3 7\n",
"10\n9 5\n7 1\n9 10\n7 2\n5 4\n9 6\n2 9\n10 8\n1 3\n",
"4\n2 4\n2 3\n2 1\n",
"4\n1 4\n3 2\n1 3\n",
"3\n1 2\n1 3\n",
"5\n1 2\n1 5\n1 3\n1 4\n",
"20\n14 9\n12 13\n10 15\n2 1\n20 19\n16 6\n16 3\n17 14\n3 5\n2 11\n3 10\n15 8\n14 2\n6 4\n3 20\n5 18\n1 7\n1 16\n4 12\n",
"20\n7 5\n14 13\n17 6\n3 8\n16 12\n18 9\n3 18\n14 1\n17 3\n15 2\n17 4\n9 11\n2 7\n15 17\n3 20\n16 10\n17 14\n2 16\n1 19\n",
"8\n1 2\n2 3\n3 4\n1 7\n1 8\n4 5\n4 6\n",
"5\n5 1\n5 2\n5 3\n5 4\n",
"50\n49 6\n43 7\n1 27\n19 35\n15 37\n16 12\n19 21\n16 28\n49 9\n48 39\n13 1\n2 48\n9 50\n44 3\n41 32\n48 31\n49 33\n6 11\n13 20\n49 22\n13 41\n48 29\n13 46\n15 47\n34 2\n49 13\n48 14\n34 24\n16 36\n13 40\n49 34\n49 17\n43 25\n11 23\n10 15\n19 26\n34 44\n16 42\n19 18\n46 8\n29 38\n1 45\n12 43\n13 16\n46 30\n15 5\n49 10\n11 19\n32 4\n",
"20\n13 1\n18 2\n3 7\n18 5\n20 16\n3 12\n18 9\n3 10\n18 11\n13 6\n3 18\n20 15\n20 17\n3 13\n3 4\n13 14\n3 20\n18 8\n3 19\n",
"10\n8 2\n5 6\n1 8\n2 9\n1 4\n8 10\n10 5\n2 7\n2 3\n"
],
"output": [
"NO",
"YES",
"NO",
"NO",
"NO",
"YES",
"NO",
"NO",
"NO",
"NO",
"YES",
"NO",
"NO",
"YES",
"NO",
"NO",
"NO",
"YES",
"NO",
"YES",
"NO"
]
}
|
IN-CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
const long long N = 1e5 + 7;
long long n;
vector<long long> add[N];
void nie() {
cout << "NO";
exit(0);
}
void yie() {
cout << "YES";
exit(0);
}
int32_t main() {
ios_base::sync_with_stdio(false);
cin.tie(0);
vector<pair<long long, long long>> v;
cin >> n;
for (long long i = 1; i < n; i++) {
long long x, y;
cin >> x >> y;
add[x].push_back(y);
add[y].push_back(x);
v.push_back({x, y});
}
if (n == 2) yie();
if (n == 3) nie();
for (auto i : v) {
if (add[i.first].size() == 1 && add[i.second].size() == 2 ||
add[i.second].size() == 1 && add[i.first].size() == 2)
nie();
}
yie();
return 0;
}
|
1189_D1. Add on a Tree
|
Note that this is the first problem of the two similar problems. You can hack this problem only if you solve both problems.
You are given a tree with n nodes. In the beginning, 0 is written on all edges. In one operation, you can choose any 2 distinct leaves u, v and any real number x and add x to values written on all edges on the simple path between u and v.
For example, on the picture below you can see the result of applying two operations to the graph: adding 2 on the path from 7 to 6, and then adding -0.5 on the path from 4 to 5.
<image>
Is it true that for any configuration of real numbers written on edges, we can achieve it with a finite number of operations?
Leaf is a node of a tree of degree 1. Simple path is a path that doesn't contain any node twice.
Input
The first line contains a single integer n (2 ≤ n ≤ 10^5) — the number of nodes.
Each of the next n-1 lines contains two integers u and v (1 ≤ u, v ≤ n, u ≠ v), meaning that there is an edge between nodes u and v. It is guaranteed that these edges form a tree.
Output
If there is a configuration of real numbers written on edges of the tree that we can't achieve by performing the operations, output "NO".
Otherwise, output "YES".
You can print each letter in any case (upper or lower).
Examples
Input
2
1 2
Output
YES
Input
3
1 2
2 3
Output
NO
Input
5
1 2
1 3
1 4
2 5
Output
NO
Input
6
1 2
1 3
1 4
2 5
2 6
Output
YES
Note
In the first example, we can add any real x to the value written on the only edge (1, 2).
<image>
In the second example, one of configurations that we can't reach is 0 written on (1, 2) and 1 written on (2, 3).
<image>
Below you can see graphs from examples 3, 4:
<image> <image>
|
{
"input": [
"2\n1 2\n",
"3\n1 2\n2 3\n",
"5\n1 2\n1 3\n1 4\n2 5\n",
"6\n1 2\n1 3\n1 4\n2 5\n2 6\n"
],
"output": [
"YES",
"NO",
"NO",
"YES"
]
}
|
{
"input": [
"50\n16 4\n17 9\n31 19\n22 10\n8 1\n40 30\n3 31\n20 29\n47 27\n22 25\n32 34\n12 15\n40 32\n10 33\n47 12\n6 24\n46 41\n14 23\n12 35\n31 42\n46 28\n31 20\n46 37\n1 39\n29 49\n37 47\n40 6\n42 36\n47 2\n24 46\n2 13\n8 45\n41 3\n32 17\n4 7\n47 26\n28 8\n41 50\n34 44\n33 21\n25 5\n16 40\n3 14\n8 18\n28 11\n32 22\n2 38\n3 48\n44 43\n",
"10\n8 1\n1 2\n8 9\n8 5\n1 3\n1 10\n1 6\n1 7\n8 4\n",
"5\n5 1\n5 4\n4 3\n1 2\n",
"7\n1 2\n2 3\n1 4\n1 5\n3 6\n3 7\n",
"3\n1 3\n2 3\n",
"60\n26 6\n59 30\n31 12\n31 3\n38 23\n59 29\n53 9\n38 56\n53 54\n29 21\n17 55\n59 38\n26 16\n24 59\n24 25\n17 35\n24 41\n30 15\n31 27\n8 44\n26 5\n26 48\n8 32\n53 17\n3 34\n3 51\n30 28\n47 10\n53 60\n36 42\n24 53\n59 22\n53 40\n26 52\n36 4\n59 8\n29 37\n36 20\n17 47\n53 18\n3 50\n30 2\n17 7\n8 58\n59 1\n31 11\n24 26\n24 43\n53 57\n59 45\n47 13\n26 46\n17 33\n30 31\n26 39\n26 19\n24 36\n8 49\n38 14\n",
"7\n1 2\n2 3\n3 4\n3 5\n1 6\n1 7\n",
"20\n19 16\n19 18\n20 7\n9 4\n6 17\n14 2\n9 15\n2 13\n5 11\n19 12\n12 20\n16 9\n11 8\n19 5\n3 1\n19 14\n5 3\n12 10\n19 6\n",
"7\n1 2\n1 3\n2 4\n2 5\n3 6\n3 7\n",
"10\n9 5\n7 1\n9 10\n7 2\n5 4\n9 6\n2 9\n10 8\n1 3\n",
"4\n2 4\n2 3\n2 1\n",
"4\n1 4\n3 2\n1 3\n",
"3\n1 2\n1 3\n",
"5\n1 2\n1 5\n1 3\n1 4\n",
"20\n14 9\n12 13\n10 15\n2 1\n20 19\n16 6\n16 3\n17 14\n3 5\n2 11\n3 10\n15 8\n14 2\n6 4\n3 20\n5 18\n1 7\n1 16\n4 12\n",
"20\n7 5\n14 13\n17 6\n3 8\n16 12\n18 9\n3 18\n14 1\n17 3\n15 2\n17 4\n9 11\n2 7\n15 17\n3 20\n16 10\n17 14\n2 16\n1 19\n",
"8\n1 2\n2 3\n3 4\n1 7\n1 8\n4 5\n4 6\n",
"5\n5 1\n5 2\n5 3\n5 4\n",
"50\n49 6\n43 7\n1 27\n19 35\n15 37\n16 12\n19 21\n16 28\n49 9\n48 39\n13 1\n2 48\n9 50\n44 3\n41 32\n48 31\n49 33\n6 11\n13 20\n49 22\n13 41\n48 29\n13 46\n15 47\n34 2\n49 13\n48 14\n34 24\n16 36\n13 40\n49 34\n49 17\n43 25\n11 23\n10 15\n19 26\n34 44\n16 42\n19 18\n46 8\n29 38\n1 45\n12 43\n13 16\n46 30\n15 5\n49 10\n11 19\n32 4\n",
"20\n13 1\n18 2\n3 7\n18 5\n20 16\n3 12\n18 9\n3 10\n18 11\n13 6\n3 18\n20 15\n20 17\n3 13\n3 4\n13 14\n3 20\n18 8\n3 19\n",
"10\n8 2\n5 6\n1 8\n2 9\n1 4\n8 10\n10 5\n2 7\n2 3\n"
],
"output": [
"NO",
"YES",
"NO",
"NO",
"NO",
"YES",
"NO",
"NO",
"NO",
"NO",
"YES",
"NO",
"NO",
"YES",
"NO",
"NO",
"NO",
"YES",
"NO",
"YES",
"NO"
]
}
|
IN-CORRECT
|
java
|
import java.util.Scanner;
public class TaskD {
public static void main(String[] args) {
Scanner s = new Scanner(System.in);
int n = s.nextInt();
int[] deg = new int[n];
for(int i=0;i<n-1;i++) {
int a = s.nextInt()-1;
int b = s.nextInt()-1;
deg[a]++;
deg[b]++;
}
for(int i=0;i<n;i++) {
if(deg[i]%2 > 2) {
System.out.println("NO");
return;
}
}
System.out.println("YES");
}
}
|
1189_D1. Add on a Tree
|
Note that this is the first problem of the two similar problems. You can hack this problem only if you solve both problems.
You are given a tree with n nodes. In the beginning, 0 is written on all edges. In one operation, you can choose any 2 distinct leaves u, v and any real number x and add x to values written on all edges on the simple path between u and v.
For example, on the picture below you can see the result of applying two operations to the graph: adding 2 on the path from 7 to 6, and then adding -0.5 on the path from 4 to 5.
<image>
Is it true that for any configuration of real numbers written on edges, we can achieve it with a finite number of operations?
Leaf is a node of a tree of degree 1. Simple path is a path that doesn't contain any node twice.
Input
The first line contains a single integer n (2 ≤ n ≤ 10^5) — the number of nodes.
Each of the next n-1 lines contains two integers u and v (1 ≤ u, v ≤ n, u ≠ v), meaning that there is an edge between nodes u and v. It is guaranteed that these edges form a tree.
Output
If there is a configuration of real numbers written on edges of the tree that we can't achieve by performing the operations, output "NO".
Otherwise, output "YES".
You can print each letter in any case (upper or lower).
Examples
Input
2
1 2
Output
YES
Input
3
1 2
2 3
Output
NO
Input
5
1 2
1 3
1 4
2 5
Output
NO
Input
6
1 2
1 3
1 4
2 5
2 6
Output
YES
Note
In the first example, we can add any real x to the value written on the only edge (1, 2).
<image>
In the second example, one of configurations that we can't reach is 0 written on (1, 2) and 1 written on (2, 3).
<image>
Below you can see graphs from examples 3, 4:
<image> <image>
|
{
"input": [
"2\n1 2\n",
"3\n1 2\n2 3\n",
"5\n1 2\n1 3\n1 4\n2 5\n",
"6\n1 2\n1 3\n1 4\n2 5\n2 6\n"
],
"output": [
"YES",
"NO",
"NO",
"YES"
]
}
|
{
"input": [
"50\n16 4\n17 9\n31 19\n22 10\n8 1\n40 30\n3 31\n20 29\n47 27\n22 25\n32 34\n12 15\n40 32\n10 33\n47 12\n6 24\n46 41\n14 23\n12 35\n31 42\n46 28\n31 20\n46 37\n1 39\n29 49\n37 47\n40 6\n42 36\n47 2\n24 46\n2 13\n8 45\n41 3\n32 17\n4 7\n47 26\n28 8\n41 50\n34 44\n33 21\n25 5\n16 40\n3 14\n8 18\n28 11\n32 22\n2 38\n3 48\n44 43\n",
"10\n8 1\n1 2\n8 9\n8 5\n1 3\n1 10\n1 6\n1 7\n8 4\n",
"5\n5 1\n5 4\n4 3\n1 2\n",
"7\n1 2\n2 3\n1 4\n1 5\n3 6\n3 7\n",
"3\n1 3\n2 3\n",
"60\n26 6\n59 30\n31 12\n31 3\n38 23\n59 29\n53 9\n38 56\n53 54\n29 21\n17 55\n59 38\n26 16\n24 59\n24 25\n17 35\n24 41\n30 15\n31 27\n8 44\n26 5\n26 48\n8 32\n53 17\n3 34\n3 51\n30 28\n47 10\n53 60\n36 42\n24 53\n59 22\n53 40\n26 52\n36 4\n59 8\n29 37\n36 20\n17 47\n53 18\n3 50\n30 2\n17 7\n8 58\n59 1\n31 11\n24 26\n24 43\n53 57\n59 45\n47 13\n26 46\n17 33\n30 31\n26 39\n26 19\n24 36\n8 49\n38 14\n",
"7\n1 2\n2 3\n3 4\n3 5\n1 6\n1 7\n",
"20\n19 16\n19 18\n20 7\n9 4\n6 17\n14 2\n9 15\n2 13\n5 11\n19 12\n12 20\n16 9\n11 8\n19 5\n3 1\n19 14\n5 3\n12 10\n19 6\n",
"7\n1 2\n1 3\n2 4\n2 5\n3 6\n3 7\n",
"10\n9 5\n7 1\n9 10\n7 2\n5 4\n9 6\n2 9\n10 8\n1 3\n",
"4\n2 4\n2 3\n2 1\n",
"4\n1 4\n3 2\n1 3\n",
"3\n1 2\n1 3\n",
"5\n1 2\n1 5\n1 3\n1 4\n",
"20\n14 9\n12 13\n10 15\n2 1\n20 19\n16 6\n16 3\n17 14\n3 5\n2 11\n3 10\n15 8\n14 2\n6 4\n3 20\n5 18\n1 7\n1 16\n4 12\n",
"20\n7 5\n14 13\n17 6\n3 8\n16 12\n18 9\n3 18\n14 1\n17 3\n15 2\n17 4\n9 11\n2 7\n15 17\n3 20\n16 10\n17 14\n2 16\n1 19\n",
"8\n1 2\n2 3\n3 4\n1 7\n1 8\n4 5\n4 6\n",
"5\n5 1\n5 2\n5 3\n5 4\n",
"50\n49 6\n43 7\n1 27\n19 35\n15 37\n16 12\n19 21\n16 28\n49 9\n48 39\n13 1\n2 48\n9 50\n44 3\n41 32\n48 31\n49 33\n6 11\n13 20\n49 22\n13 41\n48 29\n13 46\n15 47\n34 2\n49 13\n48 14\n34 24\n16 36\n13 40\n49 34\n49 17\n43 25\n11 23\n10 15\n19 26\n34 44\n16 42\n19 18\n46 8\n29 38\n1 45\n12 43\n13 16\n46 30\n15 5\n49 10\n11 19\n32 4\n",
"20\n13 1\n18 2\n3 7\n18 5\n20 16\n3 12\n18 9\n3 10\n18 11\n13 6\n3 18\n20 15\n20 17\n3 13\n3 4\n13 14\n3 20\n18 8\n3 19\n",
"10\n8 2\n5 6\n1 8\n2 9\n1 4\n8 10\n10 5\n2 7\n2 3\n"
],
"output": [
"NO",
"YES",
"NO",
"NO",
"NO",
"YES",
"NO",
"NO",
"NO",
"NO",
"YES",
"NO",
"NO",
"YES",
"NO",
"NO",
"NO",
"YES",
"NO",
"YES",
"NO"
]
}
|
IN-CORRECT
|
java
|
import java.io.*;
import java.util.*;
public class d2prac implements Runnable
{
private boolean console=false;
boolean ans=true; int sc=0;
ArrayList<Integer> adj[];
public void solve()
{
int i; int n=in.ni(); adj=new ArrayList [n];
for(i=0;i<n;i++)
adj[i]=new ArrayList();
for(i=0;i<n-1;i++)
{
int u=in.ni()-1; int v=in.ni()-1;
adj[u].add(v); adj[v].add(u);
}
dfs(0,-1);
if(ans&&sc>1)
out.println("YES");
else
out.println("NO");
}
public void dfs(int v,int p)
{
boolean not=false;
if(adj[v].size()>1)
not=true;
int count=0;
for(int node:adj[v])
{
if(adj[node].size()==1)
count++;
if(node==p)
continue;
else
dfs(node,v);
}
if(not&&count<2)
ans=false;
else if(not)
sc++;
}
@Override
public void run() {
try { init(); }
catch (FileNotFoundException e) { e.printStackTrace(); }
int t= 1;
while (t-->0) {
solve();
out.flush(); }
}
private FastInput in; private PrintWriter out;
public static void main(String[] args) throws Exception { new d2prac().run(); }
private void init() throws FileNotFoundException {
InputStream inputStream = System.in; OutputStream outputStream = System.out;
try { if (!console && System.getProperty("user.name").equals("sachan")) {
outputStream = new FileOutputStream("/home/sachan/Desktop/output.txt");
inputStream = new FileInputStream("/home/sachan/Desktop/input.txt"); }
} catch (Exception ignored) { }
out = new PrintWriter(outputStream); in = new FastInput(inputStream);
}
static class FastInput { InputStream obj;
public FastInput(InputStream obj) { this.obj = obj; }
byte inbuffer[] = new byte[1024]; int lenbuffer = 0, ptrbuffer = 0;
int readByte() { if (lenbuffer == -1) throw new InputMismatchException();
if (ptrbuffer >= lenbuffer) { ptrbuffer = 0;
try { lenbuffer = obj.read(inbuffer); }
catch (IOException e) { throw new InputMismatchException(); } }
if (lenbuffer <= 0) return -1;return inbuffer[ptrbuffer++]; }
String ns() { int b = skip();StringBuilder sb = new StringBuilder();
while (!(isSpaceChar(b))) { sb.appendCodePoint(b);b = readByte(); }return sb.toString();}
int ni() { int num = 0, b;boolean minus = false;
while ((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-')) ;
if (b == '-') { minus = true;b = readByte(); }
while (true) { if (b >= '0' && b <= '9') { num = num * 10 + (b - '0'); } else {
return minus ? -num : num; }b = readByte(); }}
long nl() { long num = 0;int b;boolean minus = false;
while ((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-')) ;
if (b == '-') { minus = true;b = readByte(); }
while (true) { if (b >= '0' && b <= '9') { num = num * 10L + (b - '0'); } else {
return minus ? -num : num; }b = readByte(); } }
boolean isSpaceChar(int c) { return (!(c >= 33 && c <= 126)); }
int skip() { int b;while ((b = readByte()) != -1 && isSpaceChar(b)) ;return b; }
float nf() {return Float.parseFloat(ns());}
double nd() {return Double.parseDouble(ns());}
char nc() {return (char) skip();}
}
}
|
1189_D1. Add on a Tree
|
Note that this is the first problem of the two similar problems. You can hack this problem only if you solve both problems.
You are given a tree with n nodes. In the beginning, 0 is written on all edges. In one operation, you can choose any 2 distinct leaves u, v and any real number x and add x to values written on all edges on the simple path between u and v.
For example, on the picture below you can see the result of applying two operations to the graph: adding 2 on the path from 7 to 6, and then adding -0.5 on the path from 4 to 5.
<image>
Is it true that for any configuration of real numbers written on edges, we can achieve it with a finite number of operations?
Leaf is a node of a tree of degree 1. Simple path is a path that doesn't contain any node twice.
Input
The first line contains a single integer n (2 ≤ n ≤ 10^5) — the number of nodes.
Each of the next n-1 lines contains two integers u and v (1 ≤ u, v ≤ n, u ≠ v), meaning that there is an edge between nodes u and v. It is guaranteed that these edges form a tree.
Output
If there is a configuration of real numbers written on edges of the tree that we can't achieve by performing the operations, output "NO".
Otherwise, output "YES".
You can print each letter in any case (upper or lower).
Examples
Input
2
1 2
Output
YES
Input
3
1 2
2 3
Output
NO
Input
5
1 2
1 3
1 4
2 5
Output
NO
Input
6
1 2
1 3
1 4
2 5
2 6
Output
YES
Note
In the first example, we can add any real x to the value written on the only edge (1, 2).
<image>
In the second example, one of configurations that we can't reach is 0 written on (1, 2) and 1 written on (2, 3).
<image>
Below you can see graphs from examples 3, 4:
<image> <image>
|
{
"input": [
"2\n1 2\n",
"3\n1 2\n2 3\n",
"5\n1 2\n1 3\n1 4\n2 5\n",
"6\n1 2\n1 3\n1 4\n2 5\n2 6\n"
],
"output": [
"YES",
"NO",
"NO",
"YES"
]
}
|
{
"input": [
"50\n16 4\n17 9\n31 19\n22 10\n8 1\n40 30\n3 31\n20 29\n47 27\n22 25\n32 34\n12 15\n40 32\n10 33\n47 12\n6 24\n46 41\n14 23\n12 35\n31 42\n46 28\n31 20\n46 37\n1 39\n29 49\n37 47\n40 6\n42 36\n47 2\n24 46\n2 13\n8 45\n41 3\n32 17\n4 7\n47 26\n28 8\n41 50\n34 44\n33 21\n25 5\n16 40\n3 14\n8 18\n28 11\n32 22\n2 38\n3 48\n44 43\n",
"10\n8 1\n1 2\n8 9\n8 5\n1 3\n1 10\n1 6\n1 7\n8 4\n",
"5\n5 1\n5 4\n4 3\n1 2\n",
"7\n1 2\n2 3\n1 4\n1 5\n3 6\n3 7\n",
"3\n1 3\n2 3\n",
"60\n26 6\n59 30\n31 12\n31 3\n38 23\n59 29\n53 9\n38 56\n53 54\n29 21\n17 55\n59 38\n26 16\n24 59\n24 25\n17 35\n24 41\n30 15\n31 27\n8 44\n26 5\n26 48\n8 32\n53 17\n3 34\n3 51\n30 28\n47 10\n53 60\n36 42\n24 53\n59 22\n53 40\n26 52\n36 4\n59 8\n29 37\n36 20\n17 47\n53 18\n3 50\n30 2\n17 7\n8 58\n59 1\n31 11\n24 26\n24 43\n53 57\n59 45\n47 13\n26 46\n17 33\n30 31\n26 39\n26 19\n24 36\n8 49\n38 14\n",
"7\n1 2\n2 3\n3 4\n3 5\n1 6\n1 7\n",
"20\n19 16\n19 18\n20 7\n9 4\n6 17\n14 2\n9 15\n2 13\n5 11\n19 12\n12 20\n16 9\n11 8\n19 5\n3 1\n19 14\n5 3\n12 10\n19 6\n",
"7\n1 2\n1 3\n2 4\n2 5\n3 6\n3 7\n",
"10\n9 5\n7 1\n9 10\n7 2\n5 4\n9 6\n2 9\n10 8\n1 3\n",
"4\n2 4\n2 3\n2 1\n",
"4\n1 4\n3 2\n1 3\n",
"3\n1 2\n1 3\n",
"5\n1 2\n1 5\n1 3\n1 4\n",
"20\n14 9\n12 13\n10 15\n2 1\n20 19\n16 6\n16 3\n17 14\n3 5\n2 11\n3 10\n15 8\n14 2\n6 4\n3 20\n5 18\n1 7\n1 16\n4 12\n",
"20\n7 5\n14 13\n17 6\n3 8\n16 12\n18 9\n3 18\n14 1\n17 3\n15 2\n17 4\n9 11\n2 7\n15 17\n3 20\n16 10\n17 14\n2 16\n1 19\n",
"8\n1 2\n2 3\n3 4\n1 7\n1 8\n4 5\n4 6\n",
"5\n5 1\n5 2\n5 3\n5 4\n",
"50\n49 6\n43 7\n1 27\n19 35\n15 37\n16 12\n19 21\n16 28\n49 9\n48 39\n13 1\n2 48\n9 50\n44 3\n41 32\n48 31\n49 33\n6 11\n13 20\n49 22\n13 41\n48 29\n13 46\n15 47\n34 2\n49 13\n48 14\n34 24\n16 36\n13 40\n49 34\n49 17\n43 25\n11 23\n10 15\n19 26\n34 44\n16 42\n19 18\n46 8\n29 38\n1 45\n12 43\n13 16\n46 30\n15 5\n49 10\n11 19\n32 4\n",
"20\n13 1\n18 2\n3 7\n18 5\n20 16\n3 12\n18 9\n3 10\n18 11\n13 6\n3 18\n20 15\n20 17\n3 13\n3 4\n13 14\n3 20\n18 8\n3 19\n",
"10\n8 2\n5 6\n1 8\n2 9\n1 4\n8 10\n10 5\n2 7\n2 3\n"
],
"output": [
"NO",
"YES",
"NO",
"NO",
"NO",
"YES",
"NO",
"NO",
"NO",
"NO",
"YES",
"NO",
"NO",
"YES",
"NO",
"NO",
"NO",
"YES",
"NO",
"YES",
"NO"
]
}
|
IN-CORRECT
|
java
|
import java.util.*;
import javax.lang.model.util.ElementScanner6;
import java.io.*;
public class Main {
public static void main(String[] args) {
InputStream inputStream = System.in;
OutputStream outputStream = System.out;
InputReader inp = new InputReader(inputStream);
PrintWriter out = new PrintWriter(outputStream);
Solver solver = new Solver();
solver.solve(inp, out);
out.close();
}
static class Solver {
class Node implements Comparable < Node > {
int i;
int cnt;
Node(int i, int cnt) {
this.i = i;
this.cnt = cnt;
}
public int compareTo(Node n) {
if (this.cnt == n.cnt) {
return Integer.compare(this.i, n.i);
}
return Integer.compare(this.cnt, n.cnt);
}
}
public boolean done(int[] sp, int[] par) {
int root;
root = findSet(sp[0], par);
for (int i = 1; i < sp.length; i++) {
if (root != findSet(sp[i], par))
return false;
}
return true;
}
public int findSet(int i, int[] par) {
int x = i;
boolean flag = false;
while (par[i] >= 0) {
flag = true;
i = par[i];
}
if (flag)
par[x] = i;
return i;
}
public void unionSet(int i, int j, int[] par) {
int x = findSet(i, par);
int y = findSet(j, par);
if (x < y) {
par[y] = x;
} else {
par[x] = y;
}
}
public long pow(long a, long b, long MOD) {
if (b == 0) {
return 1;
}
if (b == 1) {
return a;
}
long val = pow(a, b / 2, MOD);
if (b % 2 == 0) {
return val * val % MOD;
} else {
return val * (val * a % MOD) % MOD;
}
}
public void minPrimeFactor(int n, int[] s) {
boolean prime[] = new boolean[n + 1];
Arrays.fill(prime, true);
s[1] = 1;
s[2] = 2;
for (int i = 4; i <= n; i += 2) {
prime[i] = false;
s[i] = 2;
}
for (int i = 3; i <= n; i += 2) {
if (prime[i]) {
s[i] = i;
for (int j = 2 * i; j <= n; j += i) {
prime[j] = false;
s[j] = i;
}
}
}
}
public void findAllPrime(int n, ArrayList < Node > al, int s[]) {
int curr = s[n];
int cnt = 1;
while (n > 1) {
n /= s[n];
if (curr == s[n]) {
cnt++;
continue;
}
Node n1 = new Node(curr, cnt);
al.add(n1);
curr = s[n];
cnt = 1;
}
}
public int binarySearch(int n, int k) {
int left = 1;
int right = 100000000 + 5;
int ans = 0;
while (left <= right) {
int mid = (left + right) / 2;
if (n / mid >= k) {
left = mid + 1;
ans = mid;
} else {
right = mid - 1;
}
}
return ans;
}
public boolean checkPallindrom(String s) {
char ch[] = s.toCharArray();
for (int i = 0; i < s.length() / 2; i++) {
if (ch[i] != ch[s.length() - 1 - i])
return false;
}
return true;
}
public void remove(ArrayList < Integer > [] al, int x) {
for (int i = 0; i < al.length; i++) {
for (int j = 0; j < al[i].size(); j++) {
if (al[i].get(j) == x)
al[i].remove(j);
}
}
}
public long gcd(long a, long b) {
if (a == 0)
return b;
return gcd(b % a, a);
}
public void printDivisors(long n, ArrayList <Long> al) {
// Note that this loop runs till square root
for (long i = 1; i <= Math.sqrt(n); i++) {
if (n % i == 0) {
// If divisors are equal, print only one
if (n / i == i) {
al.add(i);
} else // Otherwise print both
{
al.add(i);
al.add(n / i);
}
}
}
}
public static long constructSegment(long seg[], long arr[], int low, int high, int pos) {
if (low == high) {
seg[pos] = arr[low];
return seg[pos];
}
int mid = (low + high) / 2;
long t1 = constructSegment(seg, arr, low, mid, (2 * pos) + 1);
long t2 = constructSegment(seg, arr, mid + 1, high, (2 * pos) + 2);
seg[pos] = t1 + t2;
return seg[pos];
}
public static long querySegment(long seg[], int low, int high, int qlow, int qhigh, int pos) {
if (qlow <= low && qhigh >= high) {
return seg[pos];
} else if (qlow > high || qhigh < low) {
return 0;
} else {
long ans = 0;
int mid = (low + high) / 2;
ans += querySegment(seg, low, mid, qlow, qhigh, (2 * pos) + 1);
ans += querySegment(seg, mid + 1, high, qlow, qhigh, (2 * pos) + 2);
return ans;
}
}
public static int lcs(char[] X, char[] Y, int m, int n) {
if (m == 0 || n == 0)
return 0;
if (X[m - 1] == Y[n - 1])
return 1 + lcs(X, Y, m - 1, n - 1);
else
return Integer.max(lcs(X, Y, m, n - 1), lcs(X, Y, m - 1, n));
}
public static long recursion(long start, long end, long cnt[], int a, int b) {
long min = 0;
long count = 0;
int ans1 = -1;
int ans2 = -1;
int l = 0;
int r = cnt.length - 1;
while (l <= r) {
int mid = (l + r) / 2;
if (cnt[mid] >= start) {
ans1 = mid;
r = mid - 1;
} else {
l = mid + 1;
}
}
l = 0;
r = cnt.length - 1;
while (l <= r) {
int mid = (l + r) / 2;
if (cnt[mid] <= end) {
ans2 = mid;
l = mid + 1;
} else {
r = mid - 1;
}
}
if (ans1 == -1 || ans2 == -1 || ans2 < ans1) {
// System.out.println("min1 "+min);
min = a;
return a;
} else {
min = b * (end - start + 1) * (ans2 - ans1 + 1);
}
if (start == end) {
// System.out.println("min "+min);
return min;
}
long mid = (end + start) / 2;
min = Long.min(min, recursion(start, mid, cnt, a, b) + recursion(mid + 1, end, cnt, a, b));
// System.out.println("min "+min);
return min;
}
public int dfs_util(ArrayList < Integer > [] al, boolean vis[], int x,int[] s,int lvl[]) {
vis[x] = true;
int cnt=1;
for (int i = 0; i < al[x].size(); i++) {
if (!vis[al[x].get(i)]) {
lvl[al[x].get(i)]=lvl[x]+1;
cnt+=dfs_util(al, vis, al[x].get(i),s,lvl);
}
}
s[x] = cnt;
return s[x];
}
public void dfs(ArrayList[] al,int[] s,int[] lvl) {
boolean vis[] = new boolean[al.length];
for (int i = 0; i < al.length; i++) {
if (!vis[i]) {
lvl[i]=1;
dfs_util(al, vis, i,s,lvl);
}
}
}
private void solve(InputReader inp, PrintWriter out1) {
int n = inp.nextInt();
ArrayList<Integer> al[] = new ArrayList[n];
for(int i=0;i<n;i++)
{
al[i] = new ArrayList<Integer>();
}
long cnt=0;
for(int i=0;i<n-1;i++)
{
int x1 = inp.nextInt();
int x2 = inp.nextInt();
al[x1-1].add(x2-1);
al[x2-1].add(x1-1);
}
for(int i=0;i<n;i++)
{
if(al[i].size()==1)
{
cnt++;
}
}
long temp = (cnt*(cnt-1))/2;
if(temp<(long)(n-1))
{
out1.println("NO");
}
else
{
out1.println("YES");
}
}
}
static class InputReader {
BufferedReader reader;
StringTokenizer tokenizer;
InputReader(InputStream stream) {
reader = new BufferedReader(new InputStreamReader(stream), 32768);
tokenizer = null;
}
String next() {
while (tokenizer == null || !tokenizer.hasMoreTokens()) {
try {
tokenizer = new StringTokenizer(reader.readLine());
} catch (IOException e) {
throw new RuntimeException(e);
}
}
return tokenizer.nextToken();
}
public int nextInt() {
return Integer.parseInt(next());
}
public long nextLong() {
return Long.parseLong(next());
}
}
}
class ele implements Comparable < ele > {
int value;
int i;
boolean flag;
public ele(int value, int i) {
this.value = value;
this.i = i;
this.flag = false;
}
public int compareTo(ele e) {
return Integer.compare(this.value, e.value);
}
}
|
1189_D1. Add on a Tree
|
Note that this is the first problem of the two similar problems. You can hack this problem only if you solve both problems.
You are given a tree with n nodes. In the beginning, 0 is written on all edges. In one operation, you can choose any 2 distinct leaves u, v and any real number x and add x to values written on all edges on the simple path between u and v.
For example, on the picture below you can see the result of applying two operations to the graph: adding 2 on the path from 7 to 6, and then adding -0.5 on the path from 4 to 5.
<image>
Is it true that for any configuration of real numbers written on edges, we can achieve it with a finite number of operations?
Leaf is a node of a tree of degree 1. Simple path is a path that doesn't contain any node twice.
Input
The first line contains a single integer n (2 ≤ n ≤ 10^5) — the number of nodes.
Each of the next n-1 lines contains two integers u and v (1 ≤ u, v ≤ n, u ≠ v), meaning that there is an edge between nodes u and v. It is guaranteed that these edges form a tree.
Output
If there is a configuration of real numbers written on edges of the tree that we can't achieve by performing the operations, output "NO".
Otherwise, output "YES".
You can print each letter in any case (upper or lower).
Examples
Input
2
1 2
Output
YES
Input
3
1 2
2 3
Output
NO
Input
5
1 2
1 3
1 4
2 5
Output
NO
Input
6
1 2
1 3
1 4
2 5
2 6
Output
YES
Note
In the first example, we can add any real x to the value written on the only edge (1, 2).
<image>
In the second example, one of configurations that we can't reach is 0 written on (1, 2) and 1 written on (2, 3).
<image>
Below you can see graphs from examples 3, 4:
<image> <image>
|
{
"input": [
"2\n1 2\n",
"3\n1 2\n2 3\n",
"5\n1 2\n1 3\n1 4\n2 5\n",
"6\n1 2\n1 3\n1 4\n2 5\n2 6\n"
],
"output": [
"YES",
"NO",
"NO",
"YES"
]
}
|
{
"input": [
"50\n16 4\n17 9\n31 19\n22 10\n8 1\n40 30\n3 31\n20 29\n47 27\n22 25\n32 34\n12 15\n40 32\n10 33\n47 12\n6 24\n46 41\n14 23\n12 35\n31 42\n46 28\n31 20\n46 37\n1 39\n29 49\n37 47\n40 6\n42 36\n47 2\n24 46\n2 13\n8 45\n41 3\n32 17\n4 7\n47 26\n28 8\n41 50\n34 44\n33 21\n25 5\n16 40\n3 14\n8 18\n28 11\n32 22\n2 38\n3 48\n44 43\n",
"10\n8 1\n1 2\n8 9\n8 5\n1 3\n1 10\n1 6\n1 7\n8 4\n",
"5\n5 1\n5 4\n4 3\n1 2\n",
"7\n1 2\n2 3\n1 4\n1 5\n3 6\n3 7\n",
"3\n1 3\n2 3\n",
"60\n26 6\n59 30\n31 12\n31 3\n38 23\n59 29\n53 9\n38 56\n53 54\n29 21\n17 55\n59 38\n26 16\n24 59\n24 25\n17 35\n24 41\n30 15\n31 27\n8 44\n26 5\n26 48\n8 32\n53 17\n3 34\n3 51\n30 28\n47 10\n53 60\n36 42\n24 53\n59 22\n53 40\n26 52\n36 4\n59 8\n29 37\n36 20\n17 47\n53 18\n3 50\n30 2\n17 7\n8 58\n59 1\n31 11\n24 26\n24 43\n53 57\n59 45\n47 13\n26 46\n17 33\n30 31\n26 39\n26 19\n24 36\n8 49\n38 14\n",
"7\n1 2\n2 3\n3 4\n3 5\n1 6\n1 7\n",
"20\n19 16\n19 18\n20 7\n9 4\n6 17\n14 2\n9 15\n2 13\n5 11\n19 12\n12 20\n16 9\n11 8\n19 5\n3 1\n19 14\n5 3\n12 10\n19 6\n",
"7\n1 2\n1 3\n2 4\n2 5\n3 6\n3 7\n",
"10\n9 5\n7 1\n9 10\n7 2\n5 4\n9 6\n2 9\n10 8\n1 3\n",
"4\n2 4\n2 3\n2 1\n",
"4\n1 4\n3 2\n1 3\n",
"3\n1 2\n1 3\n",
"5\n1 2\n1 5\n1 3\n1 4\n",
"20\n14 9\n12 13\n10 15\n2 1\n20 19\n16 6\n16 3\n17 14\n3 5\n2 11\n3 10\n15 8\n14 2\n6 4\n3 20\n5 18\n1 7\n1 16\n4 12\n",
"20\n7 5\n14 13\n17 6\n3 8\n16 12\n18 9\n3 18\n14 1\n17 3\n15 2\n17 4\n9 11\n2 7\n15 17\n3 20\n16 10\n17 14\n2 16\n1 19\n",
"8\n1 2\n2 3\n3 4\n1 7\n1 8\n4 5\n4 6\n",
"5\n5 1\n5 2\n5 3\n5 4\n",
"50\n49 6\n43 7\n1 27\n19 35\n15 37\n16 12\n19 21\n16 28\n49 9\n48 39\n13 1\n2 48\n9 50\n44 3\n41 32\n48 31\n49 33\n6 11\n13 20\n49 22\n13 41\n48 29\n13 46\n15 47\n34 2\n49 13\n48 14\n34 24\n16 36\n13 40\n49 34\n49 17\n43 25\n11 23\n10 15\n19 26\n34 44\n16 42\n19 18\n46 8\n29 38\n1 45\n12 43\n13 16\n46 30\n15 5\n49 10\n11 19\n32 4\n",
"20\n13 1\n18 2\n3 7\n18 5\n20 16\n3 12\n18 9\n3 10\n18 11\n13 6\n3 18\n20 15\n20 17\n3 13\n3 4\n13 14\n3 20\n18 8\n3 19\n",
"10\n8 2\n5 6\n1 8\n2 9\n1 4\n8 10\n10 5\n2 7\n2 3\n"
],
"output": [
"NO",
"YES",
"NO",
"NO",
"NO",
"YES",
"NO",
"NO",
"NO",
"NO",
"YES",
"NO",
"NO",
"YES",
"NO",
"NO",
"NO",
"YES",
"NO",
"YES",
"NO"
]
}
|
IN-CORRECT
|
cpp
|
#include <bits/stdc++.h>
#pragma GCC optimize("Ofast")
#pragma GCC target("sse4")
const long double EPS = 1e-9;
const long long INF = 1e9;
const long long mod = 1e9 + 7;
using namespace std;
mt19937 rnd(chrono::high_resolution_clock::now().time_since_epoch().count());
long long n, m, k;
long long q;
vector<vector<long long>> vec;
vector<bool> used;
vector<long long> cnt;
vector<long long> ans;
vector<long long> pref;
int main() {
ios::sync_with_stdio(false);
cin.tie(nullptr);
cin >> n;
vec.resize(n);
cnt.resize(n);
for (int i = 0; i < n - 1; ++i) {
int v, to;
cin >> v >> to;
--v;
--to;
vec[v].push_back(to);
vec[to].push_back(v);
cnt[v]++;
cnt[to]++;
}
int cntl = 0, cntnl = 0;
for (int i = 0; i < n; ++i) {
if (cnt[i] == 1)
cntl++;
else {
int cntal = 0;
for (int j = 0; j < vec[i].size(); ++j) {
if (cnt[vec[i][j]] == 1) {
cntal++;
}
}
if (cntal <= 1) {
cntnl++;
}
}
}
if ((cntl == 2 && n > 2) || cntnl) {
cout << "NO";
} else {
cout << "YES";
}
return 0;
}
|
1189_D1. Add on a Tree
|
Note that this is the first problem of the two similar problems. You can hack this problem only if you solve both problems.
You are given a tree with n nodes. In the beginning, 0 is written on all edges. In one operation, you can choose any 2 distinct leaves u, v and any real number x and add x to values written on all edges on the simple path between u and v.
For example, on the picture below you can see the result of applying two operations to the graph: adding 2 on the path from 7 to 6, and then adding -0.5 on the path from 4 to 5.
<image>
Is it true that for any configuration of real numbers written on edges, we can achieve it with a finite number of operations?
Leaf is a node of a tree of degree 1. Simple path is a path that doesn't contain any node twice.
Input
The first line contains a single integer n (2 ≤ n ≤ 10^5) — the number of nodes.
Each of the next n-1 lines contains two integers u and v (1 ≤ u, v ≤ n, u ≠ v), meaning that there is an edge between nodes u and v. It is guaranteed that these edges form a tree.
Output
If there is a configuration of real numbers written on edges of the tree that we can't achieve by performing the operations, output "NO".
Otherwise, output "YES".
You can print each letter in any case (upper or lower).
Examples
Input
2
1 2
Output
YES
Input
3
1 2
2 3
Output
NO
Input
5
1 2
1 3
1 4
2 5
Output
NO
Input
6
1 2
1 3
1 4
2 5
2 6
Output
YES
Note
In the first example, we can add any real x to the value written on the only edge (1, 2).
<image>
In the second example, one of configurations that we can't reach is 0 written on (1, 2) and 1 written on (2, 3).
<image>
Below you can see graphs from examples 3, 4:
<image> <image>
|
{
"input": [
"2\n1 2\n",
"3\n1 2\n2 3\n",
"5\n1 2\n1 3\n1 4\n2 5\n",
"6\n1 2\n1 3\n1 4\n2 5\n2 6\n"
],
"output": [
"YES",
"NO",
"NO",
"YES"
]
}
|
{
"input": [
"50\n16 4\n17 9\n31 19\n22 10\n8 1\n40 30\n3 31\n20 29\n47 27\n22 25\n32 34\n12 15\n40 32\n10 33\n47 12\n6 24\n46 41\n14 23\n12 35\n31 42\n46 28\n31 20\n46 37\n1 39\n29 49\n37 47\n40 6\n42 36\n47 2\n24 46\n2 13\n8 45\n41 3\n32 17\n4 7\n47 26\n28 8\n41 50\n34 44\n33 21\n25 5\n16 40\n3 14\n8 18\n28 11\n32 22\n2 38\n3 48\n44 43\n",
"10\n8 1\n1 2\n8 9\n8 5\n1 3\n1 10\n1 6\n1 7\n8 4\n",
"5\n5 1\n5 4\n4 3\n1 2\n",
"7\n1 2\n2 3\n1 4\n1 5\n3 6\n3 7\n",
"3\n1 3\n2 3\n",
"60\n26 6\n59 30\n31 12\n31 3\n38 23\n59 29\n53 9\n38 56\n53 54\n29 21\n17 55\n59 38\n26 16\n24 59\n24 25\n17 35\n24 41\n30 15\n31 27\n8 44\n26 5\n26 48\n8 32\n53 17\n3 34\n3 51\n30 28\n47 10\n53 60\n36 42\n24 53\n59 22\n53 40\n26 52\n36 4\n59 8\n29 37\n36 20\n17 47\n53 18\n3 50\n30 2\n17 7\n8 58\n59 1\n31 11\n24 26\n24 43\n53 57\n59 45\n47 13\n26 46\n17 33\n30 31\n26 39\n26 19\n24 36\n8 49\n38 14\n",
"7\n1 2\n2 3\n3 4\n3 5\n1 6\n1 7\n",
"20\n19 16\n19 18\n20 7\n9 4\n6 17\n14 2\n9 15\n2 13\n5 11\n19 12\n12 20\n16 9\n11 8\n19 5\n3 1\n19 14\n5 3\n12 10\n19 6\n",
"7\n1 2\n1 3\n2 4\n2 5\n3 6\n3 7\n",
"10\n9 5\n7 1\n9 10\n7 2\n5 4\n9 6\n2 9\n10 8\n1 3\n",
"4\n2 4\n2 3\n2 1\n",
"4\n1 4\n3 2\n1 3\n",
"3\n1 2\n1 3\n",
"5\n1 2\n1 5\n1 3\n1 4\n",
"20\n14 9\n12 13\n10 15\n2 1\n20 19\n16 6\n16 3\n17 14\n3 5\n2 11\n3 10\n15 8\n14 2\n6 4\n3 20\n5 18\n1 7\n1 16\n4 12\n",
"20\n7 5\n14 13\n17 6\n3 8\n16 12\n18 9\n3 18\n14 1\n17 3\n15 2\n17 4\n9 11\n2 7\n15 17\n3 20\n16 10\n17 14\n2 16\n1 19\n",
"8\n1 2\n2 3\n3 4\n1 7\n1 8\n4 5\n4 6\n",
"5\n5 1\n5 2\n5 3\n5 4\n",
"50\n49 6\n43 7\n1 27\n19 35\n15 37\n16 12\n19 21\n16 28\n49 9\n48 39\n13 1\n2 48\n9 50\n44 3\n41 32\n48 31\n49 33\n6 11\n13 20\n49 22\n13 41\n48 29\n13 46\n15 47\n34 2\n49 13\n48 14\n34 24\n16 36\n13 40\n49 34\n49 17\n43 25\n11 23\n10 15\n19 26\n34 44\n16 42\n19 18\n46 8\n29 38\n1 45\n12 43\n13 16\n46 30\n15 5\n49 10\n11 19\n32 4\n",
"20\n13 1\n18 2\n3 7\n18 5\n20 16\n3 12\n18 9\n3 10\n18 11\n13 6\n3 18\n20 15\n20 17\n3 13\n3 4\n13 14\n3 20\n18 8\n3 19\n",
"10\n8 2\n5 6\n1 8\n2 9\n1 4\n8 10\n10 5\n2 7\n2 3\n"
],
"output": [
"NO",
"YES",
"NO",
"NO",
"NO",
"YES",
"NO",
"NO",
"NO",
"NO",
"YES",
"NO",
"NO",
"YES",
"NO",
"NO",
"NO",
"YES",
"NO",
"YES",
"NO"
]
}
|
IN-CORRECT
|
java
|
import java.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.StringTokenizer;
import java.io.IOException;
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.InputStream;
/**
* Built using CHelper plug-in
* Actual solution is at the top
*/
public class Main {
public static void main(String[] args) {
InputStream inputStream = System.in;
OutputStream outputStream = System.out;
InputReader sc = new InputReader(inputStream);
PrintWriter out = new PrintWriter(outputStream);
Task solver = new Task();
solver.solve(1, sc, out);
out.close();
}
static class Task {
public void solve(int testNumber, InputReader sc, PrintWriter out) {
int n=sc.nextInt();
ArrayList<Integer>[] map=new ArrayList[n+1];
for(int i=1;i<=n;i++)
map[i]=new ArrayList<Integer>();
for(int i=0;i<n-1;i++) {
int u=sc.nextInt();
int v=sc.nextInt();
map[u].add(v);
}
if(n==2) {
out.println("YES");
return ;
}
boolean[] jud=new boolean[n+1];
for(int i=1;i<=n;i++) {
if(map[i].size()==0)
jud[i]=true;
}
for(int i=1;i<=n;i++) {
if(!jud[i]&&map[i].size()<=2) {
out.println("NO");
return ;
}
}
out.println("YES");
}
}
static class InputReader {
public BufferedReader reader;
public StringTokenizer tokenizer;
public InputReader(InputStream stream) {
reader = new BufferedReader(new InputStreamReader(stream), 32768);
tokenizer = null;
}
public String next() {
while (tokenizer == null || !tokenizer.hasMoreTokens()) {
try {
tokenizer = new StringTokenizer(reader.readLine());
} catch (IOException e) {
throw new RuntimeException(e);
}
}
return tokenizer.nextToken();
}
public int nextInt() {
return Integer.parseInt(next());
}
public long nextLong() {
return Long.parseLong(next());
}
}
}
|
1189_D1. Add on a Tree
|
Note that this is the first problem of the two similar problems. You can hack this problem only if you solve both problems.
You are given a tree with n nodes. In the beginning, 0 is written on all edges. In one operation, you can choose any 2 distinct leaves u, v and any real number x and add x to values written on all edges on the simple path between u and v.
For example, on the picture below you can see the result of applying two operations to the graph: adding 2 on the path from 7 to 6, and then adding -0.5 on the path from 4 to 5.
<image>
Is it true that for any configuration of real numbers written on edges, we can achieve it with a finite number of operations?
Leaf is a node of a tree of degree 1. Simple path is a path that doesn't contain any node twice.
Input
The first line contains a single integer n (2 ≤ n ≤ 10^5) — the number of nodes.
Each of the next n-1 lines contains two integers u and v (1 ≤ u, v ≤ n, u ≠ v), meaning that there is an edge between nodes u and v. It is guaranteed that these edges form a tree.
Output
If there is a configuration of real numbers written on edges of the tree that we can't achieve by performing the operations, output "NO".
Otherwise, output "YES".
You can print each letter in any case (upper or lower).
Examples
Input
2
1 2
Output
YES
Input
3
1 2
2 3
Output
NO
Input
5
1 2
1 3
1 4
2 5
Output
NO
Input
6
1 2
1 3
1 4
2 5
2 6
Output
YES
Note
In the first example, we can add any real x to the value written on the only edge (1, 2).
<image>
In the second example, one of configurations that we can't reach is 0 written on (1, 2) and 1 written on (2, 3).
<image>
Below you can see graphs from examples 3, 4:
<image> <image>
|
{
"input": [
"2\n1 2\n",
"3\n1 2\n2 3\n",
"5\n1 2\n1 3\n1 4\n2 5\n",
"6\n1 2\n1 3\n1 4\n2 5\n2 6\n"
],
"output": [
"YES",
"NO",
"NO",
"YES"
]
}
|
{
"input": [
"50\n16 4\n17 9\n31 19\n22 10\n8 1\n40 30\n3 31\n20 29\n47 27\n22 25\n32 34\n12 15\n40 32\n10 33\n47 12\n6 24\n46 41\n14 23\n12 35\n31 42\n46 28\n31 20\n46 37\n1 39\n29 49\n37 47\n40 6\n42 36\n47 2\n24 46\n2 13\n8 45\n41 3\n32 17\n4 7\n47 26\n28 8\n41 50\n34 44\n33 21\n25 5\n16 40\n3 14\n8 18\n28 11\n32 22\n2 38\n3 48\n44 43\n",
"10\n8 1\n1 2\n8 9\n8 5\n1 3\n1 10\n1 6\n1 7\n8 4\n",
"5\n5 1\n5 4\n4 3\n1 2\n",
"7\n1 2\n2 3\n1 4\n1 5\n3 6\n3 7\n",
"3\n1 3\n2 3\n",
"60\n26 6\n59 30\n31 12\n31 3\n38 23\n59 29\n53 9\n38 56\n53 54\n29 21\n17 55\n59 38\n26 16\n24 59\n24 25\n17 35\n24 41\n30 15\n31 27\n8 44\n26 5\n26 48\n8 32\n53 17\n3 34\n3 51\n30 28\n47 10\n53 60\n36 42\n24 53\n59 22\n53 40\n26 52\n36 4\n59 8\n29 37\n36 20\n17 47\n53 18\n3 50\n30 2\n17 7\n8 58\n59 1\n31 11\n24 26\n24 43\n53 57\n59 45\n47 13\n26 46\n17 33\n30 31\n26 39\n26 19\n24 36\n8 49\n38 14\n",
"7\n1 2\n2 3\n3 4\n3 5\n1 6\n1 7\n",
"20\n19 16\n19 18\n20 7\n9 4\n6 17\n14 2\n9 15\n2 13\n5 11\n19 12\n12 20\n16 9\n11 8\n19 5\n3 1\n19 14\n5 3\n12 10\n19 6\n",
"7\n1 2\n1 3\n2 4\n2 5\n3 6\n3 7\n",
"10\n9 5\n7 1\n9 10\n7 2\n5 4\n9 6\n2 9\n10 8\n1 3\n",
"4\n2 4\n2 3\n2 1\n",
"4\n1 4\n3 2\n1 3\n",
"3\n1 2\n1 3\n",
"5\n1 2\n1 5\n1 3\n1 4\n",
"20\n14 9\n12 13\n10 15\n2 1\n20 19\n16 6\n16 3\n17 14\n3 5\n2 11\n3 10\n15 8\n14 2\n6 4\n3 20\n5 18\n1 7\n1 16\n4 12\n",
"20\n7 5\n14 13\n17 6\n3 8\n16 12\n18 9\n3 18\n14 1\n17 3\n15 2\n17 4\n9 11\n2 7\n15 17\n3 20\n16 10\n17 14\n2 16\n1 19\n",
"8\n1 2\n2 3\n3 4\n1 7\n1 8\n4 5\n4 6\n",
"5\n5 1\n5 2\n5 3\n5 4\n",
"50\n49 6\n43 7\n1 27\n19 35\n15 37\n16 12\n19 21\n16 28\n49 9\n48 39\n13 1\n2 48\n9 50\n44 3\n41 32\n48 31\n49 33\n6 11\n13 20\n49 22\n13 41\n48 29\n13 46\n15 47\n34 2\n49 13\n48 14\n34 24\n16 36\n13 40\n49 34\n49 17\n43 25\n11 23\n10 15\n19 26\n34 44\n16 42\n19 18\n46 8\n29 38\n1 45\n12 43\n13 16\n46 30\n15 5\n49 10\n11 19\n32 4\n",
"20\n13 1\n18 2\n3 7\n18 5\n20 16\n3 12\n18 9\n3 10\n18 11\n13 6\n3 18\n20 15\n20 17\n3 13\n3 4\n13 14\n3 20\n18 8\n3 19\n",
"10\n8 2\n5 6\n1 8\n2 9\n1 4\n8 10\n10 5\n2 7\n2 3\n"
],
"output": [
"NO",
"YES",
"NO",
"NO",
"NO",
"YES",
"NO",
"NO",
"NO",
"NO",
"YES",
"NO",
"NO",
"YES",
"NO",
"NO",
"NO",
"YES",
"NO",
"YES",
"NO"
]
}
|
IN-CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
vector<long long> adj[200005];
long long c = 0;
bool vis[200005];
vector<long long> lf;
void dfs(long long v) {
if (vis[v]) {
return;
}
vis[v] = true;
if (adj[v].size() == 1) {
c++;
lf.push_back(v);
}
for (auto u : adj[v]) {
dfs(u);
}
}
int main() {
long long n;
cin >> n;
long long x, y;
for (int i = 0; i < n - 1; i++) {
cin >> x >> y;
adj[x].push_back(y);
adj[y].push_back(x);
}
dfs(x);
for (auto u : lf) {
if (adj[adj[u][0]].size() == 2) {
cout << "NO";
return 0;
}
}
cout << "YES";
return 0;
}
|
1189_D1. Add on a Tree
|
Note that this is the first problem of the two similar problems. You can hack this problem only if you solve both problems.
You are given a tree with n nodes. In the beginning, 0 is written on all edges. In one operation, you can choose any 2 distinct leaves u, v and any real number x and add x to values written on all edges on the simple path between u and v.
For example, on the picture below you can see the result of applying two operations to the graph: adding 2 on the path from 7 to 6, and then adding -0.5 on the path from 4 to 5.
<image>
Is it true that for any configuration of real numbers written on edges, we can achieve it with a finite number of operations?
Leaf is a node of a tree of degree 1. Simple path is a path that doesn't contain any node twice.
Input
The first line contains a single integer n (2 ≤ n ≤ 10^5) — the number of nodes.
Each of the next n-1 lines contains two integers u and v (1 ≤ u, v ≤ n, u ≠ v), meaning that there is an edge between nodes u and v. It is guaranteed that these edges form a tree.
Output
If there is a configuration of real numbers written on edges of the tree that we can't achieve by performing the operations, output "NO".
Otherwise, output "YES".
You can print each letter in any case (upper or lower).
Examples
Input
2
1 2
Output
YES
Input
3
1 2
2 3
Output
NO
Input
5
1 2
1 3
1 4
2 5
Output
NO
Input
6
1 2
1 3
1 4
2 5
2 6
Output
YES
Note
In the first example, we can add any real x to the value written on the only edge (1, 2).
<image>
In the second example, one of configurations that we can't reach is 0 written on (1, 2) and 1 written on (2, 3).
<image>
Below you can see graphs from examples 3, 4:
<image> <image>
|
{
"input": [
"2\n1 2\n",
"3\n1 2\n2 3\n",
"5\n1 2\n1 3\n1 4\n2 5\n",
"6\n1 2\n1 3\n1 4\n2 5\n2 6\n"
],
"output": [
"YES",
"NO",
"NO",
"YES"
]
}
|
{
"input": [
"50\n16 4\n17 9\n31 19\n22 10\n8 1\n40 30\n3 31\n20 29\n47 27\n22 25\n32 34\n12 15\n40 32\n10 33\n47 12\n6 24\n46 41\n14 23\n12 35\n31 42\n46 28\n31 20\n46 37\n1 39\n29 49\n37 47\n40 6\n42 36\n47 2\n24 46\n2 13\n8 45\n41 3\n32 17\n4 7\n47 26\n28 8\n41 50\n34 44\n33 21\n25 5\n16 40\n3 14\n8 18\n28 11\n32 22\n2 38\n3 48\n44 43\n",
"10\n8 1\n1 2\n8 9\n8 5\n1 3\n1 10\n1 6\n1 7\n8 4\n",
"5\n5 1\n5 4\n4 3\n1 2\n",
"7\n1 2\n2 3\n1 4\n1 5\n3 6\n3 7\n",
"3\n1 3\n2 3\n",
"60\n26 6\n59 30\n31 12\n31 3\n38 23\n59 29\n53 9\n38 56\n53 54\n29 21\n17 55\n59 38\n26 16\n24 59\n24 25\n17 35\n24 41\n30 15\n31 27\n8 44\n26 5\n26 48\n8 32\n53 17\n3 34\n3 51\n30 28\n47 10\n53 60\n36 42\n24 53\n59 22\n53 40\n26 52\n36 4\n59 8\n29 37\n36 20\n17 47\n53 18\n3 50\n30 2\n17 7\n8 58\n59 1\n31 11\n24 26\n24 43\n53 57\n59 45\n47 13\n26 46\n17 33\n30 31\n26 39\n26 19\n24 36\n8 49\n38 14\n",
"7\n1 2\n2 3\n3 4\n3 5\n1 6\n1 7\n",
"20\n19 16\n19 18\n20 7\n9 4\n6 17\n14 2\n9 15\n2 13\n5 11\n19 12\n12 20\n16 9\n11 8\n19 5\n3 1\n19 14\n5 3\n12 10\n19 6\n",
"7\n1 2\n1 3\n2 4\n2 5\n3 6\n3 7\n",
"10\n9 5\n7 1\n9 10\n7 2\n5 4\n9 6\n2 9\n10 8\n1 3\n",
"4\n2 4\n2 3\n2 1\n",
"4\n1 4\n3 2\n1 3\n",
"3\n1 2\n1 3\n",
"5\n1 2\n1 5\n1 3\n1 4\n",
"20\n14 9\n12 13\n10 15\n2 1\n20 19\n16 6\n16 3\n17 14\n3 5\n2 11\n3 10\n15 8\n14 2\n6 4\n3 20\n5 18\n1 7\n1 16\n4 12\n",
"20\n7 5\n14 13\n17 6\n3 8\n16 12\n18 9\n3 18\n14 1\n17 3\n15 2\n17 4\n9 11\n2 7\n15 17\n3 20\n16 10\n17 14\n2 16\n1 19\n",
"8\n1 2\n2 3\n3 4\n1 7\n1 8\n4 5\n4 6\n",
"5\n5 1\n5 2\n5 3\n5 4\n",
"50\n49 6\n43 7\n1 27\n19 35\n15 37\n16 12\n19 21\n16 28\n49 9\n48 39\n13 1\n2 48\n9 50\n44 3\n41 32\n48 31\n49 33\n6 11\n13 20\n49 22\n13 41\n48 29\n13 46\n15 47\n34 2\n49 13\n48 14\n34 24\n16 36\n13 40\n49 34\n49 17\n43 25\n11 23\n10 15\n19 26\n34 44\n16 42\n19 18\n46 8\n29 38\n1 45\n12 43\n13 16\n46 30\n15 5\n49 10\n11 19\n32 4\n",
"20\n13 1\n18 2\n3 7\n18 5\n20 16\n3 12\n18 9\n3 10\n18 11\n13 6\n3 18\n20 15\n20 17\n3 13\n3 4\n13 14\n3 20\n18 8\n3 19\n",
"10\n8 2\n5 6\n1 8\n2 9\n1 4\n8 10\n10 5\n2 7\n2 3\n"
],
"output": [
"NO",
"YES",
"NO",
"NO",
"NO",
"YES",
"NO",
"NO",
"NO",
"NO",
"YES",
"NO",
"NO",
"YES",
"NO",
"NO",
"NO",
"YES",
"NO",
"YES",
"NO"
]
}
|
IN-CORRECT
|
python3
|
n = int(input())
l = [0 for i in range(n)]
for _ in range(n-1):
a, b = list(map(int, input().split()))
l[a-1] += 1
l[b-1] += 1
for i in range(n):
if (l[i]&1 == 0):
print('NO')
break
else:
print('YES')
|
1189_D1. Add on a Tree
|
Note that this is the first problem of the two similar problems. You can hack this problem only if you solve both problems.
You are given a tree with n nodes. In the beginning, 0 is written on all edges. In one operation, you can choose any 2 distinct leaves u, v and any real number x and add x to values written on all edges on the simple path between u and v.
For example, on the picture below you can see the result of applying two operations to the graph: adding 2 on the path from 7 to 6, and then adding -0.5 on the path from 4 to 5.
<image>
Is it true that for any configuration of real numbers written on edges, we can achieve it with a finite number of operations?
Leaf is a node of a tree of degree 1. Simple path is a path that doesn't contain any node twice.
Input
The first line contains a single integer n (2 ≤ n ≤ 10^5) — the number of nodes.
Each of the next n-1 lines contains two integers u and v (1 ≤ u, v ≤ n, u ≠ v), meaning that there is an edge between nodes u and v. It is guaranteed that these edges form a tree.
Output
If there is a configuration of real numbers written on edges of the tree that we can't achieve by performing the operations, output "NO".
Otherwise, output "YES".
You can print each letter in any case (upper or lower).
Examples
Input
2
1 2
Output
YES
Input
3
1 2
2 3
Output
NO
Input
5
1 2
1 3
1 4
2 5
Output
NO
Input
6
1 2
1 3
1 4
2 5
2 6
Output
YES
Note
In the first example, we can add any real x to the value written on the only edge (1, 2).
<image>
In the second example, one of configurations that we can't reach is 0 written on (1, 2) and 1 written on (2, 3).
<image>
Below you can see graphs from examples 3, 4:
<image> <image>
|
{
"input": [
"2\n1 2\n",
"3\n1 2\n2 3\n",
"5\n1 2\n1 3\n1 4\n2 5\n",
"6\n1 2\n1 3\n1 4\n2 5\n2 6\n"
],
"output": [
"YES",
"NO",
"NO",
"YES"
]
}
|
{
"input": [
"50\n16 4\n17 9\n31 19\n22 10\n8 1\n40 30\n3 31\n20 29\n47 27\n22 25\n32 34\n12 15\n40 32\n10 33\n47 12\n6 24\n46 41\n14 23\n12 35\n31 42\n46 28\n31 20\n46 37\n1 39\n29 49\n37 47\n40 6\n42 36\n47 2\n24 46\n2 13\n8 45\n41 3\n32 17\n4 7\n47 26\n28 8\n41 50\n34 44\n33 21\n25 5\n16 40\n3 14\n8 18\n28 11\n32 22\n2 38\n3 48\n44 43\n",
"10\n8 1\n1 2\n8 9\n8 5\n1 3\n1 10\n1 6\n1 7\n8 4\n",
"5\n5 1\n5 4\n4 3\n1 2\n",
"7\n1 2\n2 3\n1 4\n1 5\n3 6\n3 7\n",
"3\n1 3\n2 3\n",
"60\n26 6\n59 30\n31 12\n31 3\n38 23\n59 29\n53 9\n38 56\n53 54\n29 21\n17 55\n59 38\n26 16\n24 59\n24 25\n17 35\n24 41\n30 15\n31 27\n8 44\n26 5\n26 48\n8 32\n53 17\n3 34\n3 51\n30 28\n47 10\n53 60\n36 42\n24 53\n59 22\n53 40\n26 52\n36 4\n59 8\n29 37\n36 20\n17 47\n53 18\n3 50\n30 2\n17 7\n8 58\n59 1\n31 11\n24 26\n24 43\n53 57\n59 45\n47 13\n26 46\n17 33\n30 31\n26 39\n26 19\n24 36\n8 49\n38 14\n",
"7\n1 2\n2 3\n3 4\n3 5\n1 6\n1 7\n",
"20\n19 16\n19 18\n20 7\n9 4\n6 17\n14 2\n9 15\n2 13\n5 11\n19 12\n12 20\n16 9\n11 8\n19 5\n3 1\n19 14\n5 3\n12 10\n19 6\n",
"7\n1 2\n1 3\n2 4\n2 5\n3 6\n3 7\n",
"10\n9 5\n7 1\n9 10\n7 2\n5 4\n9 6\n2 9\n10 8\n1 3\n",
"4\n2 4\n2 3\n2 1\n",
"4\n1 4\n3 2\n1 3\n",
"3\n1 2\n1 3\n",
"5\n1 2\n1 5\n1 3\n1 4\n",
"20\n14 9\n12 13\n10 15\n2 1\n20 19\n16 6\n16 3\n17 14\n3 5\n2 11\n3 10\n15 8\n14 2\n6 4\n3 20\n5 18\n1 7\n1 16\n4 12\n",
"20\n7 5\n14 13\n17 6\n3 8\n16 12\n18 9\n3 18\n14 1\n17 3\n15 2\n17 4\n9 11\n2 7\n15 17\n3 20\n16 10\n17 14\n2 16\n1 19\n",
"8\n1 2\n2 3\n3 4\n1 7\n1 8\n4 5\n4 6\n",
"5\n5 1\n5 2\n5 3\n5 4\n",
"50\n49 6\n43 7\n1 27\n19 35\n15 37\n16 12\n19 21\n16 28\n49 9\n48 39\n13 1\n2 48\n9 50\n44 3\n41 32\n48 31\n49 33\n6 11\n13 20\n49 22\n13 41\n48 29\n13 46\n15 47\n34 2\n49 13\n48 14\n34 24\n16 36\n13 40\n49 34\n49 17\n43 25\n11 23\n10 15\n19 26\n34 44\n16 42\n19 18\n46 8\n29 38\n1 45\n12 43\n13 16\n46 30\n15 5\n49 10\n11 19\n32 4\n",
"20\n13 1\n18 2\n3 7\n18 5\n20 16\n3 12\n18 9\n3 10\n18 11\n13 6\n3 18\n20 15\n20 17\n3 13\n3 4\n13 14\n3 20\n18 8\n3 19\n",
"10\n8 2\n5 6\n1 8\n2 9\n1 4\n8 10\n10 5\n2 7\n2 3\n"
],
"output": [
"NO",
"YES",
"NO",
"NO",
"NO",
"YES",
"NO",
"NO",
"NO",
"NO",
"YES",
"NO",
"NO",
"YES",
"NO",
"NO",
"NO",
"YES",
"NO",
"YES",
"NO"
]
}
|
IN-CORRECT
|
cpp
|
#include <bits/stdc++.h>
int main() {
long long in[100005];
long long n, i, j, m, t;
memset(in, 0, sizeof(in));
scanf("%lld", &n);
for (i = 1; i <= n - 1; i++) {
long long u, v;
scanf("%lld %lld", &u, &v);
in[u]++;
in[v]++;
}
long long num = 0;
long long ans = 0;
for (i = 1; i <= n; i++)
if (in[i] == 1) num++;
num--;
ans = (1 + num) * num / 2;
if (ans >= n - 1)
printf("YES\n");
else
printf("NO\n");
return 0;
}
|
1189_D1. Add on a Tree
|
Note that this is the first problem of the two similar problems. You can hack this problem only if you solve both problems.
You are given a tree with n nodes. In the beginning, 0 is written on all edges. In one operation, you can choose any 2 distinct leaves u, v and any real number x and add x to values written on all edges on the simple path between u and v.
For example, on the picture below you can see the result of applying two operations to the graph: adding 2 on the path from 7 to 6, and then adding -0.5 on the path from 4 to 5.
<image>
Is it true that for any configuration of real numbers written on edges, we can achieve it with a finite number of operations?
Leaf is a node of a tree of degree 1. Simple path is a path that doesn't contain any node twice.
Input
The first line contains a single integer n (2 ≤ n ≤ 10^5) — the number of nodes.
Each of the next n-1 lines contains two integers u and v (1 ≤ u, v ≤ n, u ≠ v), meaning that there is an edge between nodes u and v. It is guaranteed that these edges form a tree.
Output
If there is a configuration of real numbers written on edges of the tree that we can't achieve by performing the operations, output "NO".
Otherwise, output "YES".
You can print each letter in any case (upper or lower).
Examples
Input
2
1 2
Output
YES
Input
3
1 2
2 3
Output
NO
Input
5
1 2
1 3
1 4
2 5
Output
NO
Input
6
1 2
1 3
1 4
2 5
2 6
Output
YES
Note
In the first example, we can add any real x to the value written on the only edge (1, 2).
<image>
In the second example, one of configurations that we can't reach is 0 written on (1, 2) and 1 written on (2, 3).
<image>
Below you can see graphs from examples 3, 4:
<image> <image>
|
{
"input": [
"2\n1 2\n",
"3\n1 2\n2 3\n",
"5\n1 2\n1 3\n1 4\n2 5\n",
"6\n1 2\n1 3\n1 4\n2 5\n2 6\n"
],
"output": [
"YES",
"NO",
"NO",
"YES"
]
}
|
{
"input": [
"50\n16 4\n17 9\n31 19\n22 10\n8 1\n40 30\n3 31\n20 29\n47 27\n22 25\n32 34\n12 15\n40 32\n10 33\n47 12\n6 24\n46 41\n14 23\n12 35\n31 42\n46 28\n31 20\n46 37\n1 39\n29 49\n37 47\n40 6\n42 36\n47 2\n24 46\n2 13\n8 45\n41 3\n32 17\n4 7\n47 26\n28 8\n41 50\n34 44\n33 21\n25 5\n16 40\n3 14\n8 18\n28 11\n32 22\n2 38\n3 48\n44 43\n",
"10\n8 1\n1 2\n8 9\n8 5\n1 3\n1 10\n1 6\n1 7\n8 4\n",
"5\n5 1\n5 4\n4 3\n1 2\n",
"7\n1 2\n2 3\n1 4\n1 5\n3 6\n3 7\n",
"3\n1 3\n2 3\n",
"60\n26 6\n59 30\n31 12\n31 3\n38 23\n59 29\n53 9\n38 56\n53 54\n29 21\n17 55\n59 38\n26 16\n24 59\n24 25\n17 35\n24 41\n30 15\n31 27\n8 44\n26 5\n26 48\n8 32\n53 17\n3 34\n3 51\n30 28\n47 10\n53 60\n36 42\n24 53\n59 22\n53 40\n26 52\n36 4\n59 8\n29 37\n36 20\n17 47\n53 18\n3 50\n30 2\n17 7\n8 58\n59 1\n31 11\n24 26\n24 43\n53 57\n59 45\n47 13\n26 46\n17 33\n30 31\n26 39\n26 19\n24 36\n8 49\n38 14\n",
"7\n1 2\n2 3\n3 4\n3 5\n1 6\n1 7\n",
"20\n19 16\n19 18\n20 7\n9 4\n6 17\n14 2\n9 15\n2 13\n5 11\n19 12\n12 20\n16 9\n11 8\n19 5\n3 1\n19 14\n5 3\n12 10\n19 6\n",
"7\n1 2\n1 3\n2 4\n2 5\n3 6\n3 7\n",
"10\n9 5\n7 1\n9 10\n7 2\n5 4\n9 6\n2 9\n10 8\n1 3\n",
"4\n2 4\n2 3\n2 1\n",
"4\n1 4\n3 2\n1 3\n",
"3\n1 2\n1 3\n",
"5\n1 2\n1 5\n1 3\n1 4\n",
"20\n14 9\n12 13\n10 15\n2 1\n20 19\n16 6\n16 3\n17 14\n3 5\n2 11\n3 10\n15 8\n14 2\n6 4\n3 20\n5 18\n1 7\n1 16\n4 12\n",
"20\n7 5\n14 13\n17 6\n3 8\n16 12\n18 9\n3 18\n14 1\n17 3\n15 2\n17 4\n9 11\n2 7\n15 17\n3 20\n16 10\n17 14\n2 16\n1 19\n",
"8\n1 2\n2 3\n3 4\n1 7\n1 8\n4 5\n4 6\n",
"5\n5 1\n5 2\n5 3\n5 4\n",
"50\n49 6\n43 7\n1 27\n19 35\n15 37\n16 12\n19 21\n16 28\n49 9\n48 39\n13 1\n2 48\n9 50\n44 3\n41 32\n48 31\n49 33\n6 11\n13 20\n49 22\n13 41\n48 29\n13 46\n15 47\n34 2\n49 13\n48 14\n34 24\n16 36\n13 40\n49 34\n49 17\n43 25\n11 23\n10 15\n19 26\n34 44\n16 42\n19 18\n46 8\n29 38\n1 45\n12 43\n13 16\n46 30\n15 5\n49 10\n11 19\n32 4\n",
"20\n13 1\n18 2\n3 7\n18 5\n20 16\n3 12\n18 9\n3 10\n18 11\n13 6\n3 18\n20 15\n20 17\n3 13\n3 4\n13 14\n3 20\n18 8\n3 19\n",
"10\n8 2\n5 6\n1 8\n2 9\n1 4\n8 10\n10 5\n2 7\n2 3\n"
],
"output": [
"NO",
"YES",
"NO",
"NO",
"NO",
"YES",
"NO",
"NO",
"NO",
"NO",
"YES",
"NO",
"NO",
"YES",
"NO",
"NO",
"NO",
"YES",
"NO",
"YES",
"NO"
]
}
|
IN-CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
int32_t main() {
long long n;
cin >> n;
long long u, v;
for (long long i = 1; i < n; i++) cin >> u >> v;
if (n % 2 == 0)
cout << "YES\n";
else
cout << "NO\n";
}
|
1189_D1. Add on a Tree
|
Note that this is the first problem of the two similar problems. You can hack this problem only if you solve both problems.
You are given a tree with n nodes. In the beginning, 0 is written on all edges. In one operation, you can choose any 2 distinct leaves u, v and any real number x and add x to values written on all edges on the simple path between u and v.
For example, on the picture below you can see the result of applying two operations to the graph: adding 2 on the path from 7 to 6, and then adding -0.5 on the path from 4 to 5.
<image>
Is it true that for any configuration of real numbers written on edges, we can achieve it with a finite number of operations?
Leaf is a node of a tree of degree 1. Simple path is a path that doesn't contain any node twice.
Input
The first line contains a single integer n (2 ≤ n ≤ 10^5) — the number of nodes.
Each of the next n-1 lines contains two integers u and v (1 ≤ u, v ≤ n, u ≠ v), meaning that there is an edge between nodes u and v. It is guaranteed that these edges form a tree.
Output
If there is a configuration of real numbers written on edges of the tree that we can't achieve by performing the operations, output "NO".
Otherwise, output "YES".
You can print each letter in any case (upper or lower).
Examples
Input
2
1 2
Output
YES
Input
3
1 2
2 3
Output
NO
Input
5
1 2
1 3
1 4
2 5
Output
NO
Input
6
1 2
1 3
1 4
2 5
2 6
Output
YES
Note
In the first example, we can add any real x to the value written on the only edge (1, 2).
<image>
In the second example, one of configurations that we can't reach is 0 written on (1, 2) and 1 written on (2, 3).
<image>
Below you can see graphs from examples 3, 4:
<image> <image>
|
{
"input": [
"2\n1 2\n",
"3\n1 2\n2 3\n",
"5\n1 2\n1 3\n1 4\n2 5\n",
"6\n1 2\n1 3\n1 4\n2 5\n2 6\n"
],
"output": [
"YES",
"NO",
"NO",
"YES"
]
}
|
{
"input": [
"50\n16 4\n17 9\n31 19\n22 10\n8 1\n40 30\n3 31\n20 29\n47 27\n22 25\n32 34\n12 15\n40 32\n10 33\n47 12\n6 24\n46 41\n14 23\n12 35\n31 42\n46 28\n31 20\n46 37\n1 39\n29 49\n37 47\n40 6\n42 36\n47 2\n24 46\n2 13\n8 45\n41 3\n32 17\n4 7\n47 26\n28 8\n41 50\n34 44\n33 21\n25 5\n16 40\n3 14\n8 18\n28 11\n32 22\n2 38\n3 48\n44 43\n",
"10\n8 1\n1 2\n8 9\n8 5\n1 3\n1 10\n1 6\n1 7\n8 4\n",
"5\n5 1\n5 4\n4 3\n1 2\n",
"7\n1 2\n2 3\n1 4\n1 5\n3 6\n3 7\n",
"3\n1 3\n2 3\n",
"60\n26 6\n59 30\n31 12\n31 3\n38 23\n59 29\n53 9\n38 56\n53 54\n29 21\n17 55\n59 38\n26 16\n24 59\n24 25\n17 35\n24 41\n30 15\n31 27\n8 44\n26 5\n26 48\n8 32\n53 17\n3 34\n3 51\n30 28\n47 10\n53 60\n36 42\n24 53\n59 22\n53 40\n26 52\n36 4\n59 8\n29 37\n36 20\n17 47\n53 18\n3 50\n30 2\n17 7\n8 58\n59 1\n31 11\n24 26\n24 43\n53 57\n59 45\n47 13\n26 46\n17 33\n30 31\n26 39\n26 19\n24 36\n8 49\n38 14\n",
"7\n1 2\n2 3\n3 4\n3 5\n1 6\n1 7\n",
"20\n19 16\n19 18\n20 7\n9 4\n6 17\n14 2\n9 15\n2 13\n5 11\n19 12\n12 20\n16 9\n11 8\n19 5\n3 1\n19 14\n5 3\n12 10\n19 6\n",
"7\n1 2\n1 3\n2 4\n2 5\n3 6\n3 7\n",
"10\n9 5\n7 1\n9 10\n7 2\n5 4\n9 6\n2 9\n10 8\n1 3\n",
"4\n2 4\n2 3\n2 1\n",
"4\n1 4\n3 2\n1 3\n",
"3\n1 2\n1 3\n",
"5\n1 2\n1 5\n1 3\n1 4\n",
"20\n14 9\n12 13\n10 15\n2 1\n20 19\n16 6\n16 3\n17 14\n3 5\n2 11\n3 10\n15 8\n14 2\n6 4\n3 20\n5 18\n1 7\n1 16\n4 12\n",
"20\n7 5\n14 13\n17 6\n3 8\n16 12\n18 9\n3 18\n14 1\n17 3\n15 2\n17 4\n9 11\n2 7\n15 17\n3 20\n16 10\n17 14\n2 16\n1 19\n",
"8\n1 2\n2 3\n3 4\n1 7\n1 8\n4 5\n4 6\n",
"5\n5 1\n5 2\n5 3\n5 4\n",
"50\n49 6\n43 7\n1 27\n19 35\n15 37\n16 12\n19 21\n16 28\n49 9\n48 39\n13 1\n2 48\n9 50\n44 3\n41 32\n48 31\n49 33\n6 11\n13 20\n49 22\n13 41\n48 29\n13 46\n15 47\n34 2\n49 13\n48 14\n34 24\n16 36\n13 40\n49 34\n49 17\n43 25\n11 23\n10 15\n19 26\n34 44\n16 42\n19 18\n46 8\n29 38\n1 45\n12 43\n13 16\n46 30\n15 5\n49 10\n11 19\n32 4\n",
"20\n13 1\n18 2\n3 7\n18 5\n20 16\n3 12\n18 9\n3 10\n18 11\n13 6\n3 18\n20 15\n20 17\n3 13\n3 4\n13 14\n3 20\n18 8\n3 19\n",
"10\n8 2\n5 6\n1 8\n2 9\n1 4\n8 10\n10 5\n2 7\n2 3\n"
],
"output": [
"NO",
"YES",
"NO",
"NO",
"NO",
"YES",
"NO",
"NO",
"NO",
"NO",
"YES",
"NO",
"NO",
"YES",
"NO",
"NO",
"NO",
"YES",
"NO",
"YES",
"NO"
]
}
|
IN-CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
const long long max_n = 1e6 + 20;
long long n, m, k, ans, sum;
long long a[max_n];
long long mark[max_n];
vector<long long> v, adj[max_n], jda[max_n];
void dfs(long long v) {
mark[v] = 1;
if (jda[v].size() == 1 && adj[v].size() == 1) ans = -1;
if (jda[v].size() == 2 && jda[jda[v][0]].size() == 0 &&
jda[jda[v][1]].size() == 0 && adj[v].size() == 0)
ans = -1;
for (auto i : adj[v]) {
if (!mark[i]) {
dfs(i);
}
}
}
int32_t main() {
cin >> n;
for (long long i = 1; i < n; i++) {
long long u, v;
cin >> u >> v;
u--, v--;
adj[v].push_back(u);
jda[u].push_back(v);
}
if (n == 2) {
cout << "YES";
return 0;
}
for (long long i = 0; i < n; i++) {
if (jda[i].size() == 0) {
dfs(i);
}
}
if (ans == -1) {
cout << "NO";
} else
cout << "YES";
return 0;
}
|
1189_D1. Add on a Tree
|
Note that this is the first problem of the two similar problems. You can hack this problem only if you solve both problems.
You are given a tree with n nodes. In the beginning, 0 is written on all edges. In one operation, you can choose any 2 distinct leaves u, v and any real number x and add x to values written on all edges on the simple path between u and v.
For example, on the picture below you can see the result of applying two operations to the graph: adding 2 on the path from 7 to 6, and then adding -0.5 on the path from 4 to 5.
<image>
Is it true that for any configuration of real numbers written on edges, we can achieve it with a finite number of operations?
Leaf is a node of a tree of degree 1. Simple path is a path that doesn't contain any node twice.
Input
The first line contains a single integer n (2 ≤ n ≤ 10^5) — the number of nodes.
Each of the next n-1 lines contains two integers u and v (1 ≤ u, v ≤ n, u ≠ v), meaning that there is an edge between nodes u and v. It is guaranteed that these edges form a tree.
Output
If there is a configuration of real numbers written on edges of the tree that we can't achieve by performing the operations, output "NO".
Otherwise, output "YES".
You can print each letter in any case (upper or lower).
Examples
Input
2
1 2
Output
YES
Input
3
1 2
2 3
Output
NO
Input
5
1 2
1 3
1 4
2 5
Output
NO
Input
6
1 2
1 3
1 4
2 5
2 6
Output
YES
Note
In the first example, we can add any real x to the value written on the only edge (1, 2).
<image>
In the second example, one of configurations that we can't reach is 0 written on (1, 2) and 1 written on (2, 3).
<image>
Below you can see graphs from examples 3, 4:
<image> <image>
|
{
"input": [
"2\n1 2\n",
"3\n1 2\n2 3\n",
"5\n1 2\n1 3\n1 4\n2 5\n",
"6\n1 2\n1 3\n1 4\n2 5\n2 6\n"
],
"output": [
"YES",
"NO",
"NO",
"YES"
]
}
|
{
"input": [
"50\n16 4\n17 9\n31 19\n22 10\n8 1\n40 30\n3 31\n20 29\n47 27\n22 25\n32 34\n12 15\n40 32\n10 33\n47 12\n6 24\n46 41\n14 23\n12 35\n31 42\n46 28\n31 20\n46 37\n1 39\n29 49\n37 47\n40 6\n42 36\n47 2\n24 46\n2 13\n8 45\n41 3\n32 17\n4 7\n47 26\n28 8\n41 50\n34 44\n33 21\n25 5\n16 40\n3 14\n8 18\n28 11\n32 22\n2 38\n3 48\n44 43\n",
"10\n8 1\n1 2\n8 9\n8 5\n1 3\n1 10\n1 6\n1 7\n8 4\n",
"5\n5 1\n5 4\n4 3\n1 2\n",
"7\n1 2\n2 3\n1 4\n1 5\n3 6\n3 7\n",
"3\n1 3\n2 3\n",
"60\n26 6\n59 30\n31 12\n31 3\n38 23\n59 29\n53 9\n38 56\n53 54\n29 21\n17 55\n59 38\n26 16\n24 59\n24 25\n17 35\n24 41\n30 15\n31 27\n8 44\n26 5\n26 48\n8 32\n53 17\n3 34\n3 51\n30 28\n47 10\n53 60\n36 42\n24 53\n59 22\n53 40\n26 52\n36 4\n59 8\n29 37\n36 20\n17 47\n53 18\n3 50\n30 2\n17 7\n8 58\n59 1\n31 11\n24 26\n24 43\n53 57\n59 45\n47 13\n26 46\n17 33\n30 31\n26 39\n26 19\n24 36\n8 49\n38 14\n",
"7\n1 2\n2 3\n3 4\n3 5\n1 6\n1 7\n",
"20\n19 16\n19 18\n20 7\n9 4\n6 17\n14 2\n9 15\n2 13\n5 11\n19 12\n12 20\n16 9\n11 8\n19 5\n3 1\n19 14\n5 3\n12 10\n19 6\n",
"7\n1 2\n1 3\n2 4\n2 5\n3 6\n3 7\n",
"10\n9 5\n7 1\n9 10\n7 2\n5 4\n9 6\n2 9\n10 8\n1 3\n",
"4\n2 4\n2 3\n2 1\n",
"4\n1 4\n3 2\n1 3\n",
"3\n1 2\n1 3\n",
"5\n1 2\n1 5\n1 3\n1 4\n",
"20\n14 9\n12 13\n10 15\n2 1\n20 19\n16 6\n16 3\n17 14\n3 5\n2 11\n3 10\n15 8\n14 2\n6 4\n3 20\n5 18\n1 7\n1 16\n4 12\n",
"20\n7 5\n14 13\n17 6\n3 8\n16 12\n18 9\n3 18\n14 1\n17 3\n15 2\n17 4\n9 11\n2 7\n15 17\n3 20\n16 10\n17 14\n2 16\n1 19\n",
"8\n1 2\n2 3\n3 4\n1 7\n1 8\n4 5\n4 6\n",
"5\n5 1\n5 2\n5 3\n5 4\n",
"50\n49 6\n43 7\n1 27\n19 35\n15 37\n16 12\n19 21\n16 28\n49 9\n48 39\n13 1\n2 48\n9 50\n44 3\n41 32\n48 31\n49 33\n6 11\n13 20\n49 22\n13 41\n48 29\n13 46\n15 47\n34 2\n49 13\n48 14\n34 24\n16 36\n13 40\n49 34\n49 17\n43 25\n11 23\n10 15\n19 26\n34 44\n16 42\n19 18\n46 8\n29 38\n1 45\n12 43\n13 16\n46 30\n15 5\n49 10\n11 19\n32 4\n",
"20\n13 1\n18 2\n3 7\n18 5\n20 16\n3 12\n18 9\n3 10\n18 11\n13 6\n3 18\n20 15\n20 17\n3 13\n3 4\n13 14\n3 20\n18 8\n3 19\n",
"10\n8 2\n5 6\n1 8\n2 9\n1 4\n8 10\n10 5\n2 7\n2 3\n"
],
"output": [
"NO",
"YES",
"NO",
"NO",
"NO",
"YES",
"NO",
"NO",
"NO",
"NO",
"YES",
"NO",
"NO",
"YES",
"NO",
"NO",
"NO",
"YES",
"NO",
"YES",
"NO"
]
}
|
IN-CORRECT
|
python3
|
# https://codeforces.com/contest/1189/problem/D1
n = int(input())
g = {}
p = {}
path = {}
flg = True
for _ in range(n-1):
u,v = map(int, input().split())
if u not in g:
g[u] = []
g[u].append(v)
if v not in g:
g[v] = []
g[v].append(u)
def solve(g, p, path):
if len(g) == 2:
return False
root = None
for u in g:
if len(g[u]) >= 2:
root = u
break
S = [root]
p[root] = 0
i = 0
while i < len(S):
cur = S[i]
for x in g[cur]:
if x == p[cur]:continue
p[x] = cur
S.append(x)
i+=1
for cur in S[::-1]:
if len(g[cur]) == 1:
path[cur] = 0
else:
cnt = 0
for x in g[cur]:
if x == p[cur]:continue
if path[x] == 1:
return False
cnt += 1
path[cur] = cnt
return True
flg = solve(g, p, path)
if flg == True:
print('YES')
else:
print('NO')
|
1189_D1. Add on a Tree
|
Note that this is the first problem of the two similar problems. You can hack this problem only if you solve both problems.
You are given a tree with n nodes. In the beginning, 0 is written on all edges. In one operation, you can choose any 2 distinct leaves u, v and any real number x and add x to values written on all edges on the simple path between u and v.
For example, on the picture below you can see the result of applying two operations to the graph: adding 2 on the path from 7 to 6, and then adding -0.5 on the path from 4 to 5.
<image>
Is it true that for any configuration of real numbers written on edges, we can achieve it with a finite number of operations?
Leaf is a node of a tree of degree 1. Simple path is a path that doesn't contain any node twice.
Input
The first line contains a single integer n (2 ≤ n ≤ 10^5) — the number of nodes.
Each of the next n-1 lines contains two integers u and v (1 ≤ u, v ≤ n, u ≠ v), meaning that there is an edge between nodes u and v. It is guaranteed that these edges form a tree.
Output
If there is a configuration of real numbers written on edges of the tree that we can't achieve by performing the operations, output "NO".
Otherwise, output "YES".
You can print each letter in any case (upper or lower).
Examples
Input
2
1 2
Output
YES
Input
3
1 2
2 3
Output
NO
Input
5
1 2
1 3
1 4
2 5
Output
NO
Input
6
1 2
1 3
1 4
2 5
2 6
Output
YES
Note
In the first example, we can add any real x to the value written on the only edge (1, 2).
<image>
In the second example, one of configurations that we can't reach is 0 written on (1, 2) and 1 written on (2, 3).
<image>
Below you can see graphs from examples 3, 4:
<image> <image>
|
{
"input": [
"2\n1 2\n",
"3\n1 2\n2 3\n",
"5\n1 2\n1 3\n1 4\n2 5\n",
"6\n1 2\n1 3\n1 4\n2 5\n2 6\n"
],
"output": [
"YES",
"NO",
"NO",
"YES"
]
}
|
{
"input": [
"50\n16 4\n17 9\n31 19\n22 10\n8 1\n40 30\n3 31\n20 29\n47 27\n22 25\n32 34\n12 15\n40 32\n10 33\n47 12\n6 24\n46 41\n14 23\n12 35\n31 42\n46 28\n31 20\n46 37\n1 39\n29 49\n37 47\n40 6\n42 36\n47 2\n24 46\n2 13\n8 45\n41 3\n32 17\n4 7\n47 26\n28 8\n41 50\n34 44\n33 21\n25 5\n16 40\n3 14\n8 18\n28 11\n32 22\n2 38\n3 48\n44 43\n",
"10\n8 1\n1 2\n8 9\n8 5\n1 3\n1 10\n1 6\n1 7\n8 4\n",
"5\n5 1\n5 4\n4 3\n1 2\n",
"7\n1 2\n2 3\n1 4\n1 5\n3 6\n3 7\n",
"3\n1 3\n2 3\n",
"60\n26 6\n59 30\n31 12\n31 3\n38 23\n59 29\n53 9\n38 56\n53 54\n29 21\n17 55\n59 38\n26 16\n24 59\n24 25\n17 35\n24 41\n30 15\n31 27\n8 44\n26 5\n26 48\n8 32\n53 17\n3 34\n3 51\n30 28\n47 10\n53 60\n36 42\n24 53\n59 22\n53 40\n26 52\n36 4\n59 8\n29 37\n36 20\n17 47\n53 18\n3 50\n30 2\n17 7\n8 58\n59 1\n31 11\n24 26\n24 43\n53 57\n59 45\n47 13\n26 46\n17 33\n30 31\n26 39\n26 19\n24 36\n8 49\n38 14\n",
"7\n1 2\n2 3\n3 4\n3 5\n1 6\n1 7\n",
"20\n19 16\n19 18\n20 7\n9 4\n6 17\n14 2\n9 15\n2 13\n5 11\n19 12\n12 20\n16 9\n11 8\n19 5\n3 1\n19 14\n5 3\n12 10\n19 6\n",
"7\n1 2\n1 3\n2 4\n2 5\n3 6\n3 7\n",
"10\n9 5\n7 1\n9 10\n7 2\n5 4\n9 6\n2 9\n10 8\n1 3\n",
"4\n2 4\n2 3\n2 1\n",
"4\n1 4\n3 2\n1 3\n",
"3\n1 2\n1 3\n",
"5\n1 2\n1 5\n1 3\n1 4\n",
"20\n14 9\n12 13\n10 15\n2 1\n20 19\n16 6\n16 3\n17 14\n3 5\n2 11\n3 10\n15 8\n14 2\n6 4\n3 20\n5 18\n1 7\n1 16\n4 12\n",
"20\n7 5\n14 13\n17 6\n3 8\n16 12\n18 9\n3 18\n14 1\n17 3\n15 2\n17 4\n9 11\n2 7\n15 17\n3 20\n16 10\n17 14\n2 16\n1 19\n",
"8\n1 2\n2 3\n3 4\n1 7\n1 8\n4 5\n4 6\n",
"5\n5 1\n5 2\n5 3\n5 4\n",
"50\n49 6\n43 7\n1 27\n19 35\n15 37\n16 12\n19 21\n16 28\n49 9\n48 39\n13 1\n2 48\n9 50\n44 3\n41 32\n48 31\n49 33\n6 11\n13 20\n49 22\n13 41\n48 29\n13 46\n15 47\n34 2\n49 13\n48 14\n34 24\n16 36\n13 40\n49 34\n49 17\n43 25\n11 23\n10 15\n19 26\n34 44\n16 42\n19 18\n46 8\n29 38\n1 45\n12 43\n13 16\n46 30\n15 5\n49 10\n11 19\n32 4\n",
"20\n13 1\n18 2\n3 7\n18 5\n20 16\n3 12\n18 9\n3 10\n18 11\n13 6\n3 18\n20 15\n20 17\n3 13\n3 4\n13 14\n3 20\n18 8\n3 19\n",
"10\n8 2\n5 6\n1 8\n2 9\n1 4\n8 10\n10 5\n2 7\n2 3\n"
],
"output": [
"NO",
"YES",
"NO",
"NO",
"NO",
"YES",
"NO",
"NO",
"NO",
"NO",
"YES",
"NO",
"NO",
"YES",
"NO",
"NO",
"NO",
"YES",
"NO",
"YES",
"NO"
]
}
|
IN-CORRECT
|
java
|
import java.io.*;
import java.util.*;
public class D1 {
public void solve() throws IOException {
int n = nextInt();
ArrayList<Integer>[] g = new ArrayList[n];
for (int i = 0; i < n; i++) {
g[i] = new ArrayList<>();
}
for (int i = 0; i < n - 1; i++) {
int a = nextInt() - 1;
int b = nextInt() - 1;
g[a].add(b);
g[b].add(a);
}
boolean isGood = true;
for (int i = 0; i < n; i++) {
if (g[i].size() == 1 && g[g[i].get(0)].size() == 2) {
isGood = false;
break;
}
}
if (isGood) {
out.print("YES");
} else {
out.print("NO");
}
}
public void run() {
try {
br = new BufferedReader(new InputStreamReader(System.in));
out = new PrintWriter(System.out);
solve();
out.close();
} catch (IOException e) {
e.printStackTrace();
System.exit(1);
}
}
BufferedReader br;
StringTokenizer in;
PrintWriter out;
public String nextToken() throws IOException {
while (in == null || !in.hasMoreTokens()) {
in = new StringTokenizer(br.readLine());
}
return in.nextToken();
}
public int nextInt() throws IOException {
return Integer.parseInt(nextToken());
}
public double nextDouble() throws IOException {
return Double.parseDouble(nextToken());
}
public long nextLong() throws IOException {
return Long.parseLong(nextToken());
}
public int[] nextArr(int n) throws IOException {
int[] res = new int[n];
for (int i = 0; i < n; i++) {
res[i] = nextInt();
}
return res;
}
public static void main(String[] args) throws IOException {
Locale.setDefault(Locale.US);
new D1().run();
}
}
|
1189_D1. Add on a Tree
|
Note that this is the first problem of the two similar problems. You can hack this problem only if you solve both problems.
You are given a tree with n nodes. In the beginning, 0 is written on all edges. In one operation, you can choose any 2 distinct leaves u, v and any real number x and add x to values written on all edges on the simple path between u and v.
For example, on the picture below you can see the result of applying two operations to the graph: adding 2 on the path from 7 to 6, and then adding -0.5 on the path from 4 to 5.
<image>
Is it true that for any configuration of real numbers written on edges, we can achieve it with a finite number of operations?
Leaf is a node of a tree of degree 1. Simple path is a path that doesn't contain any node twice.
Input
The first line contains a single integer n (2 ≤ n ≤ 10^5) — the number of nodes.
Each of the next n-1 lines contains two integers u and v (1 ≤ u, v ≤ n, u ≠ v), meaning that there is an edge between nodes u and v. It is guaranteed that these edges form a tree.
Output
If there is a configuration of real numbers written on edges of the tree that we can't achieve by performing the operations, output "NO".
Otherwise, output "YES".
You can print each letter in any case (upper or lower).
Examples
Input
2
1 2
Output
YES
Input
3
1 2
2 3
Output
NO
Input
5
1 2
1 3
1 4
2 5
Output
NO
Input
6
1 2
1 3
1 4
2 5
2 6
Output
YES
Note
In the first example, we can add any real x to the value written on the only edge (1, 2).
<image>
In the second example, one of configurations that we can't reach is 0 written on (1, 2) and 1 written on (2, 3).
<image>
Below you can see graphs from examples 3, 4:
<image> <image>
|
{
"input": [
"2\n1 2\n",
"3\n1 2\n2 3\n",
"5\n1 2\n1 3\n1 4\n2 5\n",
"6\n1 2\n1 3\n1 4\n2 5\n2 6\n"
],
"output": [
"YES",
"NO",
"NO",
"YES"
]
}
|
{
"input": [
"50\n16 4\n17 9\n31 19\n22 10\n8 1\n40 30\n3 31\n20 29\n47 27\n22 25\n32 34\n12 15\n40 32\n10 33\n47 12\n6 24\n46 41\n14 23\n12 35\n31 42\n46 28\n31 20\n46 37\n1 39\n29 49\n37 47\n40 6\n42 36\n47 2\n24 46\n2 13\n8 45\n41 3\n32 17\n4 7\n47 26\n28 8\n41 50\n34 44\n33 21\n25 5\n16 40\n3 14\n8 18\n28 11\n32 22\n2 38\n3 48\n44 43\n",
"10\n8 1\n1 2\n8 9\n8 5\n1 3\n1 10\n1 6\n1 7\n8 4\n",
"5\n5 1\n5 4\n4 3\n1 2\n",
"7\n1 2\n2 3\n1 4\n1 5\n3 6\n3 7\n",
"3\n1 3\n2 3\n",
"60\n26 6\n59 30\n31 12\n31 3\n38 23\n59 29\n53 9\n38 56\n53 54\n29 21\n17 55\n59 38\n26 16\n24 59\n24 25\n17 35\n24 41\n30 15\n31 27\n8 44\n26 5\n26 48\n8 32\n53 17\n3 34\n3 51\n30 28\n47 10\n53 60\n36 42\n24 53\n59 22\n53 40\n26 52\n36 4\n59 8\n29 37\n36 20\n17 47\n53 18\n3 50\n30 2\n17 7\n8 58\n59 1\n31 11\n24 26\n24 43\n53 57\n59 45\n47 13\n26 46\n17 33\n30 31\n26 39\n26 19\n24 36\n8 49\n38 14\n",
"7\n1 2\n2 3\n3 4\n3 5\n1 6\n1 7\n",
"20\n19 16\n19 18\n20 7\n9 4\n6 17\n14 2\n9 15\n2 13\n5 11\n19 12\n12 20\n16 9\n11 8\n19 5\n3 1\n19 14\n5 3\n12 10\n19 6\n",
"7\n1 2\n1 3\n2 4\n2 5\n3 6\n3 7\n",
"10\n9 5\n7 1\n9 10\n7 2\n5 4\n9 6\n2 9\n10 8\n1 3\n",
"4\n2 4\n2 3\n2 1\n",
"4\n1 4\n3 2\n1 3\n",
"3\n1 2\n1 3\n",
"5\n1 2\n1 5\n1 3\n1 4\n",
"20\n14 9\n12 13\n10 15\n2 1\n20 19\n16 6\n16 3\n17 14\n3 5\n2 11\n3 10\n15 8\n14 2\n6 4\n3 20\n5 18\n1 7\n1 16\n4 12\n",
"20\n7 5\n14 13\n17 6\n3 8\n16 12\n18 9\n3 18\n14 1\n17 3\n15 2\n17 4\n9 11\n2 7\n15 17\n3 20\n16 10\n17 14\n2 16\n1 19\n",
"8\n1 2\n2 3\n3 4\n1 7\n1 8\n4 5\n4 6\n",
"5\n5 1\n5 2\n5 3\n5 4\n",
"50\n49 6\n43 7\n1 27\n19 35\n15 37\n16 12\n19 21\n16 28\n49 9\n48 39\n13 1\n2 48\n9 50\n44 3\n41 32\n48 31\n49 33\n6 11\n13 20\n49 22\n13 41\n48 29\n13 46\n15 47\n34 2\n49 13\n48 14\n34 24\n16 36\n13 40\n49 34\n49 17\n43 25\n11 23\n10 15\n19 26\n34 44\n16 42\n19 18\n46 8\n29 38\n1 45\n12 43\n13 16\n46 30\n15 5\n49 10\n11 19\n32 4\n",
"20\n13 1\n18 2\n3 7\n18 5\n20 16\n3 12\n18 9\n3 10\n18 11\n13 6\n3 18\n20 15\n20 17\n3 13\n3 4\n13 14\n3 20\n18 8\n3 19\n",
"10\n8 2\n5 6\n1 8\n2 9\n1 4\n8 10\n10 5\n2 7\n2 3\n"
],
"output": [
"NO",
"YES",
"NO",
"NO",
"NO",
"YES",
"NO",
"NO",
"NO",
"NO",
"YES",
"NO",
"NO",
"YES",
"NO",
"NO",
"NO",
"YES",
"NO",
"YES",
"NO"
]
}
|
IN-CORRECT
|
python3
|
# @author
import sys
class D1AddOnATree:
def dfs(self, start):
self.done[start] = 1
for x in self.adj[start]:
if self.done[x]:
continue
self.par[x] = start
self.dfs(x)
def solve(self):
from collections import defaultdict
import sys
sys.setrecursionlimit(10 ** 5 + 5)
n = int(input())
self.adj = defaultdict(list)
self.par = defaultdict(int)
self.done = [0] * (n + 1)
for i in range(n - 1):
u, v = [int(_) for _ in input().split()]
self.adj[u].append(v)
v = max(len(self.adj[p]) for p in self.adj)
start = -1
for p in self.adj:
if len(self.adj[p]) == v:
start = p
break
assert(start != -1)
self.dfs(start)
cnt = [0] * (n + 1)
for k in self.adj:
if self.par[k] == 0:
continue
if len(self.adj[k]) == 1:
cnt[self.par[k]] += 1
ans = 0
for x in cnt:
if x == 1:
ans += 1
if ans % 2 == 0:
print("YES")
else:
print("NO")
solver = D1AddOnATree()
input = sys.stdin.readline
solver.solve()
|
1189_D1. Add on a Tree
|
Note that this is the first problem of the two similar problems. You can hack this problem only if you solve both problems.
You are given a tree with n nodes. In the beginning, 0 is written on all edges. In one operation, you can choose any 2 distinct leaves u, v and any real number x and add x to values written on all edges on the simple path between u and v.
For example, on the picture below you can see the result of applying two operations to the graph: adding 2 on the path from 7 to 6, and then adding -0.5 on the path from 4 to 5.
<image>
Is it true that for any configuration of real numbers written on edges, we can achieve it with a finite number of operations?
Leaf is a node of a tree of degree 1. Simple path is a path that doesn't contain any node twice.
Input
The first line contains a single integer n (2 ≤ n ≤ 10^5) — the number of nodes.
Each of the next n-1 lines contains two integers u and v (1 ≤ u, v ≤ n, u ≠ v), meaning that there is an edge between nodes u and v. It is guaranteed that these edges form a tree.
Output
If there is a configuration of real numbers written on edges of the tree that we can't achieve by performing the operations, output "NO".
Otherwise, output "YES".
You can print each letter in any case (upper or lower).
Examples
Input
2
1 2
Output
YES
Input
3
1 2
2 3
Output
NO
Input
5
1 2
1 3
1 4
2 5
Output
NO
Input
6
1 2
1 3
1 4
2 5
2 6
Output
YES
Note
In the first example, we can add any real x to the value written on the only edge (1, 2).
<image>
In the second example, one of configurations that we can't reach is 0 written on (1, 2) and 1 written on (2, 3).
<image>
Below you can see graphs from examples 3, 4:
<image> <image>
|
{
"input": [
"2\n1 2\n",
"3\n1 2\n2 3\n",
"5\n1 2\n1 3\n1 4\n2 5\n",
"6\n1 2\n1 3\n1 4\n2 5\n2 6\n"
],
"output": [
"YES",
"NO",
"NO",
"YES"
]
}
|
{
"input": [
"50\n16 4\n17 9\n31 19\n22 10\n8 1\n40 30\n3 31\n20 29\n47 27\n22 25\n32 34\n12 15\n40 32\n10 33\n47 12\n6 24\n46 41\n14 23\n12 35\n31 42\n46 28\n31 20\n46 37\n1 39\n29 49\n37 47\n40 6\n42 36\n47 2\n24 46\n2 13\n8 45\n41 3\n32 17\n4 7\n47 26\n28 8\n41 50\n34 44\n33 21\n25 5\n16 40\n3 14\n8 18\n28 11\n32 22\n2 38\n3 48\n44 43\n",
"10\n8 1\n1 2\n8 9\n8 5\n1 3\n1 10\n1 6\n1 7\n8 4\n",
"5\n5 1\n5 4\n4 3\n1 2\n",
"7\n1 2\n2 3\n1 4\n1 5\n3 6\n3 7\n",
"3\n1 3\n2 3\n",
"60\n26 6\n59 30\n31 12\n31 3\n38 23\n59 29\n53 9\n38 56\n53 54\n29 21\n17 55\n59 38\n26 16\n24 59\n24 25\n17 35\n24 41\n30 15\n31 27\n8 44\n26 5\n26 48\n8 32\n53 17\n3 34\n3 51\n30 28\n47 10\n53 60\n36 42\n24 53\n59 22\n53 40\n26 52\n36 4\n59 8\n29 37\n36 20\n17 47\n53 18\n3 50\n30 2\n17 7\n8 58\n59 1\n31 11\n24 26\n24 43\n53 57\n59 45\n47 13\n26 46\n17 33\n30 31\n26 39\n26 19\n24 36\n8 49\n38 14\n",
"7\n1 2\n2 3\n3 4\n3 5\n1 6\n1 7\n",
"20\n19 16\n19 18\n20 7\n9 4\n6 17\n14 2\n9 15\n2 13\n5 11\n19 12\n12 20\n16 9\n11 8\n19 5\n3 1\n19 14\n5 3\n12 10\n19 6\n",
"7\n1 2\n1 3\n2 4\n2 5\n3 6\n3 7\n",
"10\n9 5\n7 1\n9 10\n7 2\n5 4\n9 6\n2 9\n10 8\n1 3\n",
"4\n2 4\n2 3\n2 1\n",
"4\n1 4\n3 2\n1 3\n",
"3\n1 2\n1 3\n",
"5\n1 2\n1 5\n1 3\n1 4\n",
"20\n14 9\n12 13\n10 15\n2 1\n20 19\n16 6\n16 3\n17 14\n3 5\n2 11\n3 10\n15 8\n14 2\n6 4\n3 20\n5 18\n1 7\n1 16\n4 12\n",
"20\n7 5\n14 13\n17 6\n3 8\n16 12\n18 9\n3 18\n14 1\n17 3\n15 2\n17 4\n9 11\n2 7\n15 17\n3 20\n16 10\n17 14\n2 16\n1 19\n",
"8\n1 2\n2 3\n3 4\n1 7\n1 8\n4 5\n4 6\n",
"5\n5 1\n5 2\n5 3\n5 4\n",
"50\n49 6\n43 7\n1 27\n19 35\n15 37\n16 12\n19 21\n16 28\n49 9\n48 39\n13 1\n2 48\n9 50\n44 3\n41 32\n48 31\n49 33\n6 11\n13 20\n49 22\n13 41\n48 29\n13 46\n15 47\n34 2\n49 13\n48 14\n34 24\n16 36\n13 40\n49 34\n49 17\n43 25\n11 23\n10 15\n19 26\n34 44\n16 42\n19 18\n46 8\n29 38\n1 45\n12 43\n13 16\n46 30\n15 5\n49 10\n11 19\n32 4\n",
"20\n13 1\n18 2\n3 7\n18 5\n20 16\n3 12\n18 9\n3 10\n18 11\n13 6\n3 18\n20 15\n20 17\n3 13\n3 4\n13 14\n3 20\n18 8\n3 19\n",
"10\n8 2\n5 6\n1 8\n2 9\n1 4\n8 10\n10 5\n2 7\n2 3\n"
],
"output": [
"NO",
"YES",
"NO",
"NO",
"NO",
"YES",
"NO",
"NO",
"NO",
"NO",
"YES",
"NO",
"NO",
"YES",
"NO",
"NO",
"NO",
"YES",
"NO",
"YES",
"NO"
]
}
|
IN-CORRECT
|
python3
|
from collections import defaultdict
n = int(input())
hash = defaultdict(list)
for i in range(n-1):
a,b = map(int,input().split())
hash[a].append(b)
hash[b].append(a)
leav = []
for i in hash.keys():
if len(hash[i]) == 1:
leav.append(i)
new_hash = defaultdict(list)
seti = set()
for i in leav:
seti.add(hash[i][0])
z = hash[i][0]
new_hash[z].append(i)
if len(seti) == 1:
if len(leav) == 1 or len(leav)>2:
print('YES')
else:
print('NO')
else:
flag = 1
for i in new_hash:
if len(new_hash[i]) == 1 :
flag = 0
break
if flag == 1:
print('YES')
else:
print('NO')
|
1189_D1. Add on a Tree
|
Note that this is the first problem of the two similar problems. You can hack this problem only if you solve both problems.
You are given a tree with n nodes. In the beginning, 0 is written on all edges. In one operation, you can choose any 2 distinct leaves u, v and any real number x and add x to values written on all edges on the simple path between u and v.
For example, on the picture below you can see the result of applying two operations to the graph: adding 2 on the path from 7 to 6, and then adding -0.5 on the path from 4 to 5.
<image>
Is it true that for any configuration of real numbers written on edges, we can achieve it with a finite number of operations?
Leaf is a node of a tree of degree 1. Simple path is a path that doesn't contain any node twice.
Input
The first line contains a single integer n (2 ≤ n ≤ 10^5) — the number of nodes.
Each of the next n-1 lines contains two integers u and v (1 ≤ u, v ≤ n, u ≠ v), meaning that there is an edge between nodes u and v. It is guaranteed that these edges form a tree.
Output
If there is a configuration of real numbers written on edges of the tree that we can't achieve by performing the operations, output "NO".
Otherwise, output "YES".
You can print each letter in any case (upper or lower).
Examples
Input
2
1 2
Output
YES
Input
3
1 2
2 3
Output
NO
Input
5
1 2
1 3
1 4
2 5
Output
NO
Input
6
1 2
1 3
1 4
2 5
2 6
Output
YES
Note
In the first example, we can add any real x to the value written on the only edge (1, 2).
<image>
In the second example, one of configurations that we can't reach is 0 written on (1, 2) and 1 written on (2, 3).
<image>
Below you can see graphs from examples 3, 4:
<image> <image>
|
{
"input": [
"2\n1 2\n",
"3\n1 2\n2 3\n",
"5\n1 2\n1 3\n1 4\n2 5\n",
"6\n1 2\n1 3\n1 4\n2 5\n2 6\n"
],
"output": [
"YES",
"NO",
"NO",
"YES"
]
}
|
{
"input": [
"50\n16 4\n17 9\n31 19\n22 10\n8 1\n40 30\n3 31\n20 29\n47 27\n22 25\n32 34\n12 15\n40 32\n10 33\n47 12\n6 24\n46 41\n14 23\n12 35\n31 42\n46 28\n31 20\n46 37\n1 39\n29 49\n37 47\n40 6\n42 36\n47 2\n24 46\n2 13\n8 45\n41 3\n32 17\n4 7\n47 26\n28 8\n41 50\n34 44\n33 21\n25 5\n16 40\n3 14\n8 18\n28 11\n32 22\n2 38\n3 48\n44 43\n",
"10\n8 1\n1 2\n8 9\n8 5\n1 3\n1 10\n1 6\n1 7\n8 4\n",
"5\n5 1\n5 4\n4 3\n1 2\n",
"7\n1 2\n2 3\n1 4\n1 5\n3 6\n3 7\n",
"3\n1 3\n2 3\n",
"60\n26 6\n59 30\n31 12\n31 3\n38 23\n59 29\n53 9\n38 56\n53 54\n29 21\n17 55\n59 38\n26 16\n24 59\n24 25\n17 35\n24 41\n30 15\n31 27\n8 44\n26 5\n26 48\n8 32\n53 17\n3 34\n3 51\n30 28\n47 10\n53 60\n36 42\n24 53\n59 22\n53 40\n26 52\n36 4\n59 8\n29 37\n36 20\n17 47\n53 18\n3 50\n30 2\n17 7\n8 58\n59 1\n31 11\n24 26\n24 43\n53 57\n59 45\n47 13\n26 46\n17 33\n30 31\n26 39\n26 19\n24 36\n8 49\n38 14\n",
"7\n1 2\n2 3\n3 4\n3 5\n1 6\n1 7\n",
"20\n19 16\n19 18\n20 7\n9 4\n6 17\n14 2\n9 15\n2 13\n5 11\n19 12\n12 20\n16 9\n11 8\n19 5\n3 1\n19 14\n5 3\n12 10\n19 6\n",
"7\n1 2\n1 3\n2 4\n2 5\n3 6\n3 7\n",
"10\n9 5\n7 1\n9 10\n7 2\n5 4\n9 6\n2 9\n10 8\n1 3\n",
"4\n2 4\n2 3\n2 1\n",
"4\n1 4\n3 2\n1 3\n",
"3\n1 2\n1 3\n",
"5\n1 2\n1 5\n1 3\n1 4\n",
"20\n14 9\n12 13\n10 15\n2 1\n20 19\n16 6\n16 3\n17 14\n3 5\n2 11\n3 10\n15 8\n14 2\n6 4\n3 20\n5 18\n1 7\n1 16\n4 12\n",
"20\n7 5\n14 13\n17 6\n3 8\n16 12\n18 9\n3 18\n14 1\n17 3\n15 2\n17 4\n9 11\n2 7\n15 17\n3 20\n16 10\n17 14\n2 16\n1 19\n",
"8\n1 2\n2 3\n3 4\n1 7\n1 8\n4 5\n4 6\n",
"5\n5 1\n5 2\n5 3\n5 4\n",
"50\n49 6\n43 7\n1 27\n19 35\n15 37\n16 12\n19 21\n16 28\n49 9\n48 39\n13 1\n2 48\n9 50\n44 3\n41 32\n48 31\n49 33\n6 11\n13 20\n49 22\n13 41\n48 29\n13 46\n15 47\n34 2\n49 13\n48 14\n34 24\n16 36\n13 40\n49 34\n49 17\n43 25\n11 23\n10 15\n19 26\n34 44\n16 42\n19 18\n46 8\n29 38\n1 45\n12 43\n13 16\n46 30\n15 5\n49 10\n11 19\n32 4\n",
"20\n13 1\n18 2\n3 7\n18 5\n20 16\n3 12\n18 9\n3 10\n18 11\n13 6\n3 18\n20 15\n20 17\n3 13\n3 4\n13 14\n3 20\n18 8\n3 19\n",
"10\n8 2\n5 6\n1 8\n2 9\n1 4\n8 10\n10 5\n2 7\n2 3\n"
],
"output": [
"NO",
"YES",
"NO",
"NO",
"NO",
"YES",
"NO",
"NO",
"NO",
"NO",
"YES",
"NO",
"NO",
"YES",
"NO",
"NO",
"NO",
"YES",
"NO",
"YES",
"NO"
]
}
|
IN-CORRECT
|
java
|
import java.io.*;
import java.math.BigInteger;
import java.util.ArrayList;
import java.util.InputMismatchException;
import java.util.StringTokenizer;
public class D {
static class FastWriter {
private final BufferedWriter bw;
public FastWriter() {
this.bw = new BufferedWriter(new OutputStreamWriter(System.out));
}
public void print(Object object) throws IOException {
bw.append("" + object);
}
public void println(Object object) throws IOException {
print(object);
bw.append("\n");
}
public void close() throws IOException {
bw.close();
}
}
static class FastReader {
BufferedReader br;
StringTokenizer st;
public FastReader() {
br = new BufferedReader(new InputStreamReader(System.in));
}
String next() {
while (st == null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt() {
return Integer.parseInt(next());
}
long nextLong() {
return Long.parseLong(next());
}
double nextDouble() {
return Double.parseDouble(next());
}
String nextLine() {
String str = "";
try {
str = br.readLine();
} catch (IOException e) {
e.printStackTrace();
}
return str;
}
BigInteger nextBigInteger() {
try {
return new BigInteger(nextLine());
} catch (NumberFormatException e) {
throw new InputMismatchException();
}
}
}
public static void main(String[] args) {
FastReader fr = new FastReader();
FastWriter fw = new FastWriter();
int n = fr.nextInt();
ArrayList<ArrayList<Integer>> g = new ArrayList<>();
for (int i = 0; i < n; i++) g.add(new ArrayList<>());
for (int i = 0; i < n - 1; i++) {
int u = fr.nextInt() - 1;
int v = fr.nextInt() - 1;
g.get(u).add(v);
g.get(v).add(u);
}
HeapMover m = new HeapMover();
m.flag = true;
boolean visited[] = new boolean[n];
helper(g, 0, -1, m, visited);
if (m.flag) System.out.println("YES");
else System.out.println("NO");
}
public static class HeapMover {
boolean flag;
}
public static void helper(ArrayList<ArrayList<Integer>> g, int c, int p, HeapMover m, boolean[] visited) {
visited[c] = true;
if (p != -1) {
if (g.get(c).size() == 1) {
if (g.get(p).size() == 2) m.flag = false;
}
}
for(int ch : g.get(c)){
if(!visited[ch]){
helper(g,ch,c,m,visited);
}
}
}
}
|
1189_D1. Add on a Tree
|
Note that this is the first problem of the two similar problems. You can hack this problem only if you solve both problems.
You are given a tree with n nodes. In the beginning, 0 is written on all edges. In one operation, you can choose any 2 distinct leaves u, v and any real number x and add x to values written on all edges on the simple path between u and v.
For example, on the picture below you can see the result of applying two operations to the graph: adding 2 on the path from 7 to 6, and then adding -0.5 on the path from 4 to 5.
<image>
Is it true that for any configuration of real numbers written on edges, we can achieve it with a finite number of operations?
Leaf is a node of a tree of degree 1. Simple path is a path that doesn't contain any node twice.
Input
The first line contains a single integer n (2 ≤ n ≤ 10^5) — the number of nodes.
Each of the next n-1 lines contains two integers u and v (1 ≤ u, v ≤ n, u ≠ v), meaning that there is an edge between nodes u and v. It is guaranteed that these edges form a tree.
Output
If there is a configuration of real numbers written on edges of the tree that we can't achieve by performing the operations, output "NO".
Otherwise, output "YES".
You can print each letter in any case (upper or lower).
Examples
Input
2
1 2
Output
YES
Input
3
1 2
2 3
Output
NO
Input
5
1 2
1 3
1 4
2 5
Output
NO
Input
6
1 2
1 3
1 4
2 5
2 6
Output
YES
Note
In the first example, we can add any real x to the value written on the only edge (1, 2).
<image>
In the second example, one of configurations that we can't reach is 0 written on (1, 2) and 1 written on (2, 3).
<image>
Below you can see graphs from examples 3, 4:
<image> <image>
|
{
"input": [
"2\n1 2\n",
"3\n1 2\n2 3\n",
"5\n1 2\n1 3\n1 4\n2 5\n",
"6\n1 2\n1 3\n1 4\n2 5\n2 6\n"
],
"output": [
"YES",
"NO",
"NO",
"YES"
]
}
|
{
"input": [
"50\n16 4\n17 9\n31 19\n22 10\n8 1\n40 30\n3 31\n20 29\n47 27\n22 25\n32 34\n12 15\n40 32\n10 33\n47 12\n6 24\n46 41\n14 23\n12 35\n31 42\n46 28\n31 20\n46 37\n1 39\n29 49\n37 47\n40 6\n42 36\n47 2\n24 46\n2 13\n8 45\n41 3\n32 17\n4 7\n47 26\n28 8\n41 50\n34 44\n33 21\n25 5\n16 40\n3 14\n8 18\n28 11\n32 22\n2 38\n3 48\n44 43\n",
"10\n8 1\n1 2\n8 9\n8 5\n1 3\n1 10\n1 6\n1 7\n8 4\n",
"5\n5 1\n5 4\n4 3\n1 2\n",
"7\n1 2\n2 3\n1 4\n1 5\n3 6\n3 7\n",
"3\n1 3\n2 3\n",
"60\n26 6\n59 30\n31 12\n31 3\n38 23\n59 29\n53 9\n38 56\n53 54\n29 21\n17 55\n59 38\n26 16\n24 59\n24 25\n17 35\n24 41\n30 15\n31 27\n8 44\n26 5\n26 48\n8 32\n53 17\n3 34\n3 51\n30 28\n47 10\n53 60\n36 42\n24 53\n59 22\n53 40\n26 52\n36 4\n59 8\n29 37\n36 20\n17 47\n53 18\n3 50\n30 2\n17 7\n8 58\n59 1\n31 11\n24 26\n24 43\n53 57\n59 45\n47 13\n26 46\n17 33\n30 31\n26 39\n26 19\n24 36\n8 49\n38 14\n",
"7\n1 2\n2 3\n3 4\n3 5\n1 6\n1 7\n",
"20\n19 16\n19 18\n20 7\n9 4\n6 17\n14 2\n9 15\n2 13\n5 11\n19 12\n12 20\n16 9\n11 8\n19 5\n3 1\n19 14\n5 3\n12 10\n19 6\n",
"7\n1 2\n1 3\n2 4\n2 5\n3 6\n3 7\n",
"10\n9 5\n7 1\n9 10\n7 2\n5 4\n9 6\n2 9\n10 8\n1 3\n",
"4\n2 4\n2 3\n2 1\n",
"4\n1 4\n3 2\n1 3\n",
"3\n1 2\n1 3\n",
"5\n1 2\n1 5\n1 3\n1 4\n",
"20\n14 9\n12 13\n10 15\n2 1\n20 19\n16 6\n16 3\n17 14\n3 5\n2 11\n3 10\n15 8\n14 2\n6 4\n3 20\n5 18\n1 7\n1 16\n4 12\n",
"20\n7 5\n14 13\n17 6\n3 8\n16 12\n18 9\n3 18\n14 1\n17 3\n15 2\n17 4\n9 11\n2 7\n15 17\n3 20\n16 10\n17 14\n2 16\n1 19\n",
"8\n1 2\n2 3\n3 4\n1 7\n1 8\n4 5\n4 6\n",
"5\n5 1\n5 2\n5 3\n5 4\n",
"50\n49 6\n43 7\n1 27\n19 35\n15 37\n16 12\n19 21\n16 28\n49 9\n48 39\n13 1\n2 48\n9 50\n44 3\n41 32\n48 31\n49 33\n6 11\n13 20\n49 22\n13 41\n48 29\n13 46\n15 47\n34 2\n49 13\n48 14\n34 24\n16 36\n13 40\n49 34\n49 17\n43 25\n11 23\n10 15\n19 26\n34 44\n16 42\n19 18\n46 8\n29 38\n1 45\n12 43\n13 16\n46 30\n15 5\n49 10\n11 19\n32 4\n",
"20\n13 1\n18 2\n3 7\n18 5\n20 16\n3 12\n18 9\n3 10\n18 11\n13 6\n3 18\n20 15\n20 17\n3 13\n3 4\n13 14\n3 20\n18 8\n3 19\n",
"10\n8 2\n5 6\n1 8\n2 9\n1 4\n8 10\n10 5\n2 7\n2 3\n"
],
"output": [
"NO",
"YES",
"NO",
"NO",
"NO",
"YES",
"NO",
"NO",
"NO",
"NO",
"YES",
"NO",
"NO",
"YES",
"NO",
"NO",
"NO",
"YES",
"NO",
"YES",
"NO"
]
}
|
IN-CORRECT
|
python3
|
from collections import defaultdict
n = int(input())
hash = defaultdict(list)
for i in range(n-1):
a,b = map(int,input().split())
hash[a].append(b)
hash[b].append(a)
leav = []
for i in hash.keys():
if len(hash[i]) == 1:
leav.append(i)
new_hash = defaultdict(list)
seti = set()
for i in leav:
seti.add(hash[i][0])
z = hash[i][0]
new_hash[z].append(i)
if len(leav) == n:
print('YES')
elif len(seti) == 1:
if len(leav) == 1 or len(leav)>2:
print('YES')
else:
print('NO')
else:
flag = 1
for i in new_hash:
if len(new_hash[i]) == 1 :
flag = 0
break
if flag == 1:
print('YES')
else:
print('NO')
|
1189_D1. Add on a Tree
|
Note that this is the first problem of the two similar problems. You can hack this problem only if you solve both problems.
You are given a tree with n nodes. In the beginning, 0 is written on all edges. In one operation, you can choose any 2 distinct leaves u, v and any real number x and add x to values written on all edges on the simple path between u and v.
For example, on the picture below you can see the result of applying two operations to the graph: adding 2 on the path from 7 to 6, and then adding -0.5 on the path from 4 to 5.
<image>
Is it true that for any configuration of real numbers written on edges, we can achieve it with a finite number of operations?
Leaf is a node of a tree of degree 1. Simple path is a path that doesn't contain any node twice.
Input
The first line contains a single integer n (2 ≤ n ≤ 10^5) — the number of nodes.
Each of the next n-1 lines contains two integers u and v (1 ≤ u, v ≤ n, u ≠ v), meaning that there is an edge between nodes u and v. It is guaranteed that these edges form a tree.
Output
If there is a configuration of real numbers written on edges of the tree that we can't achieve by performing the operations, output "NO".
Otherwise, output "YES".
You can print each letter in any case (upper or lower).
Examples
Input
2
1 2
Output
YES
Input
3
1 2
2 3
Output
NO
Input
5
1 2
1 3
1 4
2 5
Output
NO
Input
6
1 2
1 3
1 4
2 5
2 6
Output
YES
Note
In the first example, we can add any real x to the value written on the only edge (1, 2).
<image>
In the second example, one of configurations that we can't reach is 0 written on (1, 2) and 1 written on (2, 3).
<image>
Below you can see graphs from examples 3, 4:
<image> <image>
|
{
"input": [
"2\n1 2\n",
"3\n1 2\n2 3\n",
"5\n1 2\n1 3\n1 4\n2 5\n",
"6\n1 2\n1 3\n1 4\n2 5\n2 6\n"
],
"output": [
"YES",
"NO",
"NO",
"YES"
]
}
|
{
"input": [
"50\n16 4\n17 9\n31 19\n22 10\n8 1\n40 30\n3 31\n20 29\n47 27\n22 25\n32 34\n12 15\n40 32\n10 33\n47 12\n6 24\n46 41\n14 23\n12 35\n31 42\n46 28\n31 20\n46 37\n1 39\n29 49\n37 47\n40 6\n42 36\n47 2\n24 46\n2 13\n8 45\n41 3\n32 17\n4 7\n47 26\n28 8\n41 50\n34 44\n33 21\n25 5\n16 40\n3 14\n8 18\n28 11\n32 22\n2 38\n3 48\n44 43\n",
"10\n8 1\n1 2\n8 9\n8 5\n1 3\n1 10\n1 6\n1 7\n8 4\n",
"5\n5 1\n5 4\n4 3\n1 2\n",
"7\n1 2\n2 3\n1 4\n1 5\n3 6\n3 7\n",
"3\n1 3\n2 3\n",
"60\n26 6\n59 30\n31 12\n31 3\n38 23\n59 29\n53 9\n38 56\n53 54\n29 21\n17 55\n59 38\n26 16\n24 59\n24 25\n17 35\n24 41\n30 15\n31 27\n8 44\n26 5\n26 48\n8 32\n53 17\n3 34\n3 51\n30 28\n47 10\n53 60\n36 42\n24 53\n59 22\n53 40\n26 52\n36 4\n59 8\n29 37\n36 20\n17 47\n53 18\n3 50\n30 2\n17 7\n8 58\n59 1\n31 11\n24 26\n24 43\n53 57\n59 45\n47 13\n26 46\n17 33\n30 31\n26 39\n26 19\n24 36\n8 49\n38 14\n",
"7\n1 2\n2 3\n3 4\n3 5\n1 6\n1 7\n",
"20\n19 16\n19 18\n20 7\n9 4\n6 17\n14 2\n9 15\n2 13\n5 11\n19 12\n12 20\n16 9\n11 8\n19 5\n3 1\n19 14\n5 3\n12 10\n19 6\n",
"7\n1 2\n1 3\n2 4\n2 5\n3 6\n3 7\n",
"10\n9 5\n7 1\n9 10\n7 2\n5 4\n9 6\n2 9\n10 8\n1 3\n",
"4\n2 4\n2 3\n2 1\n",
"4\n1 4\n3 2\n1 3\n",
"3\n1 2\n1 3\n",
"5\n1 2\n1 5\n1 3\n1 4\n",
"20\n14 9\n12 13\n10 15\n2 1\n20 19\n16 6\n16 3\n17 14\n3 5\n2 11\n3 10\n15 8\n14 2\n6 4\n3 20\n5 18\n1 7\n1 16\n4 12\n",
"20\n7 5\n14 13\n17 6\n3 8\n16 12\n18 9\n3 18\n14 1\n17 3\n15 2\n17 4\n9 11\n2 7\n15 17\n3 20\n16 10\n17 14\n2 16\n1 19\n",
"8\n1 2\n2 3\n3 4\n1 7\n1 8\n4 5\n4 6\n",
"5\n5 1\n5 2\n5 3\n5 4\n",
"50\n49 6\n43 7\n1 27\n19 35\n15 37\n16 12\n19 21\n16 28\n49 9\n48 39\n13 1\n2 48\n9 50\n44 3\n41 32\n48 31\n49 33\n6 11\n13 20\n49 22\n13 41\n48 29\n13 46\n15 47\n34 2\n49 13\n48 14\n34 24\n16 36\n13 40\n49 34\n49 17\n43 25\n11 23\n10 15\n19 26\n34 44\n16 42\n19 18\n46 8\n29 38\n1 45\n12 43\n13 16\n46 30\n15 5\n49 10\n11 19\n32 4\n",
"20\n13 1\n18 2\n3 7\n18 5\n20 16\n3 12\n18 9\n3 10\n18 11\n13 6\n3 18\n20 15\n20 17\n3 13\n3 4\n13 14\n3 20\n18 8\n3 19\n",
"10\n8 2\n5 6\n1 8\n2 9\n1 4\n8 10\n10 5\n2 7\n2 3\n"
],
"output": [
"NO",
"YES",
"NO",
"NO",
"NO",
"YES",
"NO",
"NO",
"NO",
"NO",
"YES",
"NO",
"NO",
"YES",
"NO",
"NO",
"NO",
"YES",
"NO",
"YES",
"NO"
]
}
|
IN-CORRECT
|
python3
|
# @author
import sys
class D1AddOnATree:
def dfs(self, start):
self.done[start] = 1
for x in self.adj[start]:
if self.done[x]:
continue
self.par[x] = start
self.dfs(x)
def solve(self):
from collections import defaultdict
import sys
sys.setrecursionlimit(10 ** 5 + 5)
n = int(input())
self.adj = defaultdict(list)
self.par = defaultdict(int)
self.done = [0] * (n + 1)
for i in range(n - 1):
u, v = [int(_) for _ in input().split()]
self.adj[u].append(v)
v = max(len(self.adj[p]) for p in self.adj)
start = -1
for p in self.adj:
if len(self.adj[p]) == v:
start = p
break
assert(start != -1)
self.dfs(start)
cnt = [0] * (n + 1)
for k in self.adj:
if self.par[k] == 0:
continue
if len(self.adj[k]) == 1:
cnt[self.par[k]] += 1
for x in cnt:
if x == 1:
print("NO")
break
else:
print("YES")
solver = D1AddOnATree()
input = sys.stdin.readline
solver.solve()
|
1189_D1. Add on a Tree
|
Note that this is the first problem of the two similar problems. You can hack this problem only if you solve both problems.
You are given a tree with n nodes. In the beginning, 0 is written on all edges. In one operation, you can choose any 2 distinct leaves u, v and any real number x and add x to values written on all edges on the simple path between u and v.
For example, on the picture below you can see the result of applying two operations to the graph: adding 2 on the path from 7 to 6, and then adding -0.5 on the path from 4 to 5.
<image>
Is it true that for any configuration of real numbers written on edges, we can achieve it with a finite number of operations?
Leaf is a node of a tree of degree 1. Simple path is a path that doesn't contain any node twice.
Input
The first line contains a single integer n (2 ≤ n ≤ 10^5) — the number of nodes.
Each of the next n-1 lines contains two integers u and v (1 ≤ u, v ≤ n, u ≠ v), meaning that there is an edge between nodes u and v. It is guaranteed that these edges form a tree.
Output
If there is a configuration of real numbers written on edges of the tree that we can't achieve by performing the operations, output "NO".
Otherwise, output "YES".
You can print each letter in any case (upper or lower).
Examples
Input
2
1 2
Output
YES
Input
3
1 2
2 3
Output
NO
Input
5
1 2
1 3
1 4
2 5
Output
NO
Input
6
1 2
1 3
1 4
2 5
2 6
Output
YES
Note
In the first example, we can add any real x to the value written on the only edge (1, 2).
<image>
In the second example, one of configurations that we can't reach is 0 written on (1, 2) and 1 written on (2, 3).
<image>
Below you can see graphs from examples 3, 4:
<image> <image>
|
{
"input": [
"2\n1 2\n",
"3\n1 2\n2 3\n",
"5\n1 2\n1 3\n1 4\n2 5\n",
"6\n1 2\n1 3\n1 4\n2 5\n2 6\n"
],
"output": [
"YES",
"NO",
"NO",
"YES"
]
}
|
{
"input": [
"50\n16 4\n17 9\n31 19\n22 10\n8 1\n40 30\n3 31\n20 29\n47 27\n22 25\n32 34\n12 15\n40 32\n10 33\n47 12\n6 24\n46 41\n14 23\n12 35\n31 42\n46 28\n31 20\n46 37\n1 39\n29 49\n37 47\n40 6\n42 36\n47 2\n24 46\n2 13\n8 45\n41 3\n32 17\n4 7\n47 26\n28 8\n41 50\n34 44\n33 21\n25 5\n16 40\n3 14\n8 18\n28 11\n32 22\n2 38\n3 48\n44 43\n",
"10\n8 1\n1 2\n8 9\n8 5\n1 3\n1 10\n1 6\n1 7\n8 4\n",
"5\n5 1\n5 4\n4 3\n1 2\n",
"7\n1 2\n2 3\n1 4\n1 5\n3 6\n3 7\n",
"3\n1 3\n2 3\n",
"60\n26 6\n59 30\n31 12\n31 3\n38 23\n59 29\n53 9\n38 56\n53 54\n29 21\n17 55\n59 38\n26 16\n24 59\n24 25\n17 35\n24 41\n30 15\n31 27\n8 44\n26 5\n26 48\n8 32\n53 17\n3 34\n3 51\n30 28\n47 10\n53 60\n36 42\n24 53\n59 22\n53 40\n26 52\n36 4\n59 8\n29 37\n36 20\n17 47\n53 18\n3 50\n30 2\n17 7\n8 58\n59 1\n31 11\n24 26\n24 43\n53 57\n59 45\n47 13\n26 46\n17 33\n30 31\n26 39\n26 19\n24 36\n8 49\n38 14\n",
"7\n1 2\n2 3\n3 4\n3 5\n1 6\n1 7\n",
"20\n19 16\n19 18\n20 7\n9 4\n6 17\n14 2\n9 15\n2 13\n5 11\n19 12\n12 20\n16 9\n11 8\n19 5\n3 1\n19 14\n5 3\n12 10\n19 6\n",
"7\n1 2\n1 3\n2 4\n2 5\n3 6\n3 7\n",
"10\n9 5\n7 1\n9 10\n7 2\n5 4\n9 6\n2 9\n10 8\n1 3\n",
"4\n2 4\n2 3\n2 1\n",
"4\n1 4\n3 2\n1 3\n",
"3\n1 2\n1 3\n",
"5\n1 2\n1 5\n1 3\n1 4\n",
"20\n14 9\n12 13\n10 15\n2 1\n20 19\n16 6\n16 3\n17 14\n3 5\n2 11\n3 10\n15 8\n14 2\n6 4\n3 20\n5 18\n1 7\n1 16\n4 12\n",
"20\n7 5\n14 13\n17 6\n3 8\n16 12\n18 9\n3 18\n14 1\n17 3\n15 2\n17 4\n9 11\n2 7\n15 17\n3 20\n16 10\n17 14\n2 16\n1 19\n",
"8\n1 2\n2 3\n3 4\n1 7\n1 8\n4 5\n4 6\n",
"5\n5 1\n5 2\n5 3\n5 4\n",
"50\n49 6\n43 7\n1 27\n19 35\n15 37\n16 12\n19 21\n16 28\n49 9\n48 39\n13 1\n2 48\n9 50\n44 3\n41 32\n48 31\n49 33\n6 11\n13 20\n49 22\n13 41\n48 29\n13 46\n15 47\n34 2\n49 13\n48 14\n34 24\n16 36\n13 40\n49 34\n49 17\n43 25\n11 23\n10 15\n19 26\n34 44\n16 42\n19 18\n46 8\n29 38\n1 45\n12 43\n13 16\n46 30\n15 5\n49 10\n11 19\n32 4\n",
"20\n13 1\n18 2\n3 7\n18 5\n20 16\n3 12\n18 9\n3 10\n18 11\n13 6\n3 18\n20 15\n20 17\n3 13\n3 4\n13 14\n3 20\n18 8\n3 19\n",
"10\n8 2\n5 6\n1 8\n2 9\n1 4\n8 10\n10 5\n2 7\n2 3\n"
],
"output": [
"NO",
"YES",
"NO",
"NO",
"NO",
"YES",
"NO",
"NO",
"NO",
"NO",
"YES",
"NO",
"NO",
"YES",
"NO",
"NO",
"NO",
"YES",
"NO",
"YES",
"NO"
]
}
|
IN-CORRECT
|
cpp
|
#include<bits/stdc++.h>
using namespace std;
#define int long long
vector<int> graph[(int)(1e5+5)];
vector<int> level(1e5);
bool marked[(int)(1e5+5)];
void bfs(int x)
{
queue<int> que;
que.push(x);
level[x] = 0;
marked[x] = true;
while (!que.empty()) {
x = que.front();
que.pop();
for (int i = 0; i < graph[x].size(); i++) {
int b = graph[x][i];
if (!marked[b]) {
que.push(b);
level[b] = x;
marked[b] = true;
}
}
}
}
int32_t main()
{
int n;
cin>>n;
int u,v;
int deg[n+1]={0};
for(int i=1;i<n;i++)
{
cin>>u>>v;
deg[u]++;
deg[v]++;
graph[u].push_back(v);
graph[v].push_back(u);
}
if(n==2)
{
cout<<"YES\n";
return 0;
}
if(n==3)
{
cout<<"NO\n";
return 0;
}
bfs(1);
int leaf[(int)(1e5+5)]={0};
for(int i=1;i<=n;i++)
{
if(deg[i]==1)
{
leaf[level[i]]++;
}
}
for(int i=1;i<=(1e5+1);i++)
{
if(leaf[i]==1)
{
cout<<"NO\n";
return 0;
}
}
cout<<"YES\n";
}
|
1189_D1. Add on a Tree
|
Note that this is the first problem of the two similar problems. You can hack this problem only if you solve both problems.
You are given a tree with n nodes. In the beginning, 0 is written on all edges. In one operation, you can choose any 2 distinct leaves u, v and any real number x and add x to values written on all edges on the simple path between u and v.
For example, on the picture below you can see the result of applying two operations to the graph: adding 2 on the path from 7 to 6, and then adding -0.5 on the path from 4 to 5.
<image>
Is it true that for any configuration of real numbers written on edges, we can achieve it with a finite number of operations?
Leaf is a node of a tree of degree 1. Simple path is a path that doesn't contain any node twice.
Input
The first line contains a single integer n (2 ≤ n ≤ 10^5) — the number of nodes.
Each of the next n-1 lines contains two integers u and v (1 ≤ u, v ≤ n, u ≠ v), meaning that there is an edge between nodes u and v. It is guaranteed that these edges form a tree.
Output
If there is a configuration of real numbers written on edges of the tree that we can't achieve by performing the operations, output "NO".
Otherwise, output "YES".
You can print each letter in any case (upper or lower).
Examples
Input
2
1 2
Output
YES
Input
3
1 2
2 3
Output
NO
Input
5
1 2
1 3
1 4
2 5
Output
NO
Input
6
1 2
1 3
1 4
2 5
2 6
Output
YES
Note
In the first example, we can add any real x to the value written on the only edge (1, 2).
<image>
In the second example, one of configurations that we can't reach is 0 written on (1, 2) and 1 written on (2, 3).
<image>
Below you can see graphs from examples 3, 4:
<image> <image>
|
{
"input": [
"2\n1 2\n",
"3\n1 2\n2 3\n",
"5\n1 2\n1 3\n1 4\n2 5\n",
"6\n1 2\n1 3\n1 4\n2 5\n2 6\n"
],
"output": [
"YES",
"NO",
"NO",
"YES"
]
}
|
{
"input": [
"50\n16 4\n17 9\n31 19\n22 10\n8 1\n40 30\n3 31\n20 29\n47 27\n22 25\n32 34\n12 15\n40 32\n10 33\n47 12\n6 24\n46 41\n14 23\n12 35\n31 42\n46 28\n31 20\n46 37\n1 39\n29 49\n37 47\n40 6\n42 36\n47 2\n24 46\n2 13\n8 45\n41 3\n32 17\n4 7\n47 26\n28 8\n41 50\n34 44\n33 21\n25 5\n16 40\n3 14\n8 18\n28 11\n32 22\n2 38\n3 48\n44 43\n",
"10\n8 1\n1 2\n8 9\n8 5\n1 3\n1 10\n1 6\n1 7\n8 4\n",
"5\n5 1\n5 4\n4 3\n1 2\n",
"7\n1 2\n2 3\n1 4\n1 5\n3 6\n3 7\n",
"3\n1 3\n2 3\n",
"60\n26 6\n59 30\n31 12\n31 3\n38 23\n59 29\n53 9\n38 56\n53 54\n29 21\n17 55\n59 38\n26 16\n24 59\n24 25\n17 35\n24 41\n30 15\n31 27\n8 44\n26 5\n26 48\n8 32\n53 17\n3 34\n3 51\n30 28\n47 10\n53 60\n36 42\n24 53\n59 22\n53 40\n26 52\n36 4\n59 8\n29 37\n36 20\n17 47\n53 18\n3 50\n30 2\n17 7\n8 58\n59 1\n31 11\n24 26\n24 43\n53 57\n59 45\n47 13\n26 46\n17 33\n30 31\n26 39\n26 19\n24 36\n8 49\n38 14\n",
"7\n1 2\n2 3\n3 4\n3 5\n1 6\n1 7\n",
"20\n19 16\n19 18\n20 7\n9 4\n6 17\n14 2\n9 15\n2 13\n5 11\n19 12\n12 20\n16 9\n11 8\n19 5\n3 1\n19 14\n5 3\n12 10\n19 6\n",
"7\n1 2\n1 3\n2 4\n2 5\n3 6\n3 7\n",
"10\n9 5\n7 1\n9 10\n7 2\n5 4\n9 6\n2 9\n10 8\n1 3\n",
"4\n2 4\n2 3\n2 1\n",
"4\n1 4\n3 2\n1 3\n",
"3\n1 2\n1 3\n",
"5\n1 2\n1 5\n1 3\n1 4\n",
"20\n14 9\n12 13\n10 15\n2 1\n20 19\n16 6\n16 3\n17 14\n3 5\n2 11\n3 10\n15 8\n14 2\n6 4\n3 20\n5 18\n1 7\n1 16\n4 12\n",
"20\n7 5\n14 13\n17 6\n3 8\n16 12\n18 9\n3 18\n14 1\n17 3\n15 2\n17 4\n9 11\n2 7\n15 17\n3 20\n16 10\n17 14\n2 16\n1 19\n",
"8\n1 2\n2 3\n3 4\n1 7\n1 8\n4 5\n4 6\n",
"5\n5 1\n5 2\n5 3\n5 4\n",
"50\n49 6\n43 7\n1 27\n19 35\n15 37\n16 12\n19 21\n16 28\n49 9\n48 39\n13 1\n2 48\n9 50\n44 3\n41 32\n48 31\n49 33\n6 11\n13 20\n49 22\n13 41\n48 29\n13 46\n15 47\n34 2\n49 13\n48 14\n34 24\n16 36\n13 40\n49 34\n49 17\n43 25\n11 23\n10 15\n19 26\n34 44\n16 42\n19 18\n46 8\n29 38\n1 45\n12 43\n13 16\n46 30\n15 5\n49 10\n11 19\n32 4\n",
"20\n13 1\n18 2\n3 7\n18 5\n20 16\n3 12\n18 9\n3 10\n18 11\n13 6\n3 18\n20 15\n20 17\n3 13\n3 4\n13 14\n3 20\n18 8\n3 19\n",
"10\n8 2\n5 6\n1 8\n2 9\n1 4\n8 10\n10 5\n2 7\n2 3\n"
],
"output": [
"NO",
"YES",
"NO",
"NO",
"NO",
"YES",
"NO",
"NO",
"NO",
"NO",
"YES",
"NO",
"NO",
"YES",
"NO",
"NO",
"NO",
"YES",
"NO",
"YES",
"NO"
]
}
|
IN-CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
using namespace std;
vector<int> conn[200100];
bool oka = true;
void dfs(int u, int par) {
if (par != -1) {
if (conn[u].size() == 1) return;
if (conn[u].size() < 3 || oka == false) {
oka = false;
return;
}
}
for (int v : conn[u]) {
if (v == par) continue;
dfs(v, u);
}
}
int main() {
int n;
cin >> n;
int u, v;
for (int i = 1; i <= n - 1; i++) {
cin >> u >> v;
conn[u].push_back(v);
conn[v].push_back(u);
}
dfs(1, -1);
if (oka)
cout << "YES" << endl;
else
cout << "NO" << endl;
return 0;
}
|
1189_D1. Add on a Tree
|
Note that this is the first problem of the two similar problems. You can hack this problem only if you solve both problems.
You are given a tree with n nodes. In the beginning, 0 is written on all edges. In one operation, you can choose any 2 distinct leaves u, v and any real number x and add x to values written on all edges on the simple path between u and v.
For example, on the picture below you can see the result of applying two operations to the graph: adding 2 on the path from 7 to 6, and then adding -0.5 on the path from 4 to 5.
<image>
Is it true that for any configuration of real numbers written on edges, we can achieve it with a finite number of operations?
Leaf is a node of a tree of degree 1. Simple path is a path that doesn't contain any node twice.
Input
The first line contains a single integer n (2 ≤ n ≤ 10^5) — the number of nodes.
Each of the next n-1 lines contains two integers u and v (1 ≤ u, v ≤ n, u ≠ v), meaning that there is an edge between nodes u and v. It is guaranteed that these edges form a tree.
Output
If there is a configuration of real numbers written on edges of the tree that we can't achieve by performing the operations, output "NO".
Otherwise, output "YES".
You can print each letter in any case (upper or lower).
Examples
Input
2
1 2
Output
YES
Input
3
1 2
2 3
Output
NO
Input
5
1 2
1 3
1 4
2 5
Output
NO
Input
6
1 2
1 3
1 4
2 5
2 6
Output
YES
Note
In the first example, we can add any real x to the value written on the only edge (1, 2).
<image>
In the second example, one of configurations that we can't reach is 0 written on (1, 2) and 1 written on (2, 3).
<image>
Below you can see graphs from examples 3, 4:
<image> <image>
|
{
"input": [
"2\n1 2\n",
"3\n1 2\n2 3\n",
"5\n1 2\n1 3\n1 4\n2 5\n",
"6\n1 2\n1 3\n1 4\n2 5\n2 6\n"
],
"output": [
"YES",
"NO",
"NO",
"YES"
]
}
|
{
"input": [
"50\n16 4\n17 9\n31 19\n22 10\n8 1\n40 30\n3 31\n20 29\n47 27\n22 25\n32 34\n12 15\n40 32\n10 33\n47 12\n6 24\n46 41\n14 23\n12 35\n31 42\n46 28\n31 20\n46 37\n1 39\n29 49\n37 47\n40 6\n42 36\n47 2\n24 46\n2 13\n8 45\n41 3\n32 17\n4 7\n47 26\n28 8\n41 50\n34 44\n33 21\n25 5\n16 40\n3 14\n8 18\n28 11\n32 22\n2 38\n3 48\n44 43\n",
"10\n8 1\n1 2\n8 9\n8 5\n1 3\n1 10\n1 6\n1 7\n8 4\n",
"5\n5 1\n5 4\n4 3\n1 2\n",
"7\n1 2\n2 3\n1 4\n1 5\n3 6\n3 7\n",
"3\n1 3\n2 3\n",
"60\n26 6\n59 30\n31 12\n31 3\n38 23\n59 29\n53 9\n38 56\n53 54\n29 21\n17 55\n59 38\n26 16\n24 59\n24 25\n17 35\n24 41\n30 15\n31 27\n8 44\n26 5\n26 48\n8 32\n53 17\n3 34\n3 51\n30 28\n47 10\n53 60\n36 42\n24 53\n59 22\n53 40\n26 52\n36 4\n59 8\n29 37\n36 20\n17 47\n53 18\n3 50\n30 2\n17 7\n8 58\n59 1\n31 11\n24 26\n24 43\n53 57\n59 45\n47 13\n26 46\n17 33\n30 31\n26 39\n26 19\n24 36\n8 49\n38 14\n",
"7\n1 2\n2 3\n3 4\n3 5\n1 6\n1 7\n",
"20\n19 16\n19 18\n20 7\n9 4\n6 17\n14 2\n9 15\n2 13\n5 11\n19 12\n12 20\n16 9\n11 8\n19 5\n3 1\n19 14\n5 3\n12 10\n19 6\n",
"7\n1 2\n1 3\n2 4\n2 5\n3 6\n3 7\n",
"10\n9 5\n7 1\n9 10\n7 2\n5 4\n9 6\n2 9\n10 8\n1 3\n",
"4\n2 4\n2 3\n2 1\n",
"4\n1 4\n3 2\n1 3\n",
"3\n1 2\n1 3\n",
"5\n1 2\n1 5\n1 3\n1 4\n",
"20\n14 9\n12 13\n10 15\n2 1\n20 19\n16 6\n16 3\n17 14\n3 5\n2 11\n3 10\n15 8\n14 2\n6 4\n3 20\n5 18\n1 7\n1 16\n4 12\n",
"20\n7 5\n14 13\n17 6\n3 8\n16 12\n18 9\n3 18\n14 1\n17 3\n15 2\n17 4\n9 11\n2 7\n15 17\n3 20\n16 10\n17 14\n2 16\n1 19\n",
"8\n1 2\n2 3\n3 4\n1 7\n1 8\n4 5\n4 6\n",
"5\n5 1\n5 2\n5 3\n5 4\n",
"50\n49 6\n43 7\n1 27\n19 35\n15 37\n16 12\n19 21\n16 28\n49 9\n48 39\n13 1\n2 48\n9 50\n44 3\n41 32\n48 31\n49 33\n6 11\n13 20\n49 22\n13 41\n48 29\n13 46\n15 47\n34 2\n49 13\n48 14\n34 24\n16 36\n13 40\n49 34\n49 17\n43 25\n11 23\n10 15\n19 26\n34 44\n16 42\n19 18\n46 8\n29 38\n1 45\n12 43\n13 16\n46 30\n15 5\n49 10\n11 19\n32 4\n",
"20\n13 1\n18 2\n3 7\n18 5\n20 16\n3 12\n18 9\n3 10\n18 11\n13 6\n3 18\n20 15\n20 17\n3 13\n3 4\n13 14\n3 20\n18 8\n3 19\n",
"10\n8 2\n5 6\n1 8\n2 9\n1 4\n8 10\n10 5\n2 7\n2 3\n"
],
"output": [
"NO",
"YES",
"NO",
"NO",
"NO",
"YES",
"NO",
"NO",
"NO",
"NO",
"YES",
"NO",
"NO",
"YES",
"NO",
"NO",
"NO",
"YES",
"NO",
"YES",
"NO"
]
}
|
IN-CORRECT
|
java
|
import java.io.IOException;
import java.io.InputStream;
import java.util.InputMismatchException;
public class D572 {
public static void main(String[] args) {
FastScanner in = new FastScanner(System.in);
int N = in.nextInt();
int[] degrees = new int[N];
for(int i = 0; i < N - 1; i++) {
int u = in.nextInt() - 1;
int v = in.nextInt() - 1;
degrees[u]++;
degrees[v]++;
}
long numLeaves = 0;
for(int i : degrees)
if(i == 1)
numLeaves++;
String ans = "NO";
if((numLeaves*(numLeaves-1) / 2) >= N-1) {
ans = "YES";
}
System.out.println(ans);
}
/**
* Source: Matt Fontaine
*/
static class FastScanner {
private InputStream stream;
private byte[] buf = new byte[1024];
private int curChar;
private int chars;
public FastScanner(InputStream stream) {
this.stream = stream;
}
int read() {
if (chars == -1)
throw new InputMismatchException();
if (curChar >= chars) {
curChar = 0;
try {
chars = stream.read(buf);
} catch (IOException e) {
throw new InputMismatchException();
}
if (chars <= 0)
return -1;
}
return buf[curChar++];
}
boolean isSpaceChar(int c) {
return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
}
boolean isEndline(int c) {
return c == '\n' || c == '\r' || c == -1;
}
public int nextInt() {
return Integer.parseInt(next());
}
public long nextLong() {
return Long.parseLong(next());
}
public double nextDouble() {
return Double.parseDouble(next());
}
public String next() {
int c = read();
while (isSpaceChar(c))
c = read();
StringBuilder res = new StringBuilder();
do {
res.appendCodePoint(c);
c = read();
} while (!isSpaceChar(c));
return res.toString();
}
public String nextLine() {
int c = read();
while (isEndline(c))
c = read();
StringBuilder res = new StringBuilder();
do {
res.appendCodePoint(c);
c = read();
} while (!isEndline(c));
return res.toString();
}
}
}
/*
2
1 2
outputCopy
YES
inputCopy
3
1 2
2 3
outputCopy
NO
inputCopy
5
1 2
1 3
1 4
2 5
outputCopy
NO
inputCopy
6
1 2
1 3
1 4
2 5
2 6
outputCopy
YES
*/
|
1189_D1. Add on a Tree
|
Note that this is the first problem of the two similar problems. You can hack this problem only if you solve both problems.
You are given a tree with n nodes. In the beginning, 0 is written on all edges. In one operation, you can choose any 2 distinct leaves u, v and any real number x and add x to values written on all edges on the simple path between u and v.
For example, on the picture below you can see the result of applying two operations to the graph: adding 2 on the path from 7 to 6, and then adding -0.5 on the path from 4 to 5.
<image>
Is it true that for any configuration of real numbers written on edges, we can achieve it with a finite number of operations?
Leaf is a node of a tree of degree 1. Simple path is a path that doesn't contain any node twice.
Input
The first line contains a single integer n (2 ≤ n ≤ 10^5) — the number of nodes.
Each of the next n-1 lines contains two integers u and v (1 ≤ u, v ≤ n, u ≠ v), meaning that there is an edge between nodes u and v. It is guaranteed that these edges form a tree.
Output
If there is a configuration of real numbers written on edges of the tree that we can't achieve by performing the operations, output "NO".
Otherwise, output "YES".
You can print each letter in any case (upper or lower).
Examples
Input
2
1 2
Output
YES
Input
3
1 2
2 3
Output
NO
Input
5
1 2
1 3
1 4
2 5
Output
NO
Input
6
1 2
1 3
1 4
2 5
2 6
Output
YES
Note
In the first example, we can add any real x to the value written on the only edge (1, 2).
<image>
In the second example, one of configurations that we can't reach is 0 written on (1, 2) and 1 written on (2, 3).
<image>
Below you can see graphs from examples 3, 4:
<image> <image>
|
{
"input": [
"2\n1 2\n",
"3\n1 2\n2 3\n",
"5\n1 2\n1 3\n1 4\n2 5\n",
"6\n1 2\n1 3\n1 4\n2 5\n2 6\n"
],
"output": [
"YES",
"NO",
"NO",
"YES"
]
}
|
{
"input": [
"50\n16 4\n17 9\n31 19\n22 10\n8 1\n40 30\n3 31\n20 29\n47 27\n22 25\n32 34\n12 15\n40 32\n10 33\n47 12\n6 24\n46 41\n14 23\n12 35\n31 42\n46 28\n31 20\n46 37\n1 39\n29 49\n37 47\n40 6\n42 36\n47 2\n24 46\n2 13\n8 45\n41 3\n32 17\n4 7\n47 26\n28 8\n41 50\n34 44\n33 21\n25 5\n16 40\n3 14\n8 18\n28 11\n32 22\n2 38\n3 48\n44 43\n",
"10\n8 1\n1 2\n8 9\n8 5\n1 3\n1 10\n1 6\n1 7\n8 4\n",
"5\n5 1\n5 4\n4 3\n1 2\n",
"7\n1 2\n2 3\n1 4\n1 5\n3 6\n3 7\n",
"3\n1 3\n2 3\n",
"60\n26 6\n59 30\n31 12\n31 3\n38 23\n59 29\n53 9\n38 56\n53 54\n29 21\n17 55\n59 38\n26 16\n24 59\n24 25\n17 35\n24 41\n30 15\n31 27\n8 44\n26 5\n26 48\n8 32\n53 17\n3 34\n3 51\n30 28\n47 10\n53 60\n36 42\n24 53\n59 22\n53 40\n26 52\n36 4\n59 8\n29 37\n36 20\n17 47\n53 18\n3 50\n30 2\n17 7\n8 58\n59 1\n31 11\n24 26\n24 43\n53 57\n59 45\n47 13\n26 46\n17 33\n30 31\n26 39\n26 19\n24 36\n8 49\n38 14\n",
"7\n1 2\n2 3\n3 4\n3 5\n1 6\n1 7\n",
"20\n19 16\n19 18\n20 7\n9 4\n6 17\n14 2\n9 15\n2 13\n5 11\n19 12\n12 20\n16 9\n11 8\n19 5\n3 1\n19 14\n5 3\n12 10\n19 6\n",
"7\n1 2\n1 3\n2 4\n2 5\n3 6\n3 7\n",
"10\n9 5\n7 1\n9 10\n7 2\n5 4\n9 6\n2 9\n10 8\n1 3\n",
"4\n2 4\n2 3\n2 1\n",
"4\n1 4\n3 2\n1 3\n",
"3\n1 2\n1 3\n",
"5\n1 2\n1 5\n1 3\n1 4\n",
"20\n14 9\n12 13\n10 15\n2 1\n20 19\n16 6\n16 3\n17 14\n3 5\n2 11\n3 10\n15 8\n14 2\n6 4\n3 20\n5 18\n1 7\n1 16\n4 12\n",
"20\n7 5\n14 13\n17 6\n3 8\n16 12\n18 9\n3 18\n14 1\n17 3\n15 2\n17 4\n9 11\n2 7\n15 17\n3 20\n16 10\n17 14\n2 16\n1 19\n",
"8\n1 2\n2 3\n3 4\n1 7\n1 8\n4 5\n4 6\n",
"5\n5 1\n5 2\n5 3\n5 4\n",
"50\n49 6\n43 7\n1 27\n19 35\n15 37\n16 12\n19 21\n16 28\n49 9\n48 39\n13 1\n2 48\n9 50\n44 3\n41 32\n48 31\n49 33\n6 11\n13 20\n49 22\n13 41\n48 29\n13 46\n15 47\n34 2\n49 13\n48 14\n34 24\n16 36\n13 40\n49 34\n49 17\n43 25\n11 23\n10 15\n19 26\n34 44\n16 42\n19 18\n46 8\n29 38\n1 45\n12 43\n13 16\n46 30\n15 5\n49 10\n11 19\n32 4\n",
"20\n13 1\n18 2\n3 7\n18 5\n20 16\n3 12\n18 9\n3 10\n18 11\n13 6\n3 18\n20 15\n20 17\n3 13\n3 4\n13 14\n3 20\n18 8\n3 19\n",
"10\n8 2\n5 6\n1 8\n2 9\n1 4\n8 10\n10 5\n2 7\n2 3\n"
],
"output": [
"NO",
"YES",
"NO",
"NO",
"NO",
"YES",
"NO",
"NO",
"NO",
"NO",
"YES",
"NO",
"NO",
"YES",
"NO",
"NO",
"NO",
"YES",
"NO",
"YES",
"NO"
]
}
|
IN-CORRECT
|
python3
|
n = int(input())
a = []
for i in range(0,n+9):
a.append(0)
for i in range(1,n):
u,v = map(int,input().split())
a[u] = a[u]+1
a[v] = a[v]+1
flag = 1;
for i in range(i,1,n+1):
if a[i]==2:
flag = 0
if flag==0:
print("NO")
else:
print("YES")
|
1189_D1. Add on a Tree
|
Note that this is the first problem of the two similar problems. You can hack this problem only if you solve both problems.
You are given a tree with n nodes. In the beginning, 0 is written on all edges. In one operation, you can choose any 2 distinct leaves u, v and any real number x and add x to values written on all edges on the simple path between u and v.
For example, on the picture below you can see the result of applying two operations to the graph: adding 2 on the path from 7 to 6, and then adding -0.5 on the path from 4 to 5.
<image>
Is it true that for any configuration of real numbers written on edges, we can achieve it with a finite number of operations?
Leaf is a node of a tree of degree 1. Simple path is a path that doesn't contain any node twice.
Input
The first line contains a single integer n (2 ≤ n ≤ 10^5) — the number of nodes.
Each of the next n-1 lines contains two integers u and v (1 ≤ u, v ≤ n, u ≠ v), meaning that there is an edge between nodes u and v. It is guaranteed that these edges form a tree.
Output
If there is a configuration of real numbers written on edges of the tree that we can't achieve by performing the operations, output "NO".
Otherwise, output "YES".
You can print each letter in any case (upper or lower).
Examples
Input
2
1 2
Output
YES
Input
3
1 2
2 3
Output
NO
Input
5
1 2
1 3
1 4
2 5
Output
NO
Input
6
1 2
1 3
1 4
2 5
2 6
Output
YES
Note
In the first example, we can add any real x to the value written on the only edge (1, 2).
<image>
In the second example, one of configurations that we can't reach is 0 written on (1, 2) and 1 written on (2, 3).
<image>
Below you can see graphs from examples 3, 4:
<image> <image>
|
{
"input": [
"2\n1 2\n",
"3\n1 2\n2 3\n",
"5\n1 2\n1 3\n1 4\n2 5\n",
"6\n1 2\n1 3\n1 4\n2 5\n2 6\n"
],
"output": [
"YES",
"NO",
"NO",
"YES"
]
}
|
{
"input": [
"50\n16 4\n17 9\n31 19\n22 10\n8 1\n40 30\n3 31\n20 29\n47 27\n22 25\n32 34\n12 15\n40 32\n10 33\n47 12\n6 24\n46 41\n14 23\n12 35\n31 42\n46 28\n31 20\n46 37\n1 39\n29 49\n37 47\n40 6\n42 36\n47 2\n24 46\n2 13\n8 45\n41 3\n32 17\n4 7\n47 26\n28 8\n41 50\n34 44\n33 21\n25 5\n16 40\n3 14\n8 18\n28 11\n32 22\n2 38\n3 48\n44 43\n",
"10\n8 1\n1 2\n8 9\n8 5\n1 3\n1 10\n1 6\n1 7\n8 4\n",
"5\n5 1\n5 4\n4 3\n1 2\n",
"7\n1 2\n2 3\n1 4\n1 5\n3 6\n3 7\n",
"3\n1 3\n2 3\n",
"60\n26 6\n59 30\n31 12\n31 3\n38 23\n59 29\n53 9\n38 56\n53 54\n29 21\n17 55\n59 38\n26 16\n24 59\n24 25\n17 35\n24 41\n30 15\n31 27\n8 44\n26 5\n26 48\n8 32\n53 17\n3 34\n3 51\n30 28\n47 10\n53 60\n36 42\n24 53\n59 22\n53 40\n26 52\n36 4\n59 8\n29 37\n36 20\n17 47\n53 18\n3 50\n30 2\n17 7\n8 58\n59 1\n31 11\n24 26\n24 43\n53 57\n59 45\n47 13\n26 46\n17 33\n30 31\n26 39\n26 19\n24 36\n8 49\n38 14\n",
"7\n1 2\n2 3\n3 4\n3 5\n1 6\n1 7\n",
"20\n19 16\n19 18\n20 7\n9 4\n6 17\n14 2\n9 15\n2 13\n5 11\n19 12\n12 20\n16 9\n11 8\n19 5\n3 1\n19 14\n5 3\n12 10\n19 6\n",
"7\n1 2\n1 3\n2 4\n2 5\n3 6\n3 7\n",
"10\n9 5\n7 1\n9 10\n7 2\n5 4\n9 6\n2 9\n10 8\n1 3\n",
"4\n2 4\n2 3\n2 1\n",
"4\n1 4\n3 2\n1 3\n",
"3\n1 2\n1 3\n",
"5\n1 2\n1 5\n1 3\n1 4\n",
"20\n14 9\n12 13\n10 15\n2 1\n20 19\n16 6\n16 3\n17 14\n3 5\n2 11\n3 10\n15 8\n14 2\n6 4\n3 20\n5 18\n1 7\n1 16\n4 12\n",
"20\n7 5\n14 13\n17 6\n3 8\n16 12\n18 9\n3 18\n14 1\n17 3\n15 2\n17 4\n9 11\n2 7\n15 17\n3 20\n16 10\n17 14\n2 16\n1 19\n",
"8\n1 2\n2 3\n3 4\n1 7\n1 8\n4 5\n4 6\n",
"5\n5 1\n5 2\n5 3\n5 4\n",
"50\n49 6\n43 7\n1 27\n19 35\n15 37\n16 12\n19 21\n16 28\n49 9\n48 39\n13 1\n2 48\n9 50\n44 3\n41 32\n48 31\n49 33\n6 11\n13 20\n49 22\n13 41\n48 29\n13 46\n15 47\n34 2\n49 13\n48 14\n34 24\n16 36\n13 40\n49 34\n49 17\n43 25\n11 23\n10 15\n19 26\n34 44\n16 42\n19 18\n46 8\n29 38\n1 45\n12 43\n13 16\n46 30\n15 5\n49 10\n11 19\n32 4\n",
"20\n13 1\n18 2\n3 7\n18 5\n20 16\n3 12\n18 9\n3 10\n18 11\n13 6\n3 18\n20 15\n20 17\n3 13\n3 4\n13 14\n3 20\n18 8\n3 19\n",
"10\n8 2\n5 6\n1 8\n2 9\n1 4\n8 10\n10 5\n2 7\n2 3\n"
],
"output": [
"NO",
"YES",
"NO",
"NO",
"NO",
"YES",
"NO",
"NO",
"NO",
"NO",
"YES",
"NO",
"NO",
"YES",
"NO",
"NO",
"NO",
"YES",
"NO",
"YES",
"NO"
]
}
|
IN-CORRECT
|
java
|
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.StringTokenizer;
public class Main
{
public static void main(String[] args)
{
FastReader fr =new FastReader(); PrintWriter op =new PrintWriter(System.out);
int n =fr.nextInt() ,i ,dgre[] =new int[n] ,eg[][] =new int[n-1][2] ; boolean f =false ;
if (n == 2) op.println("YES") ;
else {
for (i =0 ; i<n-1 ; ++i) {
eg[i][0] =fr.nextInt()-1 ; eg[i][1] =fr.nextInt()-1 ; dgre[eg[i][0]]++ ; dgre[eg[i][1]]++ ;
}
for (i =0 ; i<n-1 ; ++i) {
if (dgre[eg[i][0]] < 3 && dgre[eg[i][1]] < 3) { f =true ; break; }
}
if (f) op.println("NO") ; else op.println("YES") ;
} op.flush(); op.close();
}
static class FastReader {
BufferedReader br;
StringTokenizer st;
public FastReader() {
br =new BufferedReader(new InputStreamReader(System.in));
}
String next() {
while (st==null || (!st.hasMoreElements()))
{
try
{
st =new StringTokenizer(br.readLine());
}
catch(IOException e)
{
e.printStackTrace();
}
}
return st.nextToken();
}
String nextLine() {
String str ="";
try
{
str =br.readLine();
}
catch(IOException e)
{
e.printStackTrace();
}
return str;
}
int nextInt() {
return Integer.parseInt(next());
}
long nextLong() {
return Long.parseLong(next()) ;
}
}
}
|
1189_D1. Add on a Tree
|
Note that this is the first problem of the two similar problems. You can hack this problem only if you solve both problems.
You are given a tree with n nodes. In the beginning, 0 is written on all edges. In one operation, you can choose any 2 distinct leaves u, v and any real number x and add x to values written on all edges on the simple path between u and v.
For example, on the picture below you can see the result of applying two operations to the graph: adding 2 on the path from 7 to 6, and then adding -0.5 on the path from 4 to 5.
<image>
Is it true that for any configuration of real numbers written on edges, we can achieve it with a finite number of operations?
Leaf is a node of a tree of degree 1. Simple path is a path that doesn't contain any node twice.
Input
The first line contains a single integer n (2 ≤ n ≤ 10^5) — the number of nodes.
Each of the next n-1 lines contains two integers u and v (1 ≤ u, v ≤ n, u ≠ v), meaning that there is an edge between nodes u and v. It is guaranteed that these edges form a tree.
Output
If there is a configuration of real numbers written on edges of the tree that we can't achieve by performing the operations, output "NO".
Otherwise, output "YES".
You can print each letter in any case (upper or lower).
Examples
Input
2
1 2
Output
YES
Input
3
1 2
2 3
Output
NO
Input
5
1 2
1 3
1 4
2 5
Output
NO
Input
6
1 2
1 3
1 4
2 5
2 6
Output
YES
Note
In the first example, we can add any real x to the value written on the only edge (1, 2).
<image>
In the second example, one of configurations that we can't reach is 0 written on (1, 2) and 1 written on (2, 3).
<image>
Below you can see graphs from examples 3, 4:
<image> <image>
|
{
"input": [
"2\n1 2\n",
"3\n1 2\n2 3\n",
"5\n1 2\n1 3\n1 4\n2 5\n",
"6\n1 2\n1 3\n1 4\n2 5\n2 6\n"
],
"output": [
"YES",
"NO",
"NO",
"YES"
]
}
|
{
"input": [
"50\n16 4\n17 9\n31 19\n22 10\n8 1\n40 30\n3 31\n20 29\n47 27\n22 25\n32 34\n12 15\n40 32\n10 33\n47 12\n6 24\n46 41\n14 23\n12 35\n31 42\n46 28\n31 20\n46 37\n1 39\n29 49\n37 47\n40 6\n42 36\n47 2\n24 46\n2 13\n8 45\n41 3\n32 17\n4 7\n47 26\n28 8\n41 50\n34 44\n33 21\n25 5\n16 40\n3 14\n8 18\n28 11\n32 22\n2 38\n3 48\n44 43\n",
"10\n8 1\n1 2\n8 9\n8 5\n1 3\n1 10\n1 6\n1 7\n8 4\n",
"5\n5 1\n5 4\n4 3\n1 2\n",
"7\n1 2\n2 3\n1 4\n1 5\n3 6\n3 7\n",
"3\n1 3\n2 3\n",
"60\n26 6\n59 30\n31 12\n31 3\n38 23\n59 29\n53 9\n38 56\n53 54\n29 21\n17 55\n59 38\n26 16\n24 59\n24 25\n17 35\n24 41\n30 15\n31 27\n8 44\n26 5\n26 48\n8 32\n53 17\n3 34\n3 51\n30 28\n47 10\n53 60\n36 42\n24 53\n59 22\n53 40\n26 52\n36 4\n59 8\n29 37\n36 20\n17 47\n53 18\n3 50\n30 2\n17 7\n8 58\n59 1\n31 11\n24 26\n24 43\n53 57\n59 45\n47 13\n26 46\n17 33\n30 31\n26 39\n26 19\n24 36\n8 49\n38 14\n",
"7\n1 2\n2 3\n3 4\n3 5\n1 6\n1 7\n",
"20\n19 16\n19 18\n20 7\n9 4\n6 17\n14 2\n9 15\n2 13\n5 11\n19 12\n12 20\n16 9\n11 8\n19 5\n3 1\n19 14\n5 3\n12 10\n19 6\n",
"7\n1 2\n1 3\n2 4\n2 5\n3 6\n3 7\n",
"10\n9 5\n7 1\n9 10\n7 2\n5 4\n9 6\n2 9\n10 8\n1 3\n",
"4\n2 4\n2 3\n2 1\n",
"4\n1 4\n3 2\n1 3\n",
"3\n1 2\n1 3\n",
"5\n1 2\n1 5\n1 3\n1 4\n",
"20\n14 9\n12 13\n10 15\n2 1\n20 19\n16 6\n16 3\n17 14\n3 5\n2 11\n3 10\n15 8\n14 2\n6 4\n3 20\n5 18\n1 7\n1 16\n4 12\n",
"20\n7 5\n14 13\n17 6\n3 8\n16 12\n18 9\n3 18\n14 1\n17 3\n15 2\n17 4\n9 11\n2 7\n15 17\n3 20\n16 10\n17 14\n2 16\n1 19\n",
"8\n1 2\n2 3\n3 4\n1 7\n1 8\n4 5\n4 6\n",
"5\n5 1\n5 2\n5 3\n5 4\n",
"50\n49 6\n43 7\n1 27\n19 35\n15 37\n16 12\n19 21\n16 28\n49 9\n48 39\n13 1\n2 48\n9 50\n44 3\n41 32\n48 31\n49 33\n6 11\n13 20\n49 22\n13 41\n48 29\n13 46\n15 47\n34 2\n49 13\n48 14\n34 24\n16 36\n13 40\n49 34\n49 17\n43 25\n11 23\n10 15\n19 26\n34 44\n16 42\n19 18\n46 8\n29 38\n1 45\n12 43\n13 16\n46 30\n15 5\n49 10\n11 19\n32 4\n",
"20\n13 1\n18 2\n3 7\n18 5\n20 16\n3 12\n18 9\n3 10\n18 11\n13 6\n3 18\n20 15\n20 17\n3 13\n3 4\n13 14\n3 20\n18 8\n3 19\n",
"10\n8 2\n5 6\n1 8\n2 9\n1 4\n8 10\n10 5\n2 7\n2 3\n"
],
"output": [
"NO",
"YES",
"NO",
"NO",
"NO",
"YES",
"NO",
"NO",
"NO",
"NO",
"YES",
"NO",
"NO",
"YES",
"NO",
"NO",
"NO",
"YES",
"NO",
"YES",
"NO"
]
}
|
IN-CORRECT
|
python3
|
from sys import stdin
input=stdin.readline
n=int(input())
a=[[] for i in range(n)]
for i in range(n-1):
c,d=map(int,input().split())
a[c-1].append(d-1)
a[d-1].append(c-1)
if n==2:
print('YES')
else:
k=0
for i in a:
if len(i)==1:
k+=1
if k%2==1:
print('NO')
else:
print('YES')
|
1189_D1. Add on a Tree
|
Note that this is the first problem of the two similar problems. You can hack this problem only if you solve both problems.
You are given a tree with n nodes. In the beginning, 0 is written on all edges. In one operation, you can choose any 2 distinct leaves u, v and any real number x and add x to values written on all edges on the simple path between u and v.
For example, on the picture below you can see the result of applying two operations to the graph: adding 2 on the path from 7 to 6, and then adding -0.5 on the path from 4 to 5.
<image>
Is it true that for any configuration of real numbers written on edges, we can achieve it with a finite number of operations?
Leaf is a node of a tree of degree 1. Simple path is a path that doesn't contain any node twice.
Input
The first line contains a single integer n (2 ≤ n ≤ 10^5) — the number of nodes.
Each of the next n-1 lines contains two integers u and v (1 ≤ u, v ≤ n, u ≠ v), meaning that there is an edge between nodes u and v. It is guaranteed that these edges form a tree.
Output
If there is a configuration of real numbers written on edges of the tree that we can't achieve by performing the operations, output "NO".
Otherwise, output "YES".
You can print each letter in any case (upper or lower).
Examples
Input
2
1 2
Output
YES
Input
3
1 2
2 3
Output
NO
Input
5
1 2
1 3
1 4
2 5
Output
NO
Input
6
1 2
1 3
1 4
2 5
2 6
Output
YES
Note
In the first example, we can add any real x to the value written on the only edge (1, 2).
<image>
In the second example, one of configurations that we can't reach is 0 written on (1, 2) and 1 written on (2, 3).
<image>
Below you can see graphs from examples 3, 4:
<image> <image>
|
{
"input": [
"2\n1 2\n",
"3\n1 2\n2 3\n",
"5\n1 2\n1 3\n1 4\n2 5\n",
"6\n1 2\n1 3\n1 4\n2 5\n2 6\n"
],
"output": [
"YES",
"NO",
"NO",
"YES"
]
}
|
{
"input": [
"50\n16 4\n17 9\n31 19\n22 10\n8 1\n40 30\n3 31\n20 29\n47 27\n22 25\n32 34\n12 15\n40 32\n10 33\n47 12\n6 24\n46 41\n14 23\n12 35\n31 42\n46 28\n31 20\n46 37\n1 39\n29 49\n37 47\n40 6\n42 36\n47 2\n24 46\n2 13\n8 45\n41 3\n32 17\n4 7\n47 26\n28 8\n41 50\n34 44\n33 21\n25 5\n16 40\n3 14\n8 18\n28 11\n32 22\n2 38\n3 48\n44 43\n",
"10\n8 1\n1 2\n8 9\n8 5\n1 3\n1 10\n1 6\n1 7\n8 4\n",
"5\n5 1\n5 4\n4 3\n1 2\n",
"7\n1 2\n2 3\n1 4\n1 5\n3 6\n3 7\n",
"3\n1 3\n2 3\n",
"60\n26 6\n59 30\n31 12\n31 3\n38 23\n59 29\n53 9\n38 56\n53 54\n29 21\n17 55\n59 38\n26 16\n24 59\n24 25\n17 35\n24 41\n30 15\n31 27\n8 44\n26 5\n26 48\n8 32\n53 17\n3 34\n3 51\n30 28\n47 10\n53 60\n36 42\n24 53\n59 22\n53 40\n26 52\n36 4\n59 8\n29 37\n36 20\n17 47\n53 18\n3 50\n30 2\n17 7\n8 58\n59 1\n31 11\n24 26\n24 43\n53 57\n59 45\n47 13\n26 46\n17 33\n30 31\n26 39\n26 19\n24 36\n8 49\n38 14\n",
"7\n1 2\n2 3\n3 4\n3 5\n1 6\n1 7\n",
"20\n19 16\n19 18\n20 7\n9 4\n6 17\n14 2\n9 15\n2 13\n5 11\n19 12\n12 20\n16 9\n11 8\n19 5\n3 1\n19 14\n5 3\n12 10\n19 6\n",
"7\n1 2\n1 3\n2 4\n2 5\n3 6\n3 7\n",
"10\n9 5\n7 1\n9 10\n7 2\n5 4\n9 6\n2 9\n10 8\n1 3\n",
"4\n2 4\n2 3\n2 1\n",
"4\n1 4\n3 2\n1 3\n",
"3\n1 2\n1 3\n",
"5\n1 2\n1 5\n1 3\n1 4\n",
"20\n14 9\n12 13\n10 15\n2 1\n20 19\n16 6\n16 3\n17 14\n3 5\n2 11\n3 10\n15 8\n14 2\n6 4\n3 20\n5 18\n1 7\n1 16\n4 12\n",
"20\n7 5\n14 13\n17 6\n3 8\n16 12\n18 9\n3 18\n14 1\n17 3\n15 2\n17 4\n9 11\n2 7\n15 17\n3 20\n16 10\n17 14\n2 16\n1 19\n",
"8\n1 2\n2 3\n3 4\n1 7\n1 8\n4 5\n4 6\n",
"5\n5 1\n5 2\n5 3\n5 4\n",
"50\n49 6\n43 7\n1 27\n19 35\n15 37\n16 12\n19 21\n16 28\n49 9\n48 39\n13 1\n2 48\n9 50\n44 3\n41 32\n48 31\n49 33\n6 11\n13 20\n49 22\n13 41\n48 29\n13 46\n15 47\n34 2\n49 13\n48 14\n34 24\n16 36\n13 40\n49 34\n49 17\n43 25\n11 23\n10 15\n19 26\n34 44\n16 42\n19 18\n46 8\n29 38\n1 45\n12 43\n13 16\n46 30\n15 5\n49 10\n11 19\n32 4\n",
"20\n13 1\n18 2\n3 7\n18 5\n20 16\n3 12\n18 9\n3 10\n18 11\n13 6\n3 18\n20 15\n20 17\n3 13\n3 4\n13 14\n3 20\n18 8\n3 19\n",
"10\n8 2\n5 6\n1 8\n2 9\n1 4\n8 10\n10 5\n2 7\n2 3\n"
],
"output": [
"NO",
"YES",
"NO",
"NO",
"NO",
"YES",
"NO",
"NO",
"NO",
"NO",
"YES",
"NO",
"NO",
"YES",
"NO",
"NO",
"NO",
"YES",
"NO",
"YES",
"NO"
]
}
|
IN-CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
int g[100010];
int f;
int main() {
int n, i, j, ok, x, y;
scanf("%d", &n);
for (i = 1; i < n; i++) {
scanf("%d%d", &x, &y);
g[x]++;
g[y]++;
}
for (i = 1; i <= n; i++)
if (g[i] == 1) f++;
f = (f * (f - 1)) / 2;
if (f >= n - 1)
printf("YES");
else
printf("NO");
}
|
1189_D1. Add on a Tree
|
Note that this is the first problem of the two similar problems. You can hack this problem only if you solve both problems.
You are given a tree with n nodes. In the beginning, 0 is written on all edges. In one operation, you can choose any 2 distinct leaves u, v and any real number x and add x to values written on all edges on the simple path between u and v.
For example, on the picture below you can see the result of applying two operations to the graph: adding 2 on the path from 7 to 6, and then adding -0.5 on the path from 4 to 5.
<image>
Is it true that for any configuration of real numbers written on edges, we can achieve it with a finite number of operations?
Leaf is a node of a tree of degree 1. Simple path is a path that doesn't contain any node twice.
Input
The first line contains a single integer n (2 ≤ n ≤ 10^5) — the number of nodes.
Each of the next n-1 lines contains two integers u and v (1 ≤ u, v ≤ n, u ≠ v), meaning that there is an edge between nodes u and v. It is guaranteed that these edges form a tree.
Output
If there is a configuration of real numbers written on edges of the tree that we can't achieve by performing the operations, output "NO".
Otherwise, output "YES".
You can print each letter in any case (upper or lower).
Examples
Input
2
1 2
Output
YES
Input
3
1 2
2 3
Output
NO
Input
5
1 2
1 3
1 4
2 5
Output
NO
Input
6
1 2
1 3
1 4
2 5
2 6
Output
YES
Note
In the first example, we can add any real x to the value written on the only edge (1, 2).
<image>
In the second example, one of configurations that we can't reach is 0 written on (1, 2) and 1 written on (2, 3).
<image>
Below you can see graphs from examples 3, 4:
<image> <image>
|
{
"input": [
"2\n1 2\n",
"3\n1 2\n2 3\n",
"5\n1 2\n1 3\n1 4\n2 5\n",
"6\n1 2\n1 3\n1 4\n2 5\n2 6\n"
],
"output": [
"YES",
"NO",
"NO",
"YES"
]
}
|
{
"input": [
"50\n16 4\n17 9\n31 19\n22 10\n8 1\n40 30\n3 31\n20 29\n47 27\n22 25\n32 34\n12 15\n40 32\n10 33\n47 12\n6 24\n46 41\n14 23\n12 35\n31 42\n46 28\n31 20\n46 37\n1 39\n29 49\n37 47\n40 6\n42 36\n47 2\n24 46\n2 13\n8 45\n41 3\n32 17\n4 7\n47 26\n28 8\n41 50\n34 44\n33 21\n25 5\n16 40\n3 14\n8 18\n28 11\n32 22\n2 38\n3 48\n44 43\n",
"10\n8 1\n1 2\n8 9\n8 5\n1 3\n1 10\n1 6\n1 7\n8 4\n",
"5\n5 1\n5 4\n4 3\n1 2\n",
"7\n1 2\n2 3\n1 4\n1 5\n3 6\n3 7\n",
"3\n1 3\n2 3\n",
"60\n26 6\n59 30\n31 12\n31 3\n38 23\n59 29\n53 9\n38 56\n53 54\n29 21\n17 55\n59 38\n26 16\n24 59\n24 25\n17 35\n24 41\n30 15\n31 27\n8 44\n26 5\n26 48\n8 32\n53 17\n3 34\n3 51\n30 28\n47 10\n53 60\n36 42\n24 53\n59 22\n53 40\n26 52\n36 4\n59 8\n29 37\n36 20\n17 47\n53 18\n3 50\n30 2\n17 7\n8 58\n59 1\n31 11\n24 26\n24 43\n53 57\n59 45\n47 13\n26 46\n17 33\n30 31\n26 39\n26 19\n24 36\n8 49\n38 14\n",
"7\n1 2\n2 3\n3 4\n3 5\n1 6\n1 7\n",
"20\n19 16\n19 18\n20 7\n9 4\n6 17\n14 2\n9 15\n2 13\n5 11\n19 12\n12 20\n16 9\n11 8\n19 5\n3 1\n19 14\n5 3\n12 10\n19 6\n",
"7\n1 2\n1 3\n2 4\n2 5\n3 6\n3 7\n",
"10\n9 5\n7 1\n9 10\n7 2\n5 4\n9 6\n2 9\n10 8\n1 3\n",
"4\n2 4\n2 3\n2 1\n",
"4\n1 4\n3 2\n1 3\n",
"3\n1 2\n1 3\n",
"5\n1 2\n1 5\n1 3\n1 4\n",
"20\n14 9\n12 13\n10 15\n2 1\n20 19\n16 6\n16 3\n17 14\n3 5\n2 11\n3 10\n15 8\n14 2\n6 4\n3 20\n5 18\n1 7\n1 16\n4 12\n",
"20\n7 5\n14 13\n17 6\n3 8\n16 12\n18 9\n3 18\n14 1\n17 3\n15 2\n17 4\n9 11\n2 7\n15 17\n3 20\n16 10\n17 14\n2 16\n1 19\n",
"8\n1 2\n2 3\n3 4\n1 7\n1 8\n4 5\n4 6\n",
"5\n5 1\n5 2\n5 3\n5 4\n",
"50\n49 6\n43 7\n1 27\n19 35\n15 37\n16 12\n19 21\n16 28\n49 9\n48 39\n13 1\n2 48\n9 50\n44 3\n41 32\n48 31\n49 33\n6 11\n13 20\n49 22\n13 41\n48 29\n13 46\n15 47\n34 2\n49 13\n48 14\n34 24\n16 36\n13 40\n49 34\n49 17\n43 25\n11 23\n10 15\n19 26\n34 44\n16 42\n19 18\n46 8\n29 38\n1 45\n12 43\n13 16\n46 30\n15 5\n49 10\n11 19\n32 4\n",
"20\n13 1\n18 2\n3 7\n18 5\n20 16\n3 12\n18 9\n3 10\n18 11\n13 6\n3 18\n20 15\n20 17\n3 13\n3 4\n13 14\n3 20\n18 8\n3 19\n",
"10\n8 2\n5 6\n1 8\n2 9\n1 4\n8 10\n10 5\n2 7\n2 3\n"
],
"output": [
"NO",
"YES",
"NO",
"NO",
"NO",
"YES",
"NO",
"NO",
"NO",
"NO",
"YES",
"NO",
"NO",
"YES",
"NO",
"NO",
"NO",
"YES",
"NO",
"YES",
"NO"
]
}
|
IN-CORRECT
|
java
|
import java.io.*;
import java.util.ArrayList;
import java.util.StringTokenizer;
public class D1_AddOnATree {
public static void main(String[] args) {
InputStream inputStream = System.in;
OutputStream outputStream = System.out;
InputReader inp = new InputReader(inputStream);
PrintWriter out = new PrintWriter(outputStream);
Solver solver = new Solver();
solver.solve(inp, out);
out.close();
}
private static class Solver {
private void solve(InputReader inp, PrintWriter out) {
int n = inp.nextInt();
if (n == 2) {
out.print("YES");
return;
}
ArrayList<Integer>[] tree = new ArrayList[n];
for (int i = 0; i < n; i++) tree[i] = new ArrayList<>();
for (int i = 0; i < n - 1; i++) {
int u = inp.nextInt() - 1, v = inp.nextInt() - 1;
tree[u].add(v);
tree[v].add(u);
}
for (int i = 0; i < n; i++) {
if (tree[i].size() == 1) {
if (tree[tree[i].get(0)].size() == 2) {
out.print("NO");
return;
}
}
}
out.print("YES");
}
}
static class InputReader {
BufferedReader reader;
StringTokenizer tokenizer;
InputReader(InputStream stream) {
reader = new BufferedReader(new InputStreamReader(stream), 32768);
tokenizer = null;
}
String next() {
while (tokenizer == null || !tokenizer.hasMoreTokens()) {
try {
tokenizer = new StringTokenizer(reader.readLine());
} catch (IOException e) {
throw new RuntimeException(e);
}
}
return tokenizer.nextToken();
}
public int nextInt() {
return Integer.parseInt(next());
}
public long nextLong() {
return Long.parseLong(next());
}
}
}
|
1189_D1. Add on a Tree
|
Note that this is the first problem of the two similar problems. You can hack this problem only if you solve both problems.
You are given a tree with n nodes. In the beginning, 0 is written on all edges. In one operation, you can choose any 2 distinct leaves u, v and any real number x and add x to values written on all edges on the simple path between u and v.
For example, on the picture below you can see the result of applying two operations to the graph: adding 2 on the path from 7 to 6, and then adding -0.5 on the path from 4 to 5.
<image>
Is it true that for any configuration of real numbers written on edges, we can achieve it with a finite number of operations?
Leaf is a node of a tree of degree 1. Simple path is a path that doesn't contain any node twice.
Input
The first line contains a single integer n (2 ≤ n ≤ 10^5) — the number of nodes.
Each of the next n-1 lines contains two integers u and v (1 ≤ u, v ≤ n, u ≠ v), meaning that there is an edge between nodes u and v. It is guaranteed that these edges form a tree.
Output
If there is a configuration of real numbers written on edges of the tree that we can't achieve by performing the operations, output "NO".
Otherwise, output "YES".
You can print each letter in any case (upper or lower).
Examples
Input
2
1 2
Output
YES
Input
3
1 2
2 3
Output
NO
Input
5
1 2
1 3
1 4
2 5
Output
NO
Input
6
1 2
1 3
1 4
2 5
2 6
Output
YES
Note
In the first example, we can add any real x to the value written on the only edge (1, 2).
<image>
In the second example, one of configurations that we can't reach is 0 written on (1, 2) and 1 written on (2, 3).
<image>
Below you can see graphs from examples 3, 4:
<image> <image>
|
{
"input": [
"2\n1 2\n",
"3\n1 2\n2 3\n",
"5\n1 2\n1 3\n1 4\n2 5\n",
"6\n1 2\n1 3\n1 4\n2 5\n2 6\n"
],
"output": [
"YES",
"NO",
"NO",
"YES"
]
}
|
{
"input": [
"50\n16 4\n17 9\n31 19\n22 10\n8 1\n40 30\n3 31\n20 29\n47 27\n22 25\n32 34\n12 15\n40 32\n10 33\n47 12\n6 24\n46 41\n14 23\n12 35\n31 42\n46 28\n31 20\n46 37\n1 39\n29 49\n37 47\n40 6\n42 36\n47 2\n24 46\n2 13\n8 45\n41 3\n32 17\n4 7\n47 26\n28 8\n41 50\n34 44\n33 21\n25 5\n16 40\n3 14\n8 18\n28 11\n32 22\n2 38\n3 48\n44 43\n",
"10\n8 1\n1 2\n8 9\n8 5\n1 3\n1 10\n1 6\n1 7\n8 4\n",
"5\n5 1\n5 4\n4 3\n1 2\n",
"7\n1 2\n2 3\n1 4\n1 5\n3 6\n3 7\n",
"3\n1 3\n2 3\n",
"60\n26 6\n59 30\n31 12\n31 3\n38 23\n59 29\n53 9\n38 56\n53 54\n29 21\n17 55\n59 38\n26 16\n24 59\n24 25\n17 35\n24 41\n30 15\n31 27\n8 44\n26 5\n26 48\n8 32\n53 17\n3 34\n3 51\n30 28\n47 10\n53 60\n36 42\n24 53\n59 22\n53 40\n26 52\n36 4\n59 8\n29 37\n36 20\n17 47\n53 18\n3 50\n30 2\n17 7\n8 58\n59 1\n31 11\n24 26\n24 43\n53 57\n59 45\n47 13\n26 46\n17 33\n30 31\n26 39\n26 19\n24 36\n8 49\n38 14\n",
"7\n1 2\n2 3\n3 4\n3 5\n1 6\n1 7\n",
"20\n19 16\n19 18\n20 7\n9 4\n6 17\n14 2\n9 15\n2 13\n5 11\n19 12\n12 20\n16 9\n11 8\n19 5\n3 1\n19 14\n5 3\n12 10\n19 6\n",
"7\n1 2\n1 3\n2 4\n2 5\n3 6\n3 7\n",
"10\n9 5\n7 1\n9 10\n7 2\n5 4\n9 6\n2 9\n10 8\n1 3\n",
"4\n2 4\n2 3\n2 1\n",
"4\n1 4\n3 2\n1 3\n",
"3\n1 2\n1 3\n",
"5\n1 2\n1 5\n1 3\n1 4\n",
"20\n14 9\n12 13\n10 15\n2 1\n20 19\n16 6\n16 3\n17 14\n3 5\n2 11\n3 10\n15 8\n14 2\n6 4\n3 20\n5 18\n1 7\n1 16\n4 12\n",
"20\n7 5\n14 13\n17 6\n3 8\n16 12\n18 9\n3 18\n14 1\n17 3\n15 2\n17 4\n9 11\n2 7\n15 17\n3 20\n16 10\n17 14\n2 16\n1 19\n",
"8\n1 2\n2 3\n3 4\n1 7\n1 8\n4 5\n4 6\n",
"5\n5 1\n5 2\n5 3\n5 4\n",
"50\n49 6\n43 7\n1 27\n19 35\n15 37\n16 12\n19 21\n16 28\n49 9\n48 39\n13 1\n2 48\n9 50\n44 3\n41 32\n48 31\n49 33\n6 11\n13 20\n49 22\n13 41\n48 29\n13 46\n15 47\n34 2\n49 13\n48 14\n34 24\n16 36\n13 40\n49 34\n49 17\n43 25\n11 23\n10 15\n19 26\n34 44\n16 42\n19 18\n46 8\n29 38\n1 45\n12 43\n13 16\n46 30\n15 5\n49 10\n11 19\n32 4\n",
"20\n13 1\n18 2\n3 7\n18 5\n20 16\n3 12\n18 9\n3 10\n18 11\n13 6\n3 18\n20 15\n20 17\n3 13\n3 4\n13 14\n3 20\n18 8\n3 19\n",
"10\n8 2\n5 6\n1 8\n2 9\n1 4\n8 10\n10 5\n2 7\n2 3\n"
],
"output": [
"NO",
"YES",
"NO",
"NO",
"NO",
"YES",
"NO",
"NO",
"NO",
"NO",
"YES",
"NO",
"NO",
"YES",
"NO",
"NO",
"NO",
"YES",
"NO",
"YES",
"NO"
]
}
|
IN-CORRECT
|
java
|
import java.io.*;
import java.nio.CharBuffer;
import java.util.NoSuchElementException;
public class P1189D {
public static void main(String[] args) {
SimpleScanner scanner = new SimpleScanner(System.in);
PrintWriter writer = new PrintWriter(System.out);
int n = scanner.nextInt();
int[] d = new int[n];
for (int i = 0; i < n - 1; ++i) {
int u = scanner.nextInt() - 1;
int v = scanner.nextInt() - 1;
++d[u];
++d[v];
}
boolean valid = true;
for (int i = 0; i < n - 1; ++i) {
if (d[i] == 2) {
valid = false;
break;
}
}
writer.println(valid ? "YES" : "NO");
writer.close();
}
private static class SimpleScanner {
private static final int BUFFER_SIZE = 10240;
private Readable in;
private CharBuffer buffer;
private boolean eof;
SimpleScanner(InputStream in) {
this.in = new BufferedReader(new InputStreamReader(in));
buffer = CharBuffer.allocate(BUFFER_SIZE);
buffer.limit(0);
eof = false;
}
private char read() {
if (!buffer.hasRemaining()) {
buffer.clear();
int n;
try {
n = in.read(buffer);
} catch (IOException e) {
n = -1;
}
if (n <= 0) {
eof = true;
return '\0';
}
buffer.flip();
}
return buffer.get();
}
void checkEof() {
if (eof)
throw new NoSuchElementException();
}
char nextChar() {
checkEof();
char b = read();
checkEof();
return b;
}
String next() {
char b;
do {
b = read();
checkEof();
} while (Character.isWhitespace(b));
StringBuilder sb = new StringBuilder();
do {
sb.append(b);
b = read();
} while (!eof && !Character.isWhitespace(b));
return sb.toString();
}
int nextInt() {
return Integer.valueOf(next());
}
long nextLong() {
return Long.valueOf(next());
}
double nextDouble() {
return Double.parseDouble(next());
}
}
}
|
1189_D1. Add on a Tree
|
Note that this is the first problem of the two similar problems. You can hack this problem only if you solve both problems.
You are given a tree with n nodes. In the beginning, 0 is written on all edges. In one operation, you can choose any 2 distinct leaves u, v and any real number x and add x to values written on all edges on the simple path between u and v.
For example, on the picture below you can see the result of applying two operations to the graph: adding 2 on the path from 7 to 6, and then adding -0.5 on the path from 4 to 5.
<image>
Is it true that for any configuration of real numbers written on edges, we can achieve it with a finite number of operations?
Leaf is a node of a tree of degree 1. Simple path is a path that doesn't contain any node twice.
Input
The first line contains a single integer n (2 ≤ n ≤ 10^5) — the number of nodes.
Each of the next n-1 lines contains two integers u and v (1 ≤ u, v ≤ n, u ≠ v), meaning that there is an edge between nodes u and v. It is guaranteed that these edges form a tree.
Output
If there is a configuration of real numbers written on edges of the tree that we can't achieve by performing the operations, output "NO".
Otherwise, output "YES".
You can print each letter in any case (upper or lower).
Examples
Input
2
1 2
Output
YES
Input
3
1 2
2 3
Output
NO
Input
5
1 2
1 3
1 4
2 5
Output
NO
Input
6
1 2
1 3
1 4
2 5
2 6
Output
YES
Note
In the first example, we can add any real x to the value written on the only edge (1, 2).
<image>
In the second example, one of configurations that we can't reach is 0 written on (1, 2) and 1 written on (2, 3).
<image>
Below you can see graphs from examples 3, 4:
<image> <image>
|
{
"input": [
"2\n1 2\n",
"3\n1 2\n2 3\n",
"5\n1 2\n1 3\n1 4\n2 5\n",
"6\n1 2\n1 3\n1 4\n2 5\n2 6\n"
],
"output": [
"YES",
"NO",
"NO",
"YES"
]
}
|
{
"input": [
"50\n16 4\n17 9\n31 19\n22 10\n8 1\n40 30\n3 31\n20 29\n47 27\n22 25\n32 34\n12 15\n40 32\n10 33\n47 12\n6 24\n46 41\n14 23\n12 35\n31 42\n46 28\n31 20\n46 37\n1 39\n29 49\n37 47\n40 6\n42 36\n47 2\n24 46\n2 13\n8 45\n41 3\n32 17\n4 7\n47 26\n28 8\n41 50\n34 44\n33 21\n25 5\n16 40\n3 14\n8 18\n28 11\n32 22\n2 38\n3 48\n44 43\n",
"10\n8 1\n1 2\n8 9\n8 5\n1 3\n1 10\n1 6\n1 7\n8 4\n",
"5\n5 1\n5 4\n4 3\n1 2\n",
"7\n1 2\n2 3\n1 4\n1 5\n3 6\n3 7\n",
"3\n1 3\n2 3\n",
"60\n26 6\n59 30\n31 12\n31 3\n38 23\n59 29\n53 9\n38 56\n53 54\n29 21\n17 55\n59 38\n26 16\n24 59\n24 25\n17 35\n24 41\n30 15\n31 27\n8 44\n26 5\n26 48\n8 32\n53 17\n3 34\n3 51\n30 28\n47 10\n53 60\n36 42\n24 53\n59 22\n53 40\n26 52\n36 4\n59 8\n29 37\n36 20\n17 47\n53 18\n3 50\n30 2\n17 7\n8 58\n59 1\n31 11\n24 26\n24 43\n53 57\n59 45\n47 13\n26 46\n17 33\n30 31\n26 39\n26 19\n24 36\n8 49\n38 14\n",
"7\n1 2\n2 3\n3 4\n3 5\n1 6\n1 7\n",
"20\n19 16\n19 18\n20 7\n9 4\n6 17\n14 2\n9 15\n2 13\n5 11\n19 12\n12 20\n16 9\n11 8\n19 5\n3 1\n19 14\n5 3\n12 10\n19 6\n",
"7\n1 2\n1 3\n2 4\n2 5\n3 6\n3 7\n",
"10\n9 5\n7 1\n9 10\n7 2\n5 4\n9 6\n2 9\n10 8\n1 3\n",
"4\n2 4\n2 3\n2 1\n",
"4\n1 4\n3 2\n1 3\n",
"3\n1 2\n1 3\n",
"5\n1 2\n1 5\n1 3\n1 4\n",
"20\n14 9\n12 13\n10 15\n2 1\n20 19\n16 6\n16 3\n17 14\n3 5\n2 11\n3 10\n15 8\n14 2\n6 4\n3 20\n5 18\n1 7\n1 16\n4 12\n",
"20\n7 5\n14 13\n17 6\n3 8\n16 12\n18 9\n3 18\n14 1\n17 3\n15 2\n17 4\n9 11\n2 7\n15 17\n3 20\n16 10\n17 14\n2 16\n1 19\n",
"8\n1 2\n2 3\n3 4\n1 7\n1 8\n4 5\n4 6\n",
"5\n5 1\n5 2\n5 3\n5 4\n",
"50\n49 6\n43 7\n1 27\n19 35\n15 37\n16 12\n19 21\n16 28\n49 9\n48 39\n13 1\n2 48\n9 50\n44 3\n41 32\n48 31\n49 33\n6 11\n13 20\n49 22\n13 41\n48 29\n13 46\n15 47\n34 2\n49 13\n48 14\n34 24\n16 36\n13 40\n49 34\n49 17\n43 25\n11 23\n10 15\n19 26\n34 44\n16 42\n19 18\n46 8\n29 38\n1 45\n12 43\n13 16\n46 30\n15 5\n49 10\n11 19\n32 4\n",
"20\n13 1\n18 2\n3 7\n18 5\n20 16\n3 12\n18 9\n3 10\n18 11\n13 6\n3 18\n20 15\n20 17\n3 13\n3 4\n13 14\n3 20\n18 8\n3 19\n",
"10\n8 2\n5 6\n1 8\n2 9\n1 4\n8 10\n10 5\n2 7\n2 3\n"
],
"output": [
"NO",
"YES",
"NO",
"NO",
"NO",
"YES",
"NO",
"NO",
"NO",
"NO",
"YES",
"NO",
"NO",
"YES",
"NO",
"NO",
"NO",
"YES",
"NO",
"YES",
"NO"
]
}
|
IN-CORRECT
|
java
|
import java.util.*;
import java.lang.*;
import java.io.*;
public class Main
{
PrintWriter out = new PrintWriter(System.out);
BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
StringTokenizer tok = new StringTokenizer("");
String next() throws IOException {
if (!tok.hasMoreTokens()) { tok = new StringTokenizer(in.readLine()); }
return tok.nextToken();
}
int ni() throws IOException { return Integer.parseInt(next()); }
long nl() throws IOException { return Long.parseLong(next()); }
long mod=1000000007;
int n;
ArrayList<Integer>[]A;
boolean f;
boolean[]L;
void solve() throws IOException {
n=ni();
A=new ArrayList[n+1];
for (int i=1;i<=n;i++) A[i]=new ArrayList();
for (int i=1;i<n;i++) {
int u=ni(),v=ni();
A[u].add(v);
A[v].add(u);
}
f=true;
L=new boolean[n+1];
int root=0;
for (int i=1;i<=n;i++) {
if (A[i].size()>1) { root=i; break; }
}
if (root==0) out.println("YES");
else if (n==3) out.println("NO");
else {
dfs(root,0);
if (f) out.println("YES");
else out.println("NO");
}
out.flush();
}
void dfs(int u,int p) {
int lc=0;
if (A[u].size()==1) {
L[u]=true;
return;
}
for (int v:A[u]) {
if (v==p) continue;
dfs(v,u);
if (L[v]) lc++;
}
if (lc==1) f=false;
}
int gcd(int a,int b) { return(b==0?a:gcd(b,a%b)); }
long gcd(long a,long b) { return(b==0?a:gcd(b,a%b)); }
long mp(long a,long p) { long r=1; while(p>0) { if ((p&1)==1) r=(r*a)%mod; p>>=1; a=(a*a)%mod; } return r; }
public static void main(String[] args) throws IOException {
new Main().solve();
}
}
|
1189_D1. Add on a Tree
|
Note that this is the first problem of the two similar problems. You can hack this problem only if you solve both problems.
You are given a tree with n nodes. In the beginning, 0 is written on all edges. In one operation, you can choose any 2 distinct leaves u, v and any real number x and add x to values written on all edges on the simple path between u and v.
For example, on the picture below you can see the result of applying two operations to the graph: adding 2 on the path from 7 to 6, and then adding -0.5 on the path from 4 to 5.
<image>
Is it true that for any configuration of real numbers written on edges, we can achieve it with a finite number of operations?
Leaf is a node of a tree of degree 1. Simple path is a path that doesn't contain any node twice.
Input
The first line contains a single integer n (2 ≤ n ≤ 10^5) — the number of nodes.
Each of the next n-1 lines contains two integers u and v (1 ≤ u, v ≤ n, u ≠ v), meaning that there is an edge between nodes u and v. It is guaranteed that these edges form a tree.
Output
If there is a configuration of real numbers written on edges of the tree that we can't achieve by performing the operations, output "NO".
Otherwise, output "YES".
You can print each letter in any case (upper or lower).
Examples
Input
2
1 2
Output
YES
Input
3
1 2
2 3
Output
NO
Input
5
1 2
1 3
1 4
2 5
Output
NO
Input
6
1 2
1 3
1 4
2 5
2 6
Output
YES
Note
In the first example, we can add any real x to the value written on the only edge (1, 2).
<image>
In the second example, one of configurations that we can't reach is 0 written on (1, 2) and 1 written on (2, 3).
<image>
Below you can see graphs from examples 3, 4:
<image> <image>
|
{
"input": [
"2\n1 2\n",
"3\n1 2\n2 3\n",
"5\n1 2\n1 3\n1 4\n2 5\n",
"6\n1 2\n1 3\n1 4\n2 5\n2 6\n"
],
"output": [
"YES",
"NO",
"NO",
"YES"
]
}
|
{
"input": [
"50\n16 4\n17 9\n31 19\n22 10\n8 1\n40 30\n3 31\n20 29\n47 27\n22 25\n32 34\n12 15\n40 32\n10 33\n47 12\n6 24\n46 41\n14 23\n12 35\n31 42\n46 28\n31 20\n46 37\n1 39\n29 49\n37 47\n40 6\n42 36\n47 2\n24 46\n2 13\n8 45\n41 3\n32 17\n4 7\n47 26\n28 8\n41 50\n34 44\n33 21\n25 5\n16 40\n3 14\n8 18\n28 11\n32 22\n2 38\n3 48\n44 43\n",
"10\n8 1\n1 2\n8 9\n8 5\n1 3\n1 10\n1 6\n1 7\n8 4\n",
"5\n5 1\n5 4\n4 3\n1 2\n",
"7\n1 2\n2 3\n1 4\n1 5\n3 6\n3 7\n",
"3\n1 3\n2 3\n",
"60\n26 6\n59 30\n31 12\n31 3\n38 23\n59 29\n53 9\n38 56\n53 54\n29 21\n17 55\n59 38\n26 16\n24 59\n24 25\n17 35\n24 41\n30 15\n31 27\n8 44\n26 5\n26 48\n8 32\n53 17\n3 34\n3 51\n30 28\n47 10\n53 60\n36 42\n24 53\n59 22\n53 40\n26 52\n36 4\n59 8\n29 37\n36 20\n17 47\n53 18\n3 50\n30 2\n17 7\n8 58\n59 1\n31 11\n24 26\n24 43\n53 57\n59 45\n47 13\n26 46\n17 33\n30 31\n26 39\n26 19\n24 36\n8 49\n38 14\n",
"7\n1 2\n2 3\n3 4\n3 5\n1 6\n1 7\n",
"20\n19 16\n19 18\n20 7\n9 4\n6 17\n14 2\n9 15\n2 13\n5 11\n19 12\n12 20\n16 9\n11 8\n19 5\n3 1\n19 14\n5 3\n12 10\n19 6\n",
"7\n1 2\n1 3\n2 4\n2 5\n3 6\n3 7\n",
"10\n9 5\n7 1\n9 10\n7 2\n5 4\n9 6\n2 9\n10 8\n1 3\n",
"4\n2 4\n2 3\n2 1\n",
"4\n1 4\n3 2\n1 3\n",
"3\n1 2\n1 3\n",
"5\n1 2\n1 5\n1 3\n1 4\n",
"20\n14 9\n12 13\n10 15\n2 1\n20 19\n16 6\n16 3\n17 14\n3 5\n2 11\n3 10\n15 8\n14 2\n6 4\n3 20\n5 18\n1 7\n1 16\n4 12\n",
"20\n7 5\n14 13\n17 6\n3 8\n16 12\n18 9\n3 18\n14 1\n17 3\n15 2\n17 4\n9 11\n2 7\n15 17\n3 20\n16 10\n17 14\n2 16\n1 19\n",
"8\n1 2\n2 3\n3 4\n1 7\n1 8\n4 5\n4 6\n",
"5\n5 1\n5 2\n5 3\n5 4\n",
"50\n49 6\n43 7\n1 27\n19 35\n15 37\n16 12\n19 21\n16 28\n49 9\n48 39\n13 1\n2 48\n9 50\n44 3\n41 32\n48 31\n49 33\n6 11\n13 20\n49 22\n13 41\n48 29\n13 46\n15 47\n34 2\n49 13\n48 14\n34 24\n16 36\n13 40\n49 34\n49 17\n43 25\n11 23\n10 15\n19 26\n34 44\n16 42\n19 18\n46 8\n29 38\n1 45\n12 43\n13 16\n46 30\n15 5\n49 10\n11 19\n32 4\n",
"20\n13 1\n18 2\n3 7\n18 5\n20 16\n3 12\n18 9\n3 10\n18 11\n13 6\n3 18\n20 15\n20 17\n3 13\n3 4\n13 14\n3 20\n18 8\n3 19\n",
"10\n8 2\n5 6\n1 8\n2 9\n1 4\n8 10\n10 5\n2 7\n2 3\n"
],
"output": [
"NO",
"YES",
"NO",
"NO",
"NO",
"YES",
"NO",
"NO",
"NO",
"NO",
"YES",
"NO",
"NO",
"YES",
"NO",
"NO",
"NO",
"YES",
"NO",
"YES",
"NO"
]
}
|
IN-CORRECT
|
java
|
import java.io.*;
import java.util.*;
public class Solution {
public static void main(String[] args) {
FastReader in = new FastReader();
int n = in.nextInt();
if (n == 3) {
System.out.println("NO");
return;
}
int[] outDeg = new int[n + 1];
int[] inDeg = new int[n + 1];
for (int i = 0; i < n - 1; i++) {
int s = in.nextInt();
int e = in.nextInt();
outDeg[s]++;
inDeg[e]++;
}
for (int i = 1; i < outDeg.length; i++) {
if (outDeg[i] == 1 && inDeg[i] != 0) {
System.out.println("NO");
return;
}
}
System.out.println("YES");
}
static class FastReader {
BufferedReader br;
StringTokenizer st;
public FastReader() {
br = new BufferedReader(new
InputStreamReader(System.in));
}
public int[] nextIntArray(int n) {
int[] array = new int[n];
for (int i = 0; i < n; ++i) array[i] = nextInt();
return array;
}
public int[] nextSortedIntArray(int n) {
int array[] = nextIntArray(n);
Arrays.sort(array);
return array;
}
public int[] nextSumIntArray(int n) {
int[] array = new int[n];
array[0] = nextInt();
for (int i = 1; i < n; ++i) array[i] = array[i - 1] + nextInt();
return array;
}
public long[] nextLongArray(int n) {
long[] array = new long[n];
for (int i = 0; i < n; ++i) array[i] = nextLong();
return array;
}
public long[] nextSumLongArray(int n) {
long[] array = new long[n];
array[0] = nextInt();
for (int i = 1; i < n; ++i) array[i] = array[i - 1] + nextInt();
return array;
}
public long[] nextSortedLongArray(int n) {
long array[] = nextLongArray(n);
Arrays.sort(array);
return array;
}
String next() {
while (st == null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt() {
return Integer.parseInt(next());
}
long nextLong() {
return Long.parseLong(next());
}
double nextDouble() {
return Double.parseDouble(next());
}
String nextLine() {
String str = "";
try {
str = br.readLine();
} catch (IOException e) {
e.printStackTrace();
}
return str;
}
}
}
|
1189_D1. Add on a Tree
|
Note that this is the first problem of the two similar problems. You can hack this problem only if you solve both problems.
You are given a tree with n nodes. In the beginning, 0 is written on all edges. In one operation, you can choose any 2 distinct leaves u, v and any real number x and add x to values written on all edges on the simple path between u and v.
For example, on the picture below you can see the result of applying two operations to the graph: adding 2 on the path from 7 to 6, and then adding -0.5 on the path from 4 to 5.
<image>
Is it true that for any configuration of real numbers written on edges, we can achieve it with a finite number of operations?
Leaf is a node of a tree of degree 1. Simple path is a path that doesn't contain any node twice.
Input
The first line contains a single integer n (2 ≤ n ≤ 10^5) — the number of nodes.
Each of the next n-1 lines contains two integers u and v (1 ≤ u, v ≤ n, u ≠ v), meaning that there is an edge between nodes u and v. It is guaranteed that these edges form a tree.
Output
If there is a configuration of real numbers written on edges of the tree that we can't achieve by performing the operations, output "NO".
Otherwise, output "YES".
You can print each letter in any case (upper or lower).
Examples
Input
2
1 2
Output
YES
Input
3
1 2
2 3
Output
NO
Input
5
1 2
1 3
1 4
2 5
Output
NO
Input
6
1 2
1 3
1 4
2 5
2 6
Output
YES
Note
In the first example, we can add any real x to the value written on the only edge (1, 2).
<image>
In the second example, one of configurations that we can't reach is 0 written on (1, 2) and 1 written on (2, 3).
<image>
Below you can see graphs from examples 3, 4:
<image> <image>
|
{
"input": [
"2\n1 2\n",
"3\n1 2\n2 3\n",
"5\n1 2\n1 3\n1 4\n2 5\n",
"6\n1 2\n1 3\n1 4\n2 5\n2 6\n"
],
"output": [
"YES",
"NO",
"NO",
"YES"
]
}
|
{
"input": [
"50\n16 4\n17 9\n31 19\n22 10\n8 1\n40 30\n3 31\n20 29\n47 27\n22 25\n32 34\n12 15\n40 32\n10 33\n47 12\n6 24\n46 41\n14 23\n12 35\n31 42\n46 28\n31 20\n46 37\n1 39\n29 49\n37 47\n40 6\n42 36\n47 2\n24 46\n2 13\n8 45\n41 3\n32 17\n4 7\n47 26\n28 8\n41 50\n34 44\n33 21\n25 5\n16 40\n3 14\n8 18\n28 11\n32 22\n2 38\n3 48\n44 43\n",
"10\n8 1\n1 2\n8 9\n8 5\n1 3\n1 10\n1 6\n1 7\n8 4\n",
"5\n5 1\n5 4\n4 3\n1 2\n",
"7\n1 2\n2 3\n1 4\n1 5\n3 6\n3 7\n",
"3\n1 3\n2 3\n",
"60\n26 6\n59 30\n31 12\n31 3\n38 23\n59 29\n53 9\n38 56\n53 54\n29 21\n17 55\n59 38\n26 16\n24 59\n24 25\n17 35\n24 41\n30 15\n31 27\n8 44\n26 5\n26 48\n8 32\n53 17\n3 34\n3 51\n30 28\n47 10\n53 60\n36 42\n24 53\n59 22\n53 40\n26 52\n36 4\n59 8\n29 37\n36 20\n17 47\n53 18\n3 50\n30 2\n17 7\n8 58\n59 1\n31 11\n24 26\n24 43\n53 57\n59 45\n47 13\n26 46\n17 33\n30 31\n26 39\n26 19\n24 36\n8 49\n38 14\n",
"7\n1 2\n2 3\n3 4\n3 5\n1 6\n1 7\n",
"20\n19 16\n19 18\n20 7\n9 4\n6 17\n14 2\n9 15\n2 13\n5 11\n19 12\n12 20\n16 9\n11 8\n19 5\n3 1\n19 14\n5 3\n12 10\n19 6\n",
"7\n1 2\n1 3\n2 4\n2 5\n3 6\n3 7\n",
"10\n9 5\n7 1\n9 10\n7 2\n5 4\n9 6\n2 9\n10 8\n1 3\n",
"4\n2 4\n2 3\n2 1\n",
"4\n1 4\n3 2\n1 3\n",
"3\n1 2\n1 3\n",
"5\n1 2\n1 5\n1 3\n1 4\n",
"20\n14 9\n12 13\n10 15\n2 1\n20 19\n16 6\n16 3\n17 14\n3 5\n2 11\n3 10\n15 8\n14 2\n6 4\n3 20\n5 18\n1 7\n1 16\n4 12\n",
"20\n7 5\n14 13\n17 6\n3 8\n16 12\n18 9\n3 18\n14 1\n17 3\n15 2\n17 4\n9 11\n2 7\n15 17\n3 20\n16 10\n17 14\n2 16\n1 19\n",
"8\n1 2\n2 3\n3 4\n1 7\n1 8\n4 5\n4 6\n",
"5\n5 1\n5 2\n5 3\n5 4\n",
"50\n49 6\n43 7\n1 27\n19 35\n15 37\n16 12\n19 21\n16 28\n49 9\n48 39\n13 1\n2 48\n9 50\n44 3\n41 32\n48 31\n49 33\n6 11\n13 20\n49 22\n13 41\n48 29\n13 46\n15 47\n34 2\n49 13\n48 14\n34 24\n16 36\n13 40\n49 34\n49 17\n43 25\n11 23\n10 15\n19 26\n34 44\n16 42\n19 18\n46 8\n29 38\n1 45\n12 43\n13 16\n46 30\n15 5\n49 10\n11 19\n32 4\n",
"20\n13 1\n18 2\n3 7\n18 5\n20 16\n3 12\n18 9\n3 10\n18 11\n13 6\n3 18\n20 15\n20 17\n3 13\n3 4\n13 14\n3 20\n18 8\n3 19\n",
"10\n8 2\n5 6\n1 8\n2 9\n1 4\n8 10\n10 5\n2 7\n2 3\n"
],
"output": [
"NO",
"YES",
"NO",
"NO",
"NO",
"YES",
"NO",
"NO",
"NO",
"NO",
"YES",
"NO",
"NO",
"YES",
"NO",
"NO",
"NO",
"YES",
"NO",
"YES",
"NO"
]
}
|
IN-CORRECT
|
python3
|
n = int(input())
tr = [0] * (n + 1)
for i in range(n - 1):
u, v = map(int, input().split())
tr[u] += 1
tr[v] += 1
for i in range(1, n):
if tr[i] == 2:
print('No')
break
else:
print('Yes')
|
1189_D1. Add on a Tree
|
Note that this is the first problem of the two similar problems. You can hack this problem only if you solve both problems.
You are given a tree with n nodes. In the beginning, 0 is written on all edges. In one operation, you can choose any 2 distinct leaves u, v and any real number x and add x to values written on all edges on the simple path between u and v.
For example, on the picture below you can see the result of applying two operations to the graph: adding 2 on the path from 7 to 6, and then adding -0.5 on the path from 4 to 5.
<image>
Is it true that for any configuration of real numbers written on edges, we can achieve it with a finite number of operations?
Leaf is a node of a tree of degree 1. Simple path is a path that doesn't contain any node twice.
Input
The first line contains a single integer n (2 ≤ n ≤ 10^5) — the number of nodes.
Each of the next n-1 lines contains two integers u and v (1 ≤ u, v ≤ n, u ≠ v), meaning that there is an edge between nodes u and v. It is guaranteed that these edges form a tree.
Output
If there is a configuration of real numbers written on edges of the tree that we can't achieve by performing the operations, output "NO".
Otherwise, output "YES".
You can print each letter in any case (upper or lower).
Examples
Input
2
1 2
Output
YES
Input
3
1 2
2 3
Output
NO
Input
5
1 2
1 3
1 4
2 5
Output
NO
Input
6
1 2
1 3
1 4
2 5
2 6
Output
YES
Note
In the first example, we can add any real x to the value written on the only edge (1, 2).
<image>
In the second example, one of configurations that we can't reach is 0 written on (1, 2) and 1 written on (2, 3).
<image>
Below you can see graphs from examples 3, 4:
<image> <image>
|
{
"input": [
"2\n1 2\n",
"3\n1 2\n2 3\n",
"5\n1 2\n1 3\n1 4\n2 5\n",
"6\n1 2\n1 3\n1 4\n2 5\n2 6\n"
],
"output": [
"YES",
"NO",
"NO",
"YES"
]
}
|
{
"input": [
"50\n16 4\n17 9\n31 19\n22 10\n8 1\n40 30\n3 31\n20 29\n47 27\n22 25\n32 34\n12 15\n40 32\n10 33\n47 12\n6 24\n46 41\n14 23\n12 35\n31 42\n46 28\n31 20\n46 37\n1 39\n29 49\n37 47\n40 6\n42 36\n47 2\n24 46\n2 13\n8 45\n41 3\n32 17\n4 7\n47 26\n28 8\n41 50\n34 44\n33 21\n25 5\n16 40\n3 14\n8 18\n28 11\n32 22\n2 38\n3 48\n44 43\n",
"10\n8 1\n1 2\n8 9\n8 5\n1 3\n1 10\n1 6\n1 7\n8 4\n",
"5\n5 1\n5 4\n4 3\n1 2\n",
"7\n1 2\n2 3\n1 4\n1 5\n3 6\n3 7\n",
"3\n1 3\n2 3\n",
"60\n26 6\n59 30\n31 12\n31 3\n38 23\n59 29\n53 9\n38 56\n53 54\n29 21\n17 55\n59 38\n26 16\n24 59\n24 25\n17 35\n24 41\n30 15\n31 27\n8 44\n26 5\n26 48\n8 32\n53 17\n3 34\n3 51\n30 28\n47 10\n53 60\n36 42\n24 53\n59 22\n53 40\n26 52\n36 4\n59 8\n29 37\n36 20\n17 47\n53 18\n3 50\n30 2\n17 7\n8 58\n59 1\n31 11\n24 26\n24 43\n53 57\n59 45\n47 13\n26 46\n17 33\n30 31\n26 39\n26 19\n24 36\n8 49\n38 14\n",
"7\n1 2\n2 3\n3 4\n3 5\n1 6\n1 7\n",
"20\n19 16\n19 18\n20 7\n9 4\n6 17\n14 2\n9 15\n2 13\n5 11\n19 12\n12 20\n16 9\n11 8\n19 5\n3 1\n19 14\n5 3\n12 10\n19 6\n",
"7\n1 2\n1 3\n2 4\n2 5\n3 6\n3 7\n",
"10\n9 5\n7 1\n9 10\n7 2\n5 4\n9 6\n2 9\n10 8\n1 3\n",
"4\n2 4\n2 3\n2 1\n",
"4\n1 4\n3 2\n1 3\n",
"3\n1 2\n1 3\n",
"5\n1 2\n1 5\n1 3\n1 4\n",
"20\n14 9\n12 13\n10 15\n2 1\n20 19\n16 6\n16 3\n17 14\n3 5\n2 11\n3 10\n15 8\n14 2\n6 4\n3 20\n5 18\n1 7\n1 16\n4 12\n",
"20\n7 5\n14 13\n17 6\n3 8\n16 12\n18 9\n3 18\n14 1\n17 3\n15 2\n17 4\n9 11\n2 7\n15 17\n3 20\n16 10\n17 14\n2 16\n1 19\n",
"8\n1 2\n2 3\n3 4\n1 7\n1 8\n4 5\n4 6\n",
"5\n5 1\n5 2\n5 3\n5 4\n",
"50\n49 6\n43 7\n1 27\n19 35\n15 37\n16 12\n19 21\n16 28\n49 9\n48 39\n13 1\n2 48\n9 50\n44 3\n41 32\n48 31\n49 33\n6 11\n13 20\n49 22\n13 41\n48 29\n13 46\n15 47\n34 2\n49 13\n48 14\n34 24\n16 36\n13 40\n49 34\n49 17\n43 25\n11 23\n10 15\n19 26\n34 44\n16 42\n19 18\n46 8\n29 38\n1 45\n12 43\n13 16\n46 30\n15 5\n49 10\n11 19\n32 4\n",
"20\n13 1\n18 2\n3 7\n18 5\n20 16\n3 12\n18 9\n3 10\n18 11\n13 6\n3 18\n20 15\n20 17\n3 13\n3 4\n13 14\n3 20\n18 8\n3 19\n",
"10\n8 2\n5 6\n1 8\n2 9\n1 4\n8 10\n10 5\n2 7\n2 3\n"
],
"output": [
"NO",
"YES",
"NO",
"NO",
"NO",
"YES",
"NO",
"NO",
"NO",
"NO",
"YES",
"NO",
"NO",
"YES",
"NO",
"NO",
"NO",
"YES",
"NO",
"YES",
"NO"
]
}
|
IN-CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 1000, mod = 1e9 + 7;
vector<int> vec[N];
long long bin_pow(int x, int y) {
if (y == 0) return 1;
if (y == 1) return x;
if (y % 2 == 0) {
long long z = bin_pow(x, y / 2);
return (z * z) % mod;
}
return (bin_pow(x, y - 1) * x) % mod;
}
bool check(int x) {}
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
int n;
cin >> n;
for (int i = 1; i < n; i++) {
int x, y;
cin >> x >> y;
vec[x].push_back(y);
vec[y].push_back(x);
}
int k = 0;
for (int i = 1; i <= n; i++) {
if (vec[i].size() == 1) {
k++;
}
}
if (k % 2 == 0 && n % 2 == 0) {
cout << "YES";
} else
cout << "NO";
return 0;
}
|
1189_D1. Add on a Tree
|
Note that this is the first problem of the two similar problems. You can hack this problem only if you solve both problems.
You are given a tree with n nodes. In the beginning, 0 is written on all edges. In one operation, you can choose any 2 distinct leaves u, v and any real number x and add x to values written on all edges on the simple path between u and v.
For example, on the picture below you can see the result of applying two operations to the graph: adding 2 on the path from 7 to 6, and then adding -0.5 on the path from 4 to 5.
<image>
Is it true that for any configuration of real numbers written on edges, we can achieve it with a finite number of operations?
Leaf is a node of a tree of degree 1. Simple path is a path that doesn't contain any node twice.
Input
The first line contains a single integer n (2 ≤ n ≤ 10^5) — the number of nodes.
Each of the next n-1 lines contains two integers u and v (1 ≤ u, v ≤ n, u ≠ v), meaning that there is an edge between nodes u and v. It is guaranteed that these edges form a tree.
Output
If there is a configuration of real numbers written on edges of the tree that we can't achieve by performing the operations, output "NO".
Otherwise, output "YES".
You can print each letter in any case (upper or lower).
Examples
Input
2
1 2
Output
YES
Input
3
1 2
2 3
Output
NO
Input
5
1 2
1 3
1 4
2 5
Output
NO
Input
6
1 2
1 3
1 4
2 5
2 6
Output
YES
Note
In the first example, we can add any real x to the value written on the only edge (1, 2).
<image>
In the second example, one of configurations that we can't reach is 0 written on (1, 2) and 1 written on (2, 3).
<image>
Below you can see graphs from examples 3, 4:
<image> <image>
|
{
"input": [
"2\n1 2\n",
"3\n1 2\n2 3\n",
"5\n1 2\n1 3\n1 4\n2 5\n",
"6\n1 2\n1 3\n1 4\n2 5\n2 6\n"
],
"output": [
"YES",
"NO",
"NO",
"YES"
]
}
|
{
"input": [
"50\n16 4\n17 9\n31 19\n22 10\n8 1\n40 30\n3 31\n20 29\n47 27\n22 25\n32 34\n12 15\n40 32\n10 33\n47 12\n6 24\n46 41\n14 23\n12 35\n31 42\n46 28\n31 20\n46 37\n1 39\n29 49\n37 47\n40 6\n42 36\n47 2\n24 46\n2 13\n8 45\n41 3\n32 17\n4 7\n47 26\n28 8\n41 50\n34 44\n33 21\n25 5\n16 40\n3 14\n8 18\n28 11\n32 22\n2 38\n3 48\n44 43\n",
"10\n8 1\n1 2\n8 9\n8 5\n1 3\n1 10\n1 6\n1 7\n8 4\n",
"5\n5 1\n5 4\n4 3\n1 2\n",
"7\n1 2\n2 3\n1 4\n1 5\n3 6\n3 7\n",
"3\n1 3\n2 3\n",
"60\n26 6\n59 30\n31 12\n31 3\n38 23\n59 29\n53 9\n38 56\n53 54\n29 21\n17 55\n59 38\n26 16\n24 59\n24 25\n17 35\n24 41\n30 15\n31 27\n8 44\n26 5\n26 48\n8 32\n53 17\n3 34\n3 51\n30 28\n47 10\n53 60\n36 42\n24 53\n59 22\n53 40\n26 52\n36 4\n59 8\n29 37\n36 20\n17 47\n53 18\n3 50\n30 2\n17 7\n8 58\n59 1\n31 11\n24 26\n24 43\n53 57\n59 45\n47 13\n26 46\n17 33\n30 31\n26 39\n26 19\n24 36\n8 49\n38 14\n",
"7\n1 2\n2 3\n3 4\n3 5\n1 6\n1 7\n",
"20\n19 16\n19 18\n20 7\n9 4\n6 17\n14 2\n9 15\n2 13\n5 11\n19 12\n12 20\n16 9\n11 8\n19 5\n3 1\n19 14\n5 3\n12 10\n19 6\n",
"7\n1 2\n1 3\n2 4\n2 5\n3 6\n3 7\n",
"10\n9 5\n7 1\n9 10\n7 2\n5 4\n9 6\n2 9\n10 8\n1 3\n",
"4\n2 4\n2 3\n2 1\n",
"4\n1 4\n3 2\n1 3\n",
"3\n1 2\n1 3\n",
"5\n1 2\n1 5\n1 3\n1 4\n",
"20\n14 9\n12 13\n10 15\n2 1\n20 19\n16 6\n16 3\n17 14\n3 5\n2 11\n3 10\n15 8\n14 2\n6 4\n3 20\n5 18\n1 7\n1 16\n4 12\n",
"20\n7 5\n14 13\n17 6\n3 8\n16 12\n18 9\n3 18\n14 1\n17 3\n15 2\n17 4\n9 11\n2 7\n15 17\n3 20\n16 10\n17 14\n2 16\n1 19\n",
"8\n1 2\n2 3\n3 4\n1 7\n1 8\n4 5\n4 6\n",
"5\n5 1\n5 2\n5 3\n5 4\n",
"50\n49 6\n43 7\n1 27\n19 35\n15 37\n16 12\n19 21\n16 28\n49 9\n48 39\n13 1\n2 48\n9 50\n44 3\n41 32\n48 31\n49 33\n6 11\n13 20\n49 22\n13 41\n48 29\n13 46\n15 47\n34 2\n49 13\n48 14\n34 24\n16 36\n13 40\n49 34\n49 17\n43 25\n11 23\n10 15\n19 26\n34 44\n16 42\n19 18\n46 8\n29 38\n1 45\n12 43\n13 16\n46 30\n15 5\n49 10\n11 19\n32 4\n",
"20\n13 1\n18 2\n3 7\n18 5\n20 16\n3 12\n18 9\n3 10\n18 11\n13 6\n3 18\n20 15\n20 17\n3 13\n3 4\n13 14\n3 20\n18 8\n3 19\n",
"10\n8 2\n5 6\n1 8\n2 9\n1 4\n8 10\n10 5\n2 7\n2 3\n"
],
"output": [
"NO",
"YES",
"NO",
"NO",
"NO",
"YES",
"NO",
"NO",
"NO",
"NO",
"YES",
"NO",
"NO",
"YES",
"NO",
"NO",
"NO",
"YES",
"NO",
"YES",
"NO"
]
}
|
IN-CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
const int MAXN = 1e5 + 10;
int n, u, v;
int a[MAXN];
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
cin >> n;
memset(a, 0, sizeof(a));
for (int i = 0; i < n - 1; i++) {
cin >> u >> v;
a[u]++;
a[v]++;
}
int cnt[3] = {0, 0, 0};
for (int i = 1; i <= n; i++) {
if (a[i] <= 2) cnt[a[i]]++;
}
if (cnt[2] == 0 && cnt[1] * (cnt[1] - 1) / 2 >= n - 1)
cout << "YES" << endl;
else
cout << "NO" << endl;
return 0;
}
|
1189_D1. Add on a Tree
|
Note that this is the first problem of the two similar problems. You can hack this problem only if you solve both problems.
You are given a tree with n nodes. In the beginning, 0 is written on all edges. In one operation, you can choose any 2 distinct leaves u, v and any real number x and add x to values written on all edges on the simple path between u and v.
For example, on the picture below you can see the result of applying two operations to the graph: adding 2 on the path from 7 to 6, and then adding -0.5 on the path from 4 to 5.
<image>
Is it true that for any configuration of real numbers written on edges, we can achieve it with a finite number of operations?
Leaf is a node of a tree of degree 1. Simple path is a path that doesn't contain any node twice.
Input
The first line contains a single integer n (2 ≤ n ≤ 10^5) — the number of nodes.
Each of the next n-1 lines contains two integers u and v (1 ≤ u, v ≤ n, u ≠ v), meaning that there is an edge between nodes u and v. It is guaranteed that these edges form a tree.
Output
If there is a configuration of real numbers written on edges of the tree that we can't achieve by performing the operations, output "NO".
Otherwise, output "YES".
You can print each letter in any case (upper or lower).
Examples
Input
2
1 2
Output
YES
Input
3
1 2
2 3
Output
NO
Input
5
1 2
1 3
1 4
2 5
Output
NO
Input
6
1 2
1 3
1 4
2 5
2 6
Output
YES
Note
In the first example, we can add any real x to the value written on the only edge (1, 2).
<image>
In the second example, one of configurations that we can't reach is 0 written on (1, 2) and 1 written on (2, 3).
<image>
Below you can see graphs from examples 3, 4:
<image> <image>
|
{
"input": [
"2\n1 2\n",
"3\n1 2\n2 3\n",
"5\n1 2\n1 3\n1 4\n2 5\n",
"6\n1 2\n1 3\n1 4\n2 5\n2 6\n"
],
"output": [
"YES",
"NO",
"NO",
"YES"
]
}
|
{
"input": [
"50\n16 4\n17 9\n31 19\n22 10\n8 1\n40 30\n3 31\n20 29\n47 27\n22 25\n32 34\n12 15\n40 32\n10 33\n47 12\n6 24\n46 41\n14 23\n12 35\n31 42\n46 28\n31 20\n46 37\n1 39\n29 49\n37 47\n40 6\n42 36\n47 2\n24 46\n2 13\n8 45\n41 3\n32 17\n4 7\n47 26\n28 8\n41 50\n34 44\n33 21\n25 5\n16 40\n3 14\n8 18\n28 11\n32 22\n2 38\n3 48\n44 43\n",
"10\n8 1\n1 2\n8 9\n8 5\n1 3\n1 10\n1 6\n1 7\n8 4\n",
"5\n5 1\n5 4\n4 3\n1 2\n",
"7\n1 2\n2 3\n1 4\n1 5\n3 6\n3 7\n",
"3\n1 3\n2 3\n",
"60\n26 6\n59 30\n31 12\n31 3\n38 23\n59 29\n53 9\n38 56\n53 54\n29 21\n17 55\n59 38\n26 16\n24 59\n24 25\n17 35\n24 41\n30 15\n31 27\n8 44\n26 5\n26 48\n8 32\n53 17\n3 34\n3 51\n30 28\n47 10\n53 60\n36 42\n24 53\n59 22\n53 40\n26 52\n36 4\n59 8\n29 37\n36 20\n17 47\n53 18\n3 50\n30 2\n17 7\n8 58\n59 1\n31 11\n24 26\n24 43\n53 57\n59 45\n47 13\n26 46\n17 33\n30 31\n26 39\n26 19\n24 36\n8 49\n38 14\n",
"7\n1 2\n2 3\n3 4\n3 5\n1 6\n1 7\n",
"20\n19 16\n19 18\n20 7\n9 4\n6 17\n14 2\n9 15\n2 13\n5 11\n19 12\n12 20\n16 9\n11 8\n19 5\n3 1\n19 14\n5 3\n12 10\n19 6\n",
"7\n1 2\n1 3\n2 4\n2 5\n3 6\n3 7\n",
"10\n9 5\n7 1\n9 10\n7 2\n5 4\n9 6\n2 9\n10 8\n1 3\n",
"4\n2 4\n2 3\n2 1\n",
"4\n1 4\n3 2\n1 3\n",
"3\n1 2\n1 3\n",
"5\n1 2\n1 5\n1 3\n1 4\n",
"20\n14 9\n12 13\n10 15\n2 1\n20 19\n16 6\n16 3\n17 14\n3 5\n2 11\n3 10\n15 8\n14 2\n6 4\n3 20\n5 18\n1 7\n1 16\n4 12\n",
"20\n7 5\n14 13\n17 6\n3 8\n16 12\n18 9\n3 18\n14 1\n17 3\n15 2\n17 4\n9 11\n2 7\n15 17\n3 20\n16 10\n17 14\n2 16\n1 19\n",
"8\n1 2\n2 3\n3 4\n1 7\n1 8\n4 5\n4 6\n",
"5\n5 1\n5 2\n5 3\n5 4\n",
"50\n49 6\n43 7\n1 27\n19 35\n15 37\n16 12\n19 21\n16 28\n49 9\n48 39\n13 1\n2 48\n9 50\n44 3\n41 32\n48 31\n49 33\n6 11\n13 20\n49 22\n13 41\n48 29\n13 46\n15 47\n34 2\n49 13\n48 14\n34 24\n16 36\n13 40\n49 34\n49 17\n43 25\n11 23\n10 15\n19 26\n34 44\n16 42\n19 18\n46 8\n29 38\n1 45\n12 43\n13 16\n46 30\n15 5\n49 10\n11 19\n32 4\n",
"20\n13 1\n18 2\n3 7\n18 5\n20 16\n3 12\n18 9\n3 10\n18 11\n13 6\n3 18\n20 15\n20 17\n3 13\n3 4\n13 14\n3 20\n18 8\n3 19\n",
"10\n8 2\n5 6\n1 8\n2 9\n1 4\n8 10\n10 5\n2 7\n2 3\n"
],
"output": [
"NO",
"YES",
"NO",
"NO",
"NO",
"YES",
"NO",
"NO",
"NO",
"NO",
"YES",
"NO",
"NO",
"YES",
"NO",
"NO",
"NO",
"YES",
"NO",
"YES",
"NO"
]
}
|
IN-CORRECT
|
java
|
import java.util.*;
import java.io.*;
public class addonatree {
public void run() throws Exception {
Scanner file =new Scanner(System.in);
int times =file.nextInt();
ArrayList<ArrayList<Integer>> adj = new ArrayList();
for (int i= 0; i < times; i++) {
adj.add(new ArrayList());
}
for (int i = 0;i < times - 1; i++) {
int a = file.nextInt() - 1; int b= file.nextInt() -1;
adj.get(a).add(b);
adj.get(b).add(a);
}
boolean b = true;
for (int i=0; i < times; i++) {
if (adj.get(i).size() ==1 && adj.get(adj.get(i).get(0)).size() == 2) {
b = false;
break;
}
}
System.out.println(times == 2 ? "YES" : b ? "YES" : "NO");
}
public static void main(String[] args) throws Exception {
new addonatree().run();
}
}
|
1189_D1. Add on a Tree
|
Note that this is the first problem of the two similar problems. You can hack this problem only if you solve both problems.
You are given a tree with n nodes. In the beginning, 0 is written on all edges. In one operation, you can choose any 2 distinct leaves u, v and any real number x and add x to values written on all edges on the simple path between u and v.
For example, on the picture below you can see the result of applying two operations to the graph: adding 2 on the path from 7 to 6, and then adding -0.5 on the path from 4 to 5.
<image>
Is it true that for any configuration of real numbers written on edges, we can achieve it with a finite number of operations?
Leaf is a node of a tree of degree 1. Simple path is a path that doesn't contain any node twice.
Input
The first line contains a single integer n (2 ≤ n ≤ 10^5) — the number of nodes.
Each of the next n-1 lines contains two integers u and v (1 ≤ u, v ≤ n, u ≠ v), meaning that there is an edge between nodes u and v. It is guaranteed that these edges form a tree.
Output
If there is a configuration of real numbers written on edges of the tree that we can't achieve by performing the operations, output "NO".
Otherwise, output "YES".
You can print each letter in any case (upper or lower).
Examples
Input
2
1 2
Output
YES
Input
3
1 2
2 3
Output
NO
Input
5
1 2
1 3
1 4
2 5
Output
NO
Input
6
1 2
1 3
1 4
2 5
2 6
Output
YES
Note
In the first example, we can add any real x to the value written on the only edge (1, 2).
<image>
In the second example, one of configurations that we can't reach is 0 written on (1, 2) and 1 written on (2, 3).
<image>
Below you can see graphs from examples 3, 4:
<image> <image>
|
{
"input": [
"2\n1 2\n",
"3\n1 2\n2 3\n",
"5\n1 2\n1 3\n1 4\n2 5\n",
"6\n1 2\n1 3\n1 4\n2 5\n2 6\n"
],
"output": [
"YES",
"NO",
"NO",
"YES"
]
}
|
{
"input": [
"50\n16 4\n17 9\n31 19\n22 10\n8 1\n40 30\n3 31\n20 29\n47 27\n22 25\n32 34\n12 15\n40 32\n10 33\n47 12\n6 24\n46 41\n14 23\n12 35\n31 42\n46 28\n31 20\n46 37\n1 39\n29 49\n37 47\n40 6\n42 36\n47 2\n24 46\n2 13\n8 45\n41 3\n32 17\n4 7\n47 26\n28 8\n41 50\n34 44\n33 21\n25 5\n16 40\n3 14\n8 18\n28 11\n32 22\n2 38\n3 48\n44 43\n",
"10\n8 1\n1 2\n8 9\n8 5\n1 3\n1 10\n1 6\n1 7\n8 4\n",
"5\n5 1\n5 4\n4 3\n1 2\n",
"7\n1 2\n2 3\n1 4\n1 5\n3 6\n3 7\n",
"3\n1 3\n2 3\n",
"60\n26 6\n59 30\n31 12\n31 3\n38 23\n59 29\n53 9\n38 56\n53 54\n29 21\n17 55\n59 38\n26 16\n24 59\n24 25\n17 35\n24 41\n30 15\n31 27\n8 44\n26 5\n26 48\n8 32\n53 17\n3 34\n3 51\n30 28\n47 10\n53 60\n36 42\n24 53\n59 22\n53 40\n26 52\n36 4\n59 8\n29 37\n36 20\n17 47\n53 18\n3 50\n30 2\n17 7\n8 58\n59 1\n31 11\n24 26\n24 43\n53 57\n59 45\n47 13\n26 46\n17 33\n30 31\n26 39\n26 19\n24 36\n8 49\n38 14\n",
"7\n1 2\n2 3\n3 4\n3 5\n1 6\n1 7\n",
"20\n19 16\n19 18\n20 7\n9 4\n6 17\n14 2\n9 15\n2 13\n5 11\n19 12\n12 20\n16 9\n11 8\n19 5\n3 1\n19 14\n5 3\n12 10\n19 6\n",
"7\n1 2\n1 3\n2 4\n2 5\n3 6\n3 7\n",
"10\n9 5\n7 1\n9 10\n7 2\n5 4\n9 6\n2 9\n10 8\n1 3\n",
"4\n2 4\n2 3\n2 1\n",
"4\n1 4\n3 2\n1 3\n",
"3\n1 2\n1 3\n",
"5\n1 2\n1 5\n1 3\n1 4\n",
"20\n14 9\n12 13\n10 15\n2 1\n20 19\n16 6\n16 3\n17 14\n3 5\n2 11\n3 10\n15 8\n14 2\n6 4\n3 20\n5 18\n1 7\n1 16\n4 12\n",
"20\n7 5\n14 13\n17 6\n3 8\n16 12\n18 9\n3 18\n14 1\n17 3\n15 2\n17 4\n9 11\n2 7\n15 17\n3 20\n16 10\n17 14\n2 16\n1 19\n",
"8\n1 2\n2 3\n3 4\n1 7\n1 8\n4 5\n4 6\n",
"5\n5 1\n5 2\n5 3\n5 4\n",
"50\n49 6\n43 7\n1 27\n19 35\n15 37\n16 12\n19 21\n16 28\n49 9\n48 39\n13 1\n2 48\n9 50\n44 3\n41 32\n48 31\n49 33\n6 11\n13 20\n49 22\n13 41\n48 29\n13 46\n15 47\n34 2\n49 13\n48 14\n34 24\n16 36\n13 40\n49 34\n49 17\n43 25\n11 23\n10 15\n19 26\n34 44\n16 42\n19 18\n46 8\n29 38\n1 45\n12 43\n13 16\n46 30\n15 5\n49 10\n11 19\n32 4\n",
"20\n13 1\n18 2\n3 7\n18 5\n20 16\n3 12\n18 9\n3 10\n18 11\n13 6\n3 18\n20 15\n20 17\n3 13\n3 4\n13 14\n3 20\n18 8\n3 19\n",
"10\n8 2\n5 6\n1 8\n2 9\n1 4\n8 10\n10 5\n2 7\n2 3\n"
],
"output": [
"NO",
"YES",
"NO",
"NO",
"NO",
"YES",
"NO",
"NO",
"NO",
"NO",
"YES",
"NO",
"NO",
"YES",
"NO",
"NO",
"NO",
"YES",
"NO",
"YES",
"NO"
]
}
|
IN-CORRECT
|
java
|
import java.util.Scanner;
public class TaskD {
public static void main(String[] args) {
Scanner s = new Scanner(System.in);
int n = s.nextInt();
int[] deg = new int[n];
for(int i=0;i<n-1;i++) {
int a = s.nextInt()-1;
int b = s.nextInt()-1;
deg[a]++;
deg[b]++;
}
for(int i=0;i<n;i++) {
if(deg[i]%2 == 0) {
System.out.println("NO");
return;
}
}
System.out.println("YES");
}
}
|
1189_D1. Add on a Tree
|
Note that this is the first problem of the two similar problems. You can hack this problem only if you solve both problems.
You are given a tree with n nodes. In the beginning, 0 is written on all edges. In one operation, you can choose any 2 distinct leaves u, v and any real number x and add x to values written on all edges on the simple path between u and v.
For example, on the picture below you can see the result of applying two operations to the graph: adding 2 on the path from 7 to 6, and then adding -0.5 on the path from 4 to 5.
<image>
Is it true that for any configuration of real numbers written on edges, we can achieve it with a finite number of operations?
Leaf is a node of a tree of degree 1. Simple path is a path that doesn't contain any node twice.
Input
The first line contains a single integer n (2 ≤ n ≤ 10^5) — the number of nodes.
Each of the next n-1 lines contains two integers u and v (1 ≤ u, v ≤ n, u ≠ v), meaning that there is an edge between nodes u and v. It is guaranteed that these edges form a tree.
Output
If there is a configuration of real numbers written on edges of the tree that we can't achieve by performing the operations, output "NO".
Otherwise, output "YES".
You can print each letter in any case (upper or lower).
Examples
Input
2
1 2
Output
YES
Input
3
1 2
2 3
Output
NO
Input
5
1 2
1 3
1 4
2 5
Output
NO
Input
6
1 2
1 3
1 4
2 5
2 6
Output
YES
Note
In the first example, we can add any real x to the value written on the only edge (1, 2).
<image>
In the second example, one of configurations that we can't reach is 0 written on (1, 2) and 1 written on (2, 3).
<image>
Below you can see graphs from examples 3, 4:
<image> <image>
|
{
"input": [
"2\n1 2\n",
"3\n1 2\n2 3\n",
"5\n1 2\n1 3\n1 4\n2 5\n",
"6\n1 2\n1 3\n1 4\n2 5\n2 6\n"
],
"output": [
"YES",
"NO",
"NO",
"YES"
]
}
|
{
"input": [
"50\n16 4\n17 9\n31 19\n22 10\n8 1\n40 30\n3 31\n20 29\n47 27\n22 25\n32 34\n12 15\n40 32\n10 33\n47 12\n6 24\n46 41\n14 23\n12 35\n31 42\n46 28\n31 20\n46 37\n1 39\n29 49\n37 47\n40 6\n42 36\n47 2\n24 46\n2 13\n8 45\n41 3\n32 17\n4 7\n47 26\n28 8\n41 50\n34 44\n33 21\n25 5\n16 40\n3 14\n8 18\n28 11\n32 22\n2 38\n3 48\n44 43\n",
"10\n8 1\n1 2\n8 9\n8 5\n1 3\n1 10\n1 6\n1 7\n8 4\n",
"5\n5 1\n5 4\n4 3\n1 2\n",
"7\n1 2\n2 3\n1 4\n1 5\n3 6\n3 7\n",
"3\n1 3\n2 3\n",
"60\n26 6\n59 30\n31 12\n31 3\n38 23\n59 29\n53 9\n38 56\n53 54\n29 21\n17 55\n59 38\n26 16\n24 59\n24 25\n17 35\n24 41\n30 15\n31 27\n8 44\n26 5\n26 48\n8 32\n53 17\n3 34\n3 51\n30 28\n47 10\n53 60\n36 42\n24 53\n59 22\n53 40\n26 52\n36 4\n59 8\n29 37\n36 20\n17 47\n53 18\n3 50\n30 2\n17 7\n8 58\n59 1\n31 11\n24 26\n24 43\n53 57\n59 45\n47 13\n26 46\n17 33\n30 31\n26 39\n26 19\n24 36\n8 49\n38 14\n",
"7\n1 2\n2 3\n3 4\n3 5\n1 6\n1 7\n",
"20\n19 16\n19 18\n20 7\n9 4\n6 17\n14 2\n9 15\n2 13\n5 11\n19 12\n12 20\n16 9\n11 8\n19 5\n3 1\n19 14\n5 3\n12 10\n19 6\n",
"7\n1 2\n1 3\n2 4\n2 5\n3 6\n3 7\n",
"10\n9 5\n7 1\n9 10\n7 2\n5 4\n9 6\n2 9\n10 8\n1 3\n",
"4\n2 4\n2 3\n2 1\n",
"4\n1 4\n3 2\n1 3\n",
"3\n1 2\n1 3\n",
"5\n1 2\n1 5\n1 3\n1 4\n",
"20\n14 9\n12 13\n10 15\n2 1\n20 19\n16 6\n16 3\n17 14\n3 5\n2 11\n3 10\n15 8\n14 2\n6 4\n3 20\n5 18\n1 7\n1 16\n4 12\n",
"20\n7 5\n14 13\n17 6\n3 8\n16 12\n18 9\n3 18\n14 1\n17 3\n15 2\n17 4\n9 11\n2 7\n15 17\n3 20\n16 10\n17 14\n2 16\n1 19\n",
"8\n1 2\n2 3\n3 4\n1 7\n1 8\n4 5\n4 6\n",
"5\n5 1\n5 2\n5 3\n5 4\n",
"50\n49 6\n43 7\n1 27\n19 35\n15 37\n16 12\n19 21\n16 28\n49 9\n48 39\n13 1\n2 48\n9 50\n44 3\n41 32\n48 31\n49 33\n6 11\n13 20\n49 22\n13 41\n48 29\n13 46\n15 47\n34 2\n49 13\n48 14\n34 24\n16 36\n13 40\n49 34\n49 17\n43 25\n11 23\n10 15\n19 26\n34 44\n16 42\n19 18\n46 8\n29 38\n1 45\n12 43\n13 16\n46 30\n15 5\n49 10\n11 19\n32 4\n",
"20\n13 1\n18 2\n3 7\n18 5\n20 16\n3 12\n18 9\n3 10\n18 11\n13 6\n3 18\n20 15\n20 17\n3 13\n3 4\n13 14\n3 20\n18 8\n3 19\n",
"10\n8 2\n5 6\n1 8\n2 9\n1 4\n8 10\n10 5\n2 7\n2 3\n"
],
"output": [
"NO",
"YES",
"NO",
"NO",
"NO",
"YES",
"NO",
"NO",
"NO",
"NO",
"YES",
"NO",
"NO",
"YES",
"NO",
"NO",
"NO",
"YES",
"NO",
"YES",
"NO"
]
}
|
IN-CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
vector<vector<long long> > g;
void endProg() {
cout << "NO";
exit(0);
}
void dfs(long long v, long long p) {
long long l = 0;
for (auto i : g[v]) {
if (i == p) continue;
if (g[i].size() == 1) l++;
dfs(i, v);
}
if (l == 0) return;
if (v == 0 && l == 1 && g[v].size() == 1) return;
if (v == 0 && l <= 2 && g[v].size() <= 2)
endProg();
else if (v != 0 && l <= 1)
endProg();
}
int main() {
long long n;
cin >> n;
g.resize(n);
for (long long i = 0; i < n - 1; i++) {
long long a, b;
cin >> a >> b;
g[a - 1].push_back(b - 1);
g[b - 1].push_back(a - 1);
}
dfs(0, 0);
cout << "YES";
return 0;
}
|
1189_D1. Add on a Tree
|
Note that this is the first problem of the two similar problems. You can hack this problem only if you solve both problems.
You are given a tree with n nodes. In the beginning, 0 is written on all edges. In one operation, you can choose any 2 distinct leaves u, v and any real number x and add x to values written on all edges on the simple path between u and v.
For example, on the picture below you can see the result of applying two operations to the graph: adding 2 on the path from 7 to 6, and then adding -0.5 on the path from 4 to 5.
<image>
Is it true that for any configuration of real numbers written on edges, we can achieve it with a finite number of operations?
Leaf is a node of a tree of degree 1. Simple path is a path that doesn't contain any node twice.
Input
The first line contains a single integer n (2 ≤ n ≤ 10^5) — the number of nodes.
Each of the next n-1 lines contains two integers u and v (1 ≤ u, v ≤ n, u ≠ v), meaning that there is an edge between nodes u and v. It is guaranteed that these edges form a tree.
Output
If there is a configuration of real numbers written on edges of the tree that we can't achieve by performing the operations, output "NO".
Otherwise, output "YES".
You can print each letter in any case (upper or lower).
Examples
Input
2
1 2
Output
YES
Input
3
1 2
2 3
Output
NO
Input
5
1 2
1 3
1 4
2 5
Output
NO
Input
6
1 2
1 3
1 4
2 5
2 6
Output
YES
Note
In the first example, we can add any real x to the value written on the only edge (1, 2).
<image>
In the second example, one of configurations that we can't reach is 0 written on (1, 2) and 1 written on (2, 3).
<image>
Below you can see graphs from examples 3, 4:
<image> <image>
|
{
"input": [
"2\n1 2\n",
"3\n1 2\n2 3\n",
"5\n1 2\n1 3\n1 4\n2 5\n",
"6\n1 2\n1 3\n1 4\n2 5\n2 6\n"
],
"output": [
"YES",
"NO",
"NO",
"YES"
]
}
|
{
"input": [
"50\n16 4\n17 9\n31 19\n22 10\n8 1\n40 30\n3 31\n20 29\n47 27\n22 25\n32 34\n12 15\n40 32\n10 33\n47 12\n6 24\n46 41\n14 23\n12 35\n31 42\n46 28\n31 20\n46 37\n1 39\n29 49\n37 47\n40 6\n42 36\n47 2\n24 46\n2 13\n8 45\n41 3\n32 17\n4 7\n47 26\n28 8\n41 50\n34 44\n33 21\n25 5\n16 40\n3 14\n8 18\n28 11\n32 22\n2 38\n3 48\n44 43\n",
"10\n8 1\n1 2\n8 9\n8 5\n1 3\n1 10\n1 6\n1 7\n8 4\n",
"5\n5 1\n5 4\n4 3\n1 2\n",
"7\n1 2\n2 3\n1 4\n1 5\n3 6\n3 7\n",
"3\n1 3\n2 3\n",
"60\n26 6\n59 30\n31 12\n31 3\n38 23\n59 29\n53 9\n38 56\n53 54\n29 21\n17 55\n59 38\n26 16\n24 59\n24 25\n17 35\n24 41\n30 15\n31 27\n8 44\n26 5\n26 48\n8 32\n53 17\n3 34\n3 51\n30 28\n47 10\n53 60\n36 42\n24 53\n59 22\n53 40\n26 52\n36 4\n59 8\n29 37\n36 20\n17 47\n53 18\n3 50\n30 2\n17 7\n8 58\n59 1\n31 11\n24 26\n24 43\n53 57\n59 45\n47 13\n26 46\n17 33\n30 31\n26 39\n26 19\n24 36\n8 49\n38 14\n",
"7\n1 2\n2 3\n3 4\n3 5\n1 6\n1 7\n",
"20\n19 16\n19 18\n20 7\n9 4\n6 17\n14 2\n9 15\n2 13\n5 11\n19 12\n12 20\n16 9\n11 8\n19 5\n3 1\n19 14\n5 3\n12 10\n19 6\n",
"7\n1 2\n1 3\n2 4\n2 5\n3 6\n3 7\n",
"10\n9 5\n7 1\n9 10\n7 2\n5 4\n9 6\n2 9\n10 8\n1 3\n",
"4\n2 4\n2 3\n2 1\n",
"4\n1 4\n3 2\n1 3\n",
"3\n1 2\n1 3\n",
"5\n1 2\n1 5\n1 3\n1 4\n",
"20\n14 9\n12 13\n10 15\n2 1\n20 19\n16 6\n16 3\n17 14\n3 5\n2 11\n3 10\n15 8\n14 2\n6 4\n3 20\n5 18\n1 7\n1 16\n4 12\n",
"20\n7 5\n14 13\n17 6\n3 8\n16 12\n18 9\n3 18\n14 1\n17 3\n15 2\n17 4\n9 11\n2 7\n15 17\n3 20\n16 10\n17 14\n2 16\n1 19\n",
"8\n1 2\n2 3\n3 4\n1 7\n1 8\n4 5\n4 6\n",
"5\n5 1\n5 2\n5 3\n5 4\n",
"50\n49 6\n43 7\n1 27\n19 35\n15 37\n16 12\n19 21\n16 28\n49 9\n48 39\n13 1\n2 48\n9 50\n44 3\n41 32\n48 31\n49 33\n6 11\n13 20\n49 22\n13 41\n48 29\n13 46\n15 47\n34 2\n49 13\n48 14\n34 24\n16 36\n13 40\n49 34\n49 17\n43 25\n11 23\n10 15\n19 26\n34 44\n16 42\n19 18\n46 8\n29 38\n1 45\n12 43\n13 16\n46 30\n15 5\n49 10\n11 19\n32 4\n",
"20\n13 1\n18 2\n3 7\n18 5\n20 16\n3 12\n18 9\n3 10\n18 11\n13 6\n3 18\n20 15\n20 17\n3 13\n3 4\n13 14\n3 20\n18 8\n3 19\n",
"10\n8 2\n5 6\n1 8\n2 9\n1 4\n8 10\n10 5\n2 7\n2 3\n"
],
"output": [
"NO",
"YES",
"NO",
"NO",
"NO",
"YES",
"NO",
"NO",
"NO",
"NO",
"YES",
"NO",
"NO",
"YES",
"NO",
"NO",
"NO",
"YES",
"NO",
"YES",
"NO"
]
}
|
IN-CORRECT
|
python3
|
n = int(input())
g = [[] for i in range(n)]
for i in range(n - 1):
u, v = map(int, input().split())
u -= 1
v -= 1
g[u].append(v)
g[v].append(u)
for v in range(n):
if len(g[v]) == 1:
continue
flag = True
for u in g[v]:
if len(g[u]) == 1:
flag = False
break
if flag:
print('NO')
exit(0)
print('YES')
|
1189_D1. Add on a Tree
|
Note that this is the first problem of the two similar problems. You can hack this problem only if you solve both problems.
You are given a tree with n nodes. In the beginning, 0 is written on all edges. In one operation, you can choose any 2 distinct leaves u, v and any real number x and add x to values written on all edges on the simple path between u and v.
For example, on the picture below you can see the result of applying two operations to the graph: adding 2 on the path from 7 to 6, and then adding -0.5 on the path from 4 to 5.
<image>
Is it true that for any configuration of real numbers written on edges, we can achieve it with a finite number of operations?
Leaf is a node of a tree of degree 1. Simple path is a path that doesn't contain any node twice.
Input
The first line contains a single integer n (2 ≤ n ≤ 10^5) — the number of nodes.
Each of the next n-1 lines contains two integers u and v (1 ≤ u, v ≤ n, u ≠ v), meaning that there is an edge between nodes u and v. It is guaranteed that these edges form a tree.
Output
If there is a configuration of real numbers written on edges of the tree that we can't achieve by performing the operations, output "NO".
Otherwise, output "YES".
You can print each letter in any case (upper or lower).
Examples
Input
2
1 2
Output
YES
Input
3
1 2
2 3
Output
NO
Input
5
1 2
1 3
1 4
2 5
Output
NO
Input
6
1 2
1 3
1 4
2 5
2 6
Output
YES
Note
In the first example, we can add any real x to the value written on the only edge (1, 2).
<image>
In the second example, one of configurations that we can't reach is 0 written on (1, 2) and 1 written on (2, 3).
<image>
Below you can see graphs from examples 3, 4:
<image> <image>
|
{
"input": [
"2\n1 2\n",
"3\n1 2\n2 3\n",
"5\n1 2\n1 3\n1 4\n2 5\n",
"6\n1 2\n1 3\n1 4\n2 5\n2 6\n"
],
"output": [
"YES",
"NO",
"NO",
"YES"
]
}
|
{
"input": [
"50\n16 4\n17 9\n31 19\n22 10\n8 1\n40 30\n3 31\n20 29\n47 27\n22 25\n32 34\n12 15\n40 32\n10 33\n47 12\n6 24\n46 41\n14 23\n12 35\n31 42\n46 28\n31 20\n46 37\n1 39\n29 49\n37 47\n40 6\n42 36\n47 2\n24 46\n2 13\n8 45\n41 3\n32 17\n4 7\n47 26\n28 8\n41 50\n34 44\n33 21\n25 5\n16 40\n3 14\n8 18\n28 11\n32 22\n2 38\n3 48\n44 43\n",
"10\n8 1\n1 2\n8 9\n8 5\n1 3\n1 10\n1 6\n1 7\n8 4\n",
"5\n5 1\n5 4\n4 3\n1 2\n",
"7\n1 2\n2 3\n1 4\n1 5\n3 6\n3 7\n",
"3\n1 3\n2 3\n",
"60\n26 6\n59 30\n31 12\n31 3\n38 23\n59 29\n53 9\n38 56\n53 54\n29 21\n17 55\n59 38\n26 16\n24 59\n24 25\n17 35\n24 41\n30 15\n31 27\n8 44\n26 5\n26 48\n8 32\n53 17\n3 34\n3 51\n30 28\n47 10\n53 60\n36 42\n24 53\n59 22\n53 40\n26 52\n36 4\n59 8\n29 37\n36 20\n17 47\n53 18\n3 50\n30 2\n17 7\n8 58\n59 1\n31 11\n24 26\n24 43\n53 57\n59 45\n47 13\n26 46\n17 33\n30 31\n26 39\n26 19\n24 36\n8 49\n38 14\n",
"7\n1 2\n2 3\n3 4\n3 5\n1 6\n1 7\n",
"20\n19 16\n19 18\n20 7\n9 4\n6 17\n14 2\n9 15\n2 13\n5 11\n19 12\n12 20\n16 9\n11 8\n19 5\n3 1\n19 14\n5 3\n12 10\n19 6\n",
"7\n1 2\n1 3\n2 4\n2 5\n3 6\n3 7\n",
"10\n9 5\n7 1\n9 10\n7 2\n5 4\n9 6\n2 9\n10 8\n1 3\n",
"4\n2 4\n2 3\n2 1\n",
"4\n1 4\n3 2\n1 3\n",
"3\n1 2\n1 3\n",
"5\n1 2\n1 5\n1 3\n1 4\n",
"20\n14 9\n12 13\n10 15\n2 1\n20 19\n16 6\n16 3\n17 14\n3 5\n2 11\n3 10\n15 8\n14 2\n6 4\n3 20\n5 18\n1 7\n1 16\n4 12\n",
"20\n7 5\n14 13\n17 6\n3 8\n16 12\n18 9\n3 18\n14 1\n17 3\n15 2\n17 4\n9 11\n2 7\n15 17\n3 20\n16 10\n17 14\n2 16\n1 19\n",
"8\n1 2\n2 3\n3 4\n1 7\n1 8\n4 5\n4 6\n",
"5\n5 1\n5 2\n5 3\n5 4\n",
"50\n49 6\n43 7\n1 27\n19 35\n15 37\n16 12\n19 21\n16 28\n49 9\n48 39\n13 1\n2 48\n9 50\n44 3\n41 32\n48 31\n49 33\n6 11\n13 20\n49 22\n13 41\n48 29\n13 46\n15 47\n34 2\n49 13\n48 14\n34 24\n16 36\n13 40\n49 34\n49 17\n43 25\n11 23\n10 15\n19 26\n34 44\n16 42\n19 18\n46 8\n29 38\n1 45\n12 43\n13 16\n46 30\n15 5\n49 10\n11 19\n32 4\n",
"20\n13 1\n18 2\n3 7\n18 5\n20 16\n3 12\n18 9\n3 10\n18 11\n13 6\n3 18\n20 15\n20 17\n3 13\n3 4\n13 14\n3 20\n18 8\n3 19\n",
"10\n8 2\n5 6\n1 8\n2 9\n1 4\n8 10\n10 5\n2 7\n2 3\n"
],
"output": [
"NO",
"YES",
"NO",
"NO",
"NO",
"YES",
"NO",
"NO",
"NO",
"NO",
"YES",
"NO",
"NO",
"YES",
"NO",
"NO",
"NO",
"YES",
"NO",
"YES",
"NO"
]
}
|
IN-CORRECT
|
java
|
import java.util.*;
import java.io.*;
public class D572 {
public static void main(String[] args) {
MyScanner sc = new MyScanner();
PrintWriter out = new PrintWriter(new BufferedOutputStream(System.out));
int n = sc.nextInt();
if (n == 2) {
out.println("YES");
} else {
ArrayList<Integer> [] adj = new ArrayList[n + 1];
for (int i = 1; i <= n; i++) adj[i] = new ArrayList<>();
for (int i = 0; i < n - 1; i++) {
int x = sc.nextInt(); int y = sc.nextInt();
adj[x].add(y); adj[y].add(x);
}
Set<Integer> par = new HashSet<>();
for (int i = 1; i <= n; i++) {
if (adj[i].size() == 1) par.add(adj[i].get(0));
}
boolean ok = true;
for (Integer i: par) {
if (adj[i].size() == 2) ok = false;
}
out.println(ok ? "YES" :"NO");
}
out.close();
}
//-----------MyScanner class for faster input----------
public static class MyScanner {
BufferedReader br;
StringTokenizer st;
public MyScanner() {
br = new BufferedReader(new InputStreamReader(System.in));
}
String next() {
while (st == null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt() {
return Integer.parseInt(next());
}
long nextLong() {
return Long.parseLong(next());
}
double nextDouble() {
return Double.parseDouble(next());
}
String nextLine() {
String str = "";
try {
str = br.readLine();
} catch (IOException e) {
e.printStackTrace();
}
return str;
}
}
}
|
1189_D1. Add on a Tree
|
Note that this is the first problem of the two similar problems. You can hack this problem only if you solve both problems.
You are given a tree with n nodes. In the beginning, 0 is written on all edges. In one operation, you can choose any 2 distinct leaves u, v and any real number x and add x to values written on all edges on the simple path between u and v.
For example, on the picture below you can see the result of applying two operations to the graph: adding 2 on the path from 7 to 6, and then adding -0.5 on the path from 4 to 5.
<image>
Is it true that for any configuration of real numbers written on edges, we can achieve it with a finite number of operations?
Leaf is a node of a tree of degree 1. Simple path is a path that doesn't contain any node twice.
Input
The first line contains a single integer n (2 ≤ n ≤ 10^5) — the number of nodes.
Each of the next n-1 lines contains two integers u and v (1 ≤ u, v ≤ n, u ≠ v), meaning that there is an edge between nodes u and v. It is guaranteed that these edges form a tree.
Output
If there is a configuration of real numbers written on edges of the tree that we can't achieve by performing the operations, output "NO".
Otherwise, output "YES".
You can print each letter in any case (upper or lower).
Examples
Input
2
1 2
Output
YES
Input
3
1 2
2 3
Output
NO
Input
5
1 2
1 3
1 4
2 5
Output
NO
Input
6
1 2
1 3
1 4
2 5
2 6
Output
YES
Note
In the first example, we can add any real x to the value written on the only edge (1, 2).
<image>
In the second example, one of configurations that we can't reach is 0 written on (1, 2) and 1 written on (2, 3).
<image>
Below you can see graphs from examples 3, 4:
<image> <image>
|
{
"input": [
"2\n1 2\n",
"3\n1 2\n2 3\n",
"5\n1 2\n1 3\n1 4\n2 5\n",
"6\n1 2\n1 3\n1 4\n2 5\n2 6\n"
],
"output": [
"YES",
"NO",
"NO",
"YES"
]
}
|
{
"input": [
"50\n16 4\n17 9\n31 19\n22 10\n8 1\n40 30\n3 31\n20 29\n47 27\n22 25\n32 34\n12 15\n40 32\n10 33\n47 12\n6 24\n46 41\n14 23\n12 35\n31 42\n46 28\n31 20\n46 37\n1 39\n29 49\n37 47\n40 6\n42 36\n47 2\n24 46\n2 13\n8 45\n41 3\n32 17\n4 7\n47 26\n28 8\n41 50\n34 44\n33 21\n25 5\n16 40\n3 14\n8 18\n28 11\n32 22\n2 38\n3 48\n44 43\n",
"10\n8 1\n1 2\n8 9\n8 5\n1 3\n1 10\n1 6\n1 7\n8 4\n",
"5\n5 1\n5 4\n4 3\n1 2\n",
"7\n1 2\n2 3\n1 4\n1 5\n3 6\n3 7\n",
"3\n1 3\n2 3\n",
"60\n26 6\n59 30\n31 12\n31 3\n38 23\n59 29\n53 9\n38 56\n53 54\n29 21\n17 55\n59 38\n26 16\n24 59\n24 25\n17 35\n24 41\n30 15\n31 27\n8 44\n26 5\n26 48\n8 32\n53 17\n3 34\n3 51\n30 28\n47 10\n53 60\n36 42\n24 53\n59 22\n53 40\n26 52\n36 4\n59 8\n29 37\n36 20\n17 47\n53 18\n3 50\n30 2\n17 7\n8 58\n59 1\n31 11\n24 26\n24 43\n53 57\n59 45\n47 13\n26 46\n17 33\n30 31\n26 39\n26 19\n24 36\n8 49\n38 14\n",
"7\n1 2\n2 3\n3 4\n3 5\n1 6\n1 7\n",
"20\n19 16\n19 18\n20 7\n9 4\n6 17\n14 2\n9 15\n2 13\n5 11\n19 12\n12 20\n16 9\n11 8\n19 5\n3 1\n19 14\n5 3\n12 10\n19 6\n",
"7\n1 2\n1 3\n2 4\n2 5\n3 6\n3 7\n",
"10\n9 5\n7 1\n9 10\n7 2\n5 4\n9 6\n2 9\n10 8\n1 3\n",
"4\n2 4\n2 3\n2 1\n",
"4\n1 4\n3 2\n1 3\n",
"3\n1 2\n1 3\n",
"5\n1 2\n1 5\n1 3\n1 4\n",
"20\n14 9\n12 13\n10 15\n2 1\n20 19\n16 6\n16 3\n17 14\n3 5\n2 11\n3 10\n15 8\n14 2\n6 4\n3 20\n5 18\n1 7\n1 16\n4 12\n",
"20\n7 5\n14 13\n17 6\n3 8\n16 12\n18 9\n3 18\n14 1\n17 3\n15 2\n17 4\n9 11\n2 7\n15 17\n3 20\n16 10\n17 14\n2 16\n1 19\n",
"8\n1 2\n2 3\n3 4\n1 7\n1 8\n4 5\n4 6\n",
"5\n5 1\n5 2\n5 3\n5 4\n",
"50\n49 6\n43 7\n1 27\n19 35\n15 37\n16 12\n19 21\n16 28\n49 9\n48 39\n13 1\n2 48\n9 50\n44 3\n41 32\n48 31\n49 33\n6 11\n13 20\n49 22\n13 41\n48 29\n13 46\n15 47\n34 2\n49 13\n48 14\n34 24\n16 36\n13 40\n49 34\n49 17\n43 25\n11 23\n10 15\n19 26\n34 44\n16 42\n19 18\n46 8\n29 38\n1 45\n12 43\n13 16\n46 30\n15 5\n49 10\n11 19\n32 4\n",
"20\n13 1\n18 2\n3 7\n18 5\n20 16\n3 12\n18 9\n3 10\n18 11\n13 6\n3 18\n20 15\n20 17\n3 13\n3 4\n13 14\n3 20\n18 8\n3 19\n",
"10\n8 2\n5 6\n1 8\n2 9\n1 4\n8 10\n10 5\n2 7\n2 3\n"
],
"output": [
"NO",
"YES",
"NO",
"NO",
"NO",
"YES",
"NO",
"NO",
"NO",
"NO",
"YES",
"NO",
"NO",
"YES",
"NO",
"NO",
"NO",
"YES",
"NO",
"YES",
"NO"
]
}
|
IN-CORRECT
|
java
|
import java.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.InputMismatchException;
import java.io.IOException;
import java.io.InputStream;
/**
* Built using CHelper plug-in
* Actual solution is at the top
*
* @author KharYusuf
*/
public class Main {
public static void main(String[] args) {
InputStream inputStream = System.in;
OutputStream outputStream = System.out;
FastReader in = new FastReader(inputStream);
PrintWriter out = new PrintWriter(outputStream);
D1AddOnATree solver = new D1AddOnATree();
solver.solve(1, in, out);
out.close();
}
static class D1AddOnATree {
public void solve(int testNumber, FastReader s, PrintWriter w) {
int n = s.nextInt(), f[] = new int[n];
long cnt = 0;
/*ArrayList<Integer>[] a=new ArrayList[n];
for(int i=0;i<n;i++){
a[i]=new ArrayList<>();
}*/
for (int i = 1; i < n; i++) {
f[s.nextInt() - 1]++;
f[s.nextInt() - 1]++;
}
for (int i = 0; i < n; i++) {
if (f[i] == 1) {
cnt++;
}
}
if ((cnt * (cnt - 1) >> 1) >= n - 1) w.println("YES");
else w.println("NO");
}
}
static class FastReader {
private InputStream stream;
private byte[] buf = new byte[1024];
private int curChar;
private int numChars;
private FastReader.SpaceCharFilter filter;
public FastReader(InputStream stream) {
this.stream = stream;
}
public int read() {
if (numChars == -1)
throw new InputMismatchException();
if (curChar >= numChars) {
curChar = 0;
try {
numChars = stream.read(buf);
} catch (IOException e) {
throw new InputMismatchException();
}
if (numChars <= 0)
return -1;
}
return buf[curChar++];
}
public int nextInt() {
int c = read();
while (isSpaceChar(c))
c = read();
int sgn = 1;
if (c == '-') {
sgn = -1;
c = read();
}
int res = 0;
do {
if (c < '0' || c > '9')
throw new InputMismatchException();
res *= 10;
res += c - '0';
c = read();
}
while (!isSpaceChar(c));
return res * sgn;
}
public boolean isSpaceChar(int c) {
if (filter != null)
return filter.isSpaceChar(c);
return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
}
public interface SpaceCharFilter {
public boolean isSpaceChar(int ch);
}
}
}
|
1189_D1. Add on a Tree
|
Note that this is the first problem of the two similar problems. You can hack this problem only if you solve both problems.
You are given a tree with n nodes. In the beginning, 0 is written on all edges. In one operation, you can choose any 2 distinct leaves u, v and any real number x and add x to values written on all edges on the simple path between u and v.
For example, on the picture below you can see the result of applying two operations to the graph: adding 2 on the path from 7 to 6, and then adding -0.5 on the path from 4 to 5.
<image>
Is it true that for any configuration of real numbers written on edges, we can achieve it with a finite number of operations?
Leaf is a node of a tree of degree 1. Simple path is a path that doesn't contain any node twice.
Input
The first line contains a single integer n (2 ≤ n ≤ 10^5) — the number of nodes.
Each of the next n-1 lines contains two integers u and v (1 ≤ u, v ≤ n, u ≠ v), meaning that there is an edge between nodes u and v. It is guaranteed that these edges form a tree.
Output
If there is a configuration of real numbers written on edges of the tree that we can't achieve by performing the operations, output "NO".
Otherwise, output "YES".
You can print each letter in any case (upper or lower).
Examples
Input
2
1 2
Output
YES
Input
3
1 2
2 3
Output
NO
Input
5
1 2
1 3
1 4
2 5
Output
NO
Input
6
1 2
1 3
1 4
2 5
2 6
Output
YES
Note
In the first example, we can add any real x to the value written on the only edge (1, 2).
<image>
In the second example, one of configurations that we can't reach is 0 written on (1, 2) and 1 written on (2, 3).
<image>
Below you can see graphs from examples 3, 4:
<image> <image>
|
{
"input": [
"2\n1 2\n",
"3\n1 2\n2 3\n",
"5\n1 2\n1 3\n1 4\n2 5\n",
"6\n1 2\n1 3\n1 4\n2 5\n2 6\n"
],
"output": [
"YES",
"NO",
"NO",
"YES"
]
}
|
{
"input": [
"50\n16 4\n17 9\n31 19\n22 10\n8 1\n40 30\n3 31\n20 29\n47 27\n22 25\n32 34\n12 15\n40 32\n10 33\n47 12\n6 24\n46 41\n14 23\n12 35\n31 42\n46 28\n31 20\n46 37\n1 39\n29 49\n37 47\n40 6\n42 36\n47 2\n24 46\n2 13\n8 45\n41 3\n32 17\n4 7\n47 26\n28 8\n41 50\n34 44\n33 21\n25 5\n16 40\n3 14\n8 18\n28 11\n32 22\n2 38\n3 48\n44 43\n",
"10\n8 1\n1 2\n8 9\n8 5\n1 3\n1 10\n1 6\n1 7\n8 4\n",
"5\n5 1\n5 4\n4 3\n1 2\n",
"7\n1 2\n2 3\n1 4\n1 5\n3 6\n3 7\n",
"3\n1 3\n2 3\n",
"60\n26 6\n59 30\n31 12\n31 3\n38 23\n59 29\n53 9\n38 56\n53 54\n29 21\n17 55\n59 38\n26 16\n24 59\n24 25\n17 35\n24 41\n30 15\n31 27\n8 44\n26 5\n26 48\n8 32\n53 17\n3 34\n3 51\n30 28\n47 10\n53 60\n36 42\n24 53\n59 22\n53 40\n26 52\n36 4\n59 8\n29 37\n36 20\n17 47\n53 18\n3 50\n30 2\n17 7\n8 58\n59 1\n31 11\n24 26\n24 43\n53 57\n59 45\n47 13\n26 46\n17 33\n30 31\n26 39\n26 19\n24 36\n8 49\n38 14\n",
"7\n1 2\n2 3\n3 4\n3 5\n1 6\n1 7\n",
"20\n19 16\n19 18\n20 7\n9 4\n6 17\n14 2\n9 15\n2 13\n5 11\n19 12\n12 20\n16 9\n11 8\n19 5\n3 1\n19 14\n5 3\n12 10\n19 6\n",
"7\n1 2\n1 3\n2 4\n2 5\n3 6\n3 7\n",
"10\n9 5\n7 1\n9 10\n7 2\n5 4\n9 6\n2 9\n10 8\n1 3\n",
"4\n2 4\n2 3\n2 1\n",
"4\n1 4\n3 2\n1 3\n",
"3\n1 2\n1 3\n",
"5\n1 2\n1 5\n1 3\n1 4\n",
"20\n14 9\n12 13\n10 15\n2 1\n20 19\n16 6\n16 3\n17 14\n3 5\n2 11\n3 10\n15 8\n14 2\n6 4\n3 20\n5 18\n1 7\n1 16\n4 12\n",
"20\n7 5\n14 13\n17 6\n3 8\n16 12\n18 9\n3 18\n14 1\n17 3\n15 2\n17 4\n9 11\n2 7\n15 17\n3 20\n16 10\n17 14\n2 16\n1 19\n",
"8\n1 2\n2 3\n3 4\n1 7\n1 8\n4 5\n4 6\n",
"5\n5 1\n5 2\n5 3\n5 4\n",
"50\n49 6\n43 7\n1 27\n19 35\n15 37\n16 12\n19 21\n16 28\n49 9\n48 39\n13 1\n2 48\n9 50\n44 3\n41 32\n48 31\n49 33\n6 11\n13 20\n49 22\n13 41\n48 29\n13 46\n15 47\n34 2\n49 13\n48 14\n34 24\n16 36\n13 40\n49 34\n49 17\n43 25\n11 23\n10 15\n19 26\n34 44\n16 42\n19 18\n46 8\n29 38\n1 45\n12 43\n13 16\n46 30\n15 5\n49 10\n11 19\n32 4\n",
"20\n13 1\n18 2\n3 7\n18 5\n20 16\n3 12\n18 9\n3 10\n18 11\n13 6\n3 18\n20 15\n20 17\n3 13\n3 4\n13 14\n3 20\n18 8\n3 19\n",
"10\n8 2\n5 6\n1 8\n2 9\n1 4\n8 10\n10 5\n2 7\n2 3\n"
],
"output": [
"NO",
"YES",
"NO",
"NO",
"NO",
"YES",
"NO",
"NO",
"NO",
"NO",
"YES",
"NO",
"NO",
"YES",
"NO",
"NO",
"NO",
"YES",
"NO",
"YES",
"NO"
]
}
|
IN-CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
vector<vector<int>> adj;
int main() {
int n;
cin >> n;
adj = vector<vector<int>>(n + 1);
for (int i = 0; i < n - 1; i++) {
int a, b;
cin >> a >> b;
adj[a].push_back(b);
adj[b].push_back(a);
}
string ans = "YES\n";
int flag = 1;
for (int i = 1; i <= n; i++) {
if (adj[i].size() <= 2) {
ans = "NO\n";
flag = 0;
break;
}
}
cout << ans;
return 0;
}
|
1189_D1. Add on a Tree
|
Note that this is the first problem of the two similar problems. You can hack this problem only if you solve both problems.
You are given a tree with n nodes. In the beginning, 0 is written on all edges. In one operation, you can choose any 2 distinct leaves u, v and any real number x and add x to values written on all edges on the simple path between u and v.
For example, on the picture below you can see the result of applying two operations to the graph: adding 2 on the path from 7 to 6, and then adding -0.5 on the path from 4 to 5.
<image>
Is it true that for any configuration of real numbers written on edges, we can achieve it with a finite number of operations?
Leaf is a node of a tree of degree 1. Simple path is a path that doesn't contain any node twice.
Input
The first line contains a single integer n (2 ≤ n ≤ 10^5) — the number of nodes.
Each of the next n-1 lines contains two integers u and v (1 ≤ u, v ≤ n, u ≠ v), meaning that there is an edge between nodes u and v. It is guaranteed that these edges form a tree.
Output
If there is a configuration of real numbers written on edges of the tree that we can't achieve by performing the operations, output "NO".
Otherwise, output "YES".
You can print each letter in any case (upper or lower).
Examples
Input
2
1 2
Output
YES
Input
3
1 2
2 3
Output
NO
Input
5
1 2
1 3
1 4
2 5
Output
NO
Input
6
1 2
1 3
1 4
2 5
2 6
Output
YES
Note
In the first example, we can add any real x to the value written on the only edge (1, 2).
<image>
In the second example, one of configurations that we can't reach is 0 written on (1, 2) and 1 written on (2, 3).
<image>
Below you can see graphs from examples 3, 4:
<image> <image>
|
{
"input": [
"2\n1 2\n",
"3\n1 2\n2 3\n",
"5\n1 2\n1 3\n1 4\n2 5\n",
"6\n1 2\n1 3\n1 4\n2 5\n2 6\n"
],
"output": [
"YES",
"NO",
"NO",
"YES"
]
}
|
{
"input": [
"50\n16 4\n17 9\n31 19\n22 10\n8 1\n40 30\n3 31\n20 29\n47 27\n22 25\n32 34\n12 15\n40 32\n10 33\n47 12\n6 24\n46 41\n14 23\n12 35\n31 42\n46 28\n31 20\n46 37\n1 39\n29 49\n37 47\n40 6\n42 36\n47 2\n24 46\n2 13\n8 45\n41 3\n32 17\n4 7\n47 26\n28 8\n41 50\n34 44\n33 21\n25 5\n16 40\n3 14\n8 18\n28 11\n32 22\n2 38\n3 48\n44 43\n",
"10\n8 1\n1 2\n8 9\n8 5\n1 3\n1 10\n1 6\n1 7\n8 4\n",
"5\n5 1\n5 4\n4 3\n1 2\n",
"7\n1 2\n2 3\n1 4\n1 5\n3 6\n3 7\n",
"3\n1 3\n2 3\n",
"60\n26 6\n59 30\n31 12\n31 3\n38 23\n59 29\n53 9\n38 56\n53 54\n29 21\n17 55\n59 38\n26 16\n24 59\n24 25\n17 35\n24 41\n30 15\n31 27\n8 44\n26 5\n26 48\n8 32\n53 17\n3 34\n3 51\n30 28\n47 10\n53 60\n36 42\n24 53\n59 22\n53 40\n26 52\n36 4\n59 8\n29 37\n36 20\n17 47\n53 18\n3 50\n30 2\n17 7\n8 58\n59 1\n31 11\n24 26\n24 43\n53 57\n59 45\n47 13\n26 46\n17 33\n30 31\n26 39\n26 19\n24 36\n8 49\n38 14\n",
"7\n1 2\n2 3\n3 4\n3 5\n1 6\n1 7\n",
"20\n19 16\n19 18\n20 7\n9 4\n6 17\n14 2\n9 15\n2 13\n5 11\n19 12\n12 20\n16 9\n11 8\n19 5\n3 1\n19 14\n5 3\n12 10\n19 6\n",
"7\n1 2\n1 3\n2 4\n2 5\n3 6\n3 7\n",
"10\n9 5\n7 1\n9 10\n7 2\n5 4\n9 6\n2 9\n10 8\n1 3\n",
"4\n2 4\n2 3\n2 1\n",
"4\n1 4\n3 2\n1 3\n",
"3\n1 2\n1 3\n",
"5\n1 2\n1 5\n1 3\n1 4\n",
"20\n14 9\n12 13\n10 15\n2 1\n20 19\n16 6\n16 3\n17 14\n3 5\n2 11\n3 10\n15 8\n14 2\n6 4\n3 20\n5 18\n1 7\n1 16\n4 12\n",
"20\n7 5\n14 13\n17 6\n3 8\n16 12\n18 9\n3 18\n14 1\n17 3\n15 2\n17 4\n9 11\n2 7\n15 17\n3 20\n16 10\n17 14\n2 16\n1 19\n",
"8\n1 2\n2 3\n3 4\n1 7\n1 8\n4 5\n4 6\n",
"5\n5 1\n5 2\n5 3\n5 4\n",
"50\n49 6\n43 7\n1 27\n19 35\n15 37\n16 12\n19 21\n16 28\n49 9\n48 39\n13 1\n2 48\n9 50\n44 3\n41 32\n48 31\n49 33\n6 11\n13 20\n49 22\n13 41\n48 29\n13 46\n15 47\n34 2\n49 13\n48 14\n34 24\n16 36\n13 40\n49 34\n49 17\n43 25\n11 23\n10 15\n19 26\n34 44\n16 42\n19 18\n46 8\n29 38\n1 45\n12 43\n13 16\n46 30\n15 5\n49 10\n11 19\n32 4\n",
"20\n13 1\n18 2\n3 7\n18 5\n20 16\n3 12\n18 9\n3 10\n18 11\n13 6\n3 18\n20 15\n20 17\n3 13\n3 4\n13 14\n3 20\n18 8\n3 19\n",
"10\n8 2\n5 6\n1 8\n2 9\n1 4\n8 10\n10 5\n2 7\n2 3\n"
],
"output": [
"NO",
"YES",
"NO",
"NO",
"NO",
"YES",
"NO",
"NO",
"NO",
"NO",
"YES",
"NO",
"NO",
"YES",
"NO",
"NO",
"NO",
"YES",
"NO",
"YES",
"NO"
]
}
|
IN-CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
vector<vector<long long> > g;
int main() {
long long n;
cin >> n;
g.resize(n);
for (long long i = 0; i < n - 1; i++) {
long long a, b;
cin >> a >> b;
g[a - 1].push_back(b - 1);
g[b - 1].push_back(a - 1);
}
for (auto i : g[0]) {
if (g[i].size() >= 2) {
return cout << "NO", 0;
}
}
cout << "YES";
return 0;
}
|
1189_D1. Add on a Tree
|
Note that this is the first problem of the two similar problems. You can hack this problem only if you solve both problems.
You are given a tree with n nodes. In the beginning, 0 is written on all edges. In one operation, you can choose any 2 distinct leaves u, v and any real number x and add x to values written on all edges on the simple path between u and v.
For example, on the picture below you can see the result of applying two operations to the graph: adding 2 on the path from 7 to 6, and then adding -0.5 on the path from 4 to 5.
<image>
Is it true that for any configuration of real numbers written on edges, we can achieve it with a finite number of operations?
Leaf is a node of a tree of degree 1. Simple path is a path that doesn't contain any node twice.
Input
The first line contains a single integer n (2 ≤ n ≤ 10^5) — the number of nodes.
Each of the next n-1 lines contains two integers u and v (1 ≤ u, v ≤ n, u ≠ v), meaning that there is an edge between nodes u and v. It is guaranteed that these edges form a tree.
Output
If there is a configuration of real numbers written on edges of the tree that we can't achieve by performing the operations, output "NO".
Otherwise, output "YES".
You can print each letter in any case (upper or lower).
Examples
Input
2
1 2
Output
YES
Input
3
1 2
2 3
Output
NO
Input
5
1 2
1 3
1 4
2 5
Output
NO
Input
6
1 2
1 3
1 4
2 5
2 6
Output
YES
Note
In the first example, we can add any real x to the value written on the only edge (1, 2).
<image>
In the second example, one of configurations that we can't reach is 0 written on (1, 2) and 1 written on (2, 3).
<image>
Below you can see graphs from examples 3, 4:
<image> <image>
|
{
"input": [
"2\n1 2\n",
"3\n1 2\n2 3\n",
"5\n1 2\n1 3\n1 4\n2 5\n",
"6\n1 2\n1 3\n1 4\n2 5\n2 6\n"
],
"output": [
"YES",
"NO",
"NO",
"YES"
]
}
|
{
"input": [
"50\n16 4\n17 9\n31 19\n22 10\n8 1\n40 30\n3 31\n20 29\n47 27\n22 25\n32 34\n12 15\n40 32\n10 33\n47 12\n6 24\n46 41\n14 23\n12 35\n31 42\n46 28\n31 20\n46 37\n1 39\n29 49\n37 47\n40 6\n42 36\n47 2\n24 46\n2 13\n8 45\n41 3\n32 17\n4 7\n47 26\n28 8\n41 50\n34 44\n33 21\n25 5\n16 40\n3 14\n8 18\n28 11\n32 22\n2 38\n3 48\n44 43\n",
"10\n8 1\n1 2\n8 9\n8 5\n1 3\n1 10\n1 6\n1 7\n8 4\n",
"5\n5 1\n5 4\n4 3\n1 2\n",
"7\n1 2\n2 3\n1 4\n1 5\n3 6\n3 7\n",
"3\n1 3\n2 3\n",
"60\n26 6\n59 30\n31 12\n31 3\n38 23\n59 29\n53 9\n38 56\n53 54\n29 21\n17 55\n59 38\n26 16\n24 59\n24 25\n17 35\n24 41\n30 15\n31 27\n8 44\n26 5\n26 48\n8 32\n53 17\n3 34\n3 51\n30 28\n47 10\n53 60\n36 42\n24 53\n59 22\n53 40\n26 52\n36 4\n59 8\n29 37\n36 20\n17 47\n53 18\n3 50\n30 2\n17 7\n8 58\n59 1\n31 11\n24 26\n24 43\n53 57\n59 45\n47 13\n26 46\n17 33\n30 31\n26 39\n26 19\n24 36\n8 49\n38 14\n",
"7\n1 2\n2 3\n3 4\n3 5\n1 6\n1 7\n",
"20\n19 16\n19 18\n20 7\n9 4\n6 17\n14 2\n9 15\n2 13\n5 11\n19 12\n12 20\n16 9\n11 8\n19 5\n3 1\n19 14\n5 3\n12 10\n19 6\n",
"7\n1 2\n1 3\n2 4\n2 5\n3 6\n3 7\n",
"10\n9 5\n7 1\n9 10\n7 2\n5 4\n9 6\n2 9\n10 8\n1 3\n",
"4\n2 4\n2 3\n2 1\n",
"4\n1 4\n3 2\n1 3\n",
"3\n1 2\n1 3\n",
"5\n1 2\n1 5\n1 3\n1 4\n",
"20\n14 9\n12 13\n10 15\n2 1\n20 19\n16 6\n16 3\n17 14\n3 5\n2 11\n3 10\n15 8\n14 2\n6 4\n3 20\n5 18\n1 7\n1 16\n4 12\n",
"20\n7 5\n14 13\n17 6\n3 8\n16 12\n18 9\n3 18\n14 1\n17 3\n15 2\n17 4\n9 11\n2 7\n15 17\n3 20\n16 10\n17 14\n2 16\n1 19\n",
"8\n1 2\n2 3\n3 4\n1 7\n1 8\n4 5\n4 6\n",
"5\n5 1\n5 2\n5 3\n5 4\n",
"50\n49 6\n43 7\n1 27\n19 35\n15 37\n16 12\n19 21\n16 28\n49 9\n48 39\n13 1\n2 48\n9 50\n44 3\n41 32\n48 31\n49 33\n6 11\n13 20\n49 22\n13 41\n48 29\n13 46\n15 47\n34 2\n49 13\n48 14\n34 24\n16 36\n13 40\n49 34\n49 17\n43 25\n11 23\n10 15\n19 26\n34 44\n16 42\n19 18\n46 8\n29 38\n1 45\n12 43\n13 16\n46 30\n15 5\n49 10\n11 19\n32 4\n",
"20\n13 1\n18 2\n3 7\n18 5\n20 16\n3 12\n18 9\n3 10\n18 11\n13 6\n3 18\n20 15\n20 17\n3 13\n3 4\n13 14\n3 20\n18 8\n3 19\n",
"10\n8 2\n5 6\n1 8\n2 9\n1 4\n8 10\n10 5\n2 7\n2 3\n"
],
"output": [
"NO",
"YES",
"NO",
"NO",
"NO",
"YES",
"NO",
"NO",
"NO",
"NO",
"YES",
"NO",
"NO",
"YES",
"NO",
"NO",
"NO",
"YES",
"NO",
"YES",
"NO"
]
}
|
IN-CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
void dfs(vector<int> AdjL[], int visited[], int source, long long int &ans) {
if (visited[source] == 1) return;
visited[source] = 1;
if (AdjL[source].size() == 1) ans++;
for (int i = 0; i < AdjL[source].size(); ++i) {
if (!visited[AdjL[source][i]]) {
dfs(AdjL, visited, AdjL[source][i], ans);
}
}
}
int main() {
long long int n;
cin >> n;
vector<int> AdjL[n + 1];
for (int i = 0; i < n - 1; ++i) {
int x, y;
cin >> x >> y;
AdjL[x].push_back(y);
AdjL[y].push_back(x);
}
long long int ans = 0, path;
int visited[n + 1];
fill_n(visited, n + 1, 0);
dfs(AdjL, visited, 1, ans);
if (AdjL[1].size() == 1) ans--;
if (ans == 1)
path = 1;
else
path = (ans * (ans - 1)) / 2;
if (path >= n - 1)
cout << "YES";
else
cout << "NO";
return 0;
}
|
1189_D1. Add on a Tree
|
Note that this is the first problem of the two similar problems. You can hack this problem only if you solve both problems.
You are given a tree with n nodes. In the beginning, 0 is written on all edges. In one operation, you can choose any 2 distinct leaves u, v and any real number x and add x to values written on all edges on the simple path between u and v.
For example, on the picture below you can see the result of applying two operations to the graph: adding 2 on the path from 7 to 6, and then adding -0.5 on the path from 4 to 5.
<image>
Is it true that for any configuration of real numbers written on edges, we can achieve it with a finite number of operations?
Leaf is a node of a tree of degree 1. Simple path is a path that doesn't contain any node twice.
Input
The first line contains a single integer n (2 ≤ n ≤ 10^5) — the number of nodes.
Each of the next n-1 lines contains two integers u and v (1 ≤ u, v ≤ n, u ≠ v), meaning that there is an edge between nodes u and v. It is guaranteed that these edges form a tree.
Output
If there is a configuration of real numbers written on edges of the tree that we can't achieve by performing the operations, output "NO".
Otherwise, output "YES".
You can print each letter in any case (upper or lower).
Examples
Input
2
1 2
Output
YES
Input
3
1 2
2 3
Output
NO
Input
5
1 2
1 3
1 4
2 5
Output
NO
Input
6
1 2
1 3
1 4
2 5
2 6
Output
YES
Note
In the first example, we can add any real x to the value written on the only edge (1, 2).
<image>
In the second example, one of configurations that we can't reach is 0 written on (1, 2) and 1 written on (2, 3).
<image>
Below you can see graphs from examples 3, 4:
<image> <image>
|
{
"input": [
"2\n1 2\n",
"3\n1 2\n2 3\n",
"5\n1 2\n1 3\n1 4\n2 5\n",
"6\n1 2\n1 3\n1 4\n2 5\n2 6\n"
],
"output": [
"YES",
"NO",
"NO",
"YES"
]
}
|
{
"input": [
"50\n16 4\n17 9\n31 19\n22 10\n8 1\n40 30\n3 31\n20 29\n47 27\n22 25\n32 34\n12 15\n40 32\n10 33\n47 12\n6 24\n46 41\n14 23\n12 35\n31 42\n46 28\n31 20\n46 37\n1 39\n29 49\n37 47\n40 6\n42 36\n47 2\n24 46\n2 13\n8 45\n41 3\n32 17\n4 7\n47 26\n28 8\n41 50\n34 44\n33 21\n25 5\n16 40\n3 14\n8 18\n28 11\n32 22\n2 38\n3 48\n44 43\n",
"10\n8 1\n1 2\n8 9\n8 5\n1 3\n1 10\n1 6\n1 7\n8 4\n",
"5\n5 1\n5 4\n4 3\n1 2\n",
"7\n1 2\n2 3\n1 4\n1 5\n3 6\n3 7\n",
"3\n1 3\n2 3\n",
"60\n26 6\n59 30\n31 12\n31 3\n38 23\n59 29\n53 9\n38 56\n53 54\n29 21\n17 55\n59 38\n26 16\n24 59\n24 25\n17 35\n24 41\n30 15\n31 27\n8 44\n26 5\n26 48\n8 32\n53 17\n3 34\n3 51\n30 28\n47 10\n53 60\n36 42\n24 53\n59 22\n53 40\n26 52\n36 4\n59 8\n29 37\n36 20\n17 47\n53 18\n3 50\n30 2\n17 7\n8 58\n59 1\n31 11\n24 26\n24 43\n53 57\n59 45\n47 13\n26 46\n17 33\n30 31\n26 39\n26 19\n24 36\n8 49\n38 14\n",
"7\n1 2\n2 3\n3 4\n3 5\n1 6\n1 7\n",
"20\n19 16\n19 18\n20 7\n9 4\n6 17\n14 2\n9 15\n2 13\n5 11\n19 12\n12 20\n16 9\n11 8\n19 5\n3 1\n19 14\n5 3\n12 10\n19 6\n",
"7\n1 2\n1 3\n2 4\n2 5\n3 6\n3 7\n",
"10\n9 5\n7 1\n9 10\n7 2\n5 4\n9 6\n2 9\n10 8\n1 3\n",
"4\n2 4\n2 3\n2 1\n",
"4\n1 4\n3 2\n1 3\n",
"3\n1 2\n1 3\n",
"5\n1 2\n1 5\n1 3\n1 4\n",
"20\n14 9\n12 13\n10 15\n2 1\n20 19\n16 6\n16 3\n17 14\n3 5\n2 11\n3 10\n15 8\n14 2\n6 4\n3 20\n5 18\n1 7\n1 16\n4 12\n",
"20\n7 5\n14 13\n17 6\n3 8\n16 12\n18 9\n3 18\n14 1\n17 3\n15 2\n17 4\n9 11\n2 7\n15 17\n3 20\n16 10\n17 14\n2 16\n1 19\n",
"8\n1 2\n2 3\n3 4\n1 7\n1 8\n4 5\n4 6\n",
"5\n5 1\n5 2\n5 3\n5 4\n",
"50\n49 6\n43 7\n1 27\n19 35\n15 37\n16 12\n19 21\n16 28\n49 9\n48 39\n13 1\n2 48\n9 50\n44 3\n41 32\n48 31\n49 33\n6 11\n13 20\n49 22\n13 41\n48 29\n13 46\n15 47\n34 2\n49 13\n48 14\n34 24\n16 36\n13 40\n49 34\n49 17\n43 25\n11 23\n10 15\n19 26\n34 44\n16 42\n19 18\n46 8\n29 38\n1 45\n12 43\n13 16\n46 30\n15 5\n49 10\n11 19\n32 4\n",
"20\n13 1\n18 2\n3 7\n18 5\n20 16\n3 12\n18 9\n3 10\n18 11\n13 6\n3 18\n20 15\n20 17\n3 13\n3 4\n13 14\n3 20\n18 8\n3 19\n",
"10\n8 2\n5 6\n1 8\n2 9\n1 4\n8 10\n10 5\n2 7\n2 3\n"
],
"output": [
"NO",
"YES",
"NO",
"NO",
"NO",
"YES",
"NO",
"NO",
"NO",
"NO",
"YES",
"NO",
"NO",
"YES",
"NO",
"NO",
"NO",
"YES",
"NO",
"YES",
"NO"
]
}
|
IN-CORRECT
|
cpp
|
#include <bits/stdc++.h>
long long mod = 998244353;
using namespace std;
const long long N = 1e5 + 5;
vector<long long> g[N];
long long n;
void fail() {
cout << "NO\n";
exit(0);
}
long long cnt;
void dfs(long long node, long long p = 0) {
if (g[node].size() == 1) cnt++;
for (long long nx : g[node]) {
if (nx == p) continue;
dfs(nx, node);
}
}
int32_t main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
cin >> n;
if (n == 2) {
cout << "YES\n";
return 0;
}
for (long long i = 1, u, v; i < n; i++) {
cin >> u >> v;
g[u].push_back(v);
g[v].push_back(u);
}
dfs(1);
if (cnt >= 4)
cout << "YES";
else
cout << "NO";
return 0;
}
|
1189_D1. Add on a Tree
|
Note that this is the first problem of the two similar problems. You can hack this problem only if you solve both problems.
You are given a tree with n nodes. In the beginning, 0 is written on all edges. In one operation, you can choose any 2 distinct leaves u, v and any real number x and add x to values written on all edges on the simple path between u and v.
For example, on the picture below you can see the result of applying two operations to the graph: adding 2 on the path from 7 to 6, and then adding -0.5 on the path from 4 to 5.
<image>
Is it true that for any configuration of real numbers written on edges, we can achieve it with a finite number of operations?
Leaf is a node of a tree of degree 1. Simple path is a path that doesn't contain any node twice.
Input
The first line contains a single integer n (2 ≤ n ≤ 10^5) — the number of nodes.
Each of the next n-1 lines contains two integers u and v (1 ≤ u, v ≤ n, u ≠ v), meaning that there is an edge between nodes u and v. It is guaranteed that these edges form a tree.
Output
If there is a configuration of real numbers written on edges of the tree that we can't achieve by performing the operations, output "NO".
Otherwise, output "YES".
You can print each letter in any case (upper or lower).
Examples
Input
2
1 2
Output
YES
Input
3
1 2
2 3
Output
NO
Input
5
1 2
1 3
1 4
2 5
Output
NO
Input
6
1 2
1 3
1 4
2 5
2 6
Output
YES
Note
In the first example, we can add any real x to the value written on the only edge (1, 2).
<image>
In the second example, one of configurations that we can't reach is 0 written on (1, 2) and 1 written on (2, 3).
<image>
Below you can see graphs from examples 3, 4:
<image> <image>
|
{
"input": [
"2\n1 2\n",
"3\n1 2\n2 3\n",
"5\n1 2\n1 3\n1 4\n2 5\n",
"6\n1 2\n1 3\n1 4\n2 5\n2 6\n"
],
"output": [
"YES",
"NO",
"NO",
"YES"
]
}
|
{
"input": [
"50\n16 4\n17 9\n31 19\n22 10\n8 1\n40 30\n3 31\n20 29\n47 27\n22 25\n32 34\n12 15\n40 32\n10 33\n47 12\n6 24\n46 41\n14 23\n12 35\n31 42\n46 28\n31 20\n46 37\n1 39\n29 49\n37 47\n40 6\n42 36\n47 2\n24 46\n2 13\n8 45\n41 3\n32 17\n4 7\n47 26\n28 8\n41 50\n34 44\n33 21\n25 5\n16 40\n3 14\n8 18\n28 11\n32 22\n2 38\n3 48\n44 43\n",
"10\n8 1\n1 2\n8 9\n8 5\n1 3\n1 10\n1 6\n1 7\n8 4\n",
"5\n5 1\n5 4\n4 3\n1 2\n",
"7\n1 2\n2 3\n1 4\n1 5\n3 6\n3 7\n",
"3\n1 3\n2 3\n",
"60\n26 6\n59 30\n31 12\n31 3\n38 23\n59 29\n53 9\n38 56\n53 54\n29 21\n17 55\n59 38\n26 16\n24 59\n24 25\n17 35\n24 41\n30 15\n31 27\n8 44\n26 5\n26 48\n8 32\n53 17\n3 34\n3 51\n30 28\n47 10\n53 60\n36 42\n24 53\n59 22\n53 40\n26 52\n36 4\n59 8\n29 37\n36 20\n17 47\n53 18\n3 50\n30 2\n17 7\n8 58\n59 1\n31 11\n24 26\n24 43\n53 57\n59 45\n47 13\n26 46\n17 33\n30 31\n26 39\n26 19\n24 36\n8 49\n38 14\n",
"7\n1 2\n2 3\n3 4\n3 5\n1 6\n1 7\n",
"20\n19 16\n19 18\n20 7\n9 4\n6 17\n14 2\n9 15\n2 13\n5 11\n19 12\n12 20\n16 9\n11 8\n19 5\n3 1\n19 14\n5 3\n12 10\n19 6\n",
"7\n1 2\n1 3\n2 4\n2 5\n3 6\n3 7\n",
"10\n9 5\n7 1\n9 10\n7 2\n5 4\n9 6\n2 9\n10 8\n1 3\n",
"4\n2 4\n2 3\n2 1\n",
"4\n1 4\n3 2\n1 3\n",
"3\n1 2\n1 3\n",
"5\n1 2\n1 5\n1 3\n1 4\n",
"20\n14 9\n12 13\n10 15\n2 1\n20 19\n16 6\n16 3\n17 14\n3 5\n2 11\n3 10\n15 8\n14 2\n6 4\n3 20\n5 18\n1 7\n1 16\n4 12\n",
"20\n7 5\n14 13\n17 6\n3 8\n16 12\n18 9\n3 18\n14 1\n17 3\n15 2\n17 4\n9 11\n2 7\n15 17\n3 20\n16 10\n17 14\n2 16\n1 19\n",
"8\n1 2\n2 3\n3 4\n1 7\n1 8\n4 5\n4 6\n",
"5\n5 1\n5 2\n5 3\n5 4\n",
"50\n49 6\n43 7\n1 27\n19 35\n15 37\n16 12\n19 21\n16 28\n49 9\n48 39\n13 1\n2 48\n9 50\n44 3\n41 32\n48 31\n49 33\n6 11\n13 20\n49 22\n13 41\n48 29\n13 46\n15 47\n34 2\n49 13\n48 14\n34 24\n16 36\n13 40\n49 34\n49 17\n43 25\n11 23\n10 15\n19 26\n34 44\n16 42\n19 18\n46 8\n29 38\n1 45\n12 43\n13 16\n46 30\n15 5\n49 10\n11 19\n32 4\n",
"20\n13 1\n18 2\n3 7\n18 5\n20 16\n3 12\n18 9\n3 10\n18 11\n13 6\n3 18\n20 15\n20 17\n3 13\n3 4\n13 14\n3 20\n18 8\n3 19\n",
"10\n8 2\n5 6\n1 8\n2 9\n1 4\n8 10\n10 5\n2 7\n2 3\n"
],
"output": [
"NO",
"YES",
"NO",
"NO",
"NO",
"YES",
"NO",
"NO",
"NO",
"NO",
"YES",
"NO",
"NO",
"YES",
"NO",
"NO",
"NO",
"YES",
"NO",
"YES",
"NO"
]
}
|
IN-CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
int power(long long int x, long long int y) {
long long int temp;
if (y == 0) return 1;
temp = power(x, y / 2);
if (y % 2 == 0)
return temp * temp;
else
return x * temp * temp;
}
bool prime_check(long long int n) {
long long int i, j;
if (n == 1) {
return false;
}
for (i = 2; i <= sqrt(n); i++) {
if (n % i == 0) {
return false;
}
}
return true;
}
long long int fact(long long int n) {
long long int prod = 1;
for (long long int i = 1; i <= n; i++) {
prod = (prod * i) % 1000000007;
}
return prod;
}
bool vowl(char c) {
return c == 'a' || c == 'e' || c == 'i' || c == 'o' || c == 'u';
}
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
cout.tie(NULL);
int n, i, u, v;
cin >> n;
int freq[100003] = {};
for (i = 0; i < n - 1; i++) {
cin >> u >> v;
u--;
v--;
freq[u]++;
freq[v]++;
}
for (i = 0; i < n; i++) {
if (freq[i] != 0 && freq[i] & 2 == 0) break;
}
if (i == n)
cout << "YES\n";
else
cout << "NO\n";
return 0;
}
|
1189_D1. Add on a Tree
|
Note that this is the first problem of the two similar problems. You can hack this problem only if you solve both problems.
You are given a tree with n nodes. In the beginning, 0 is written on all edges. In one operation, you can choose any 2 distinct leaves u, v and any real number x and add x to values written on all edges on the simple path between u and v.
For example, on the picture below you can see the result of applying two operations to the graph: adding 2 on the path from 7 to 6, and then adding -0.5 on the path from 4 to 5.
<image>
Is it true that for any configuration of real numbers written on edges, we can achieve it with a finite number of operations?
Leaf is a node of a tree of degree 1. Simple path is a path that doesn't contain any node twice.
Input
The first line contains a single integer n (2 ≤ n ≤ 10^5) — the number of nodes.
Each of the next n-1 lines contains two integers u and v (1 ≤ u, v ≤ n, u ≠ v), meaning that there is an edge between nodes u and v. It is guaranteed that these edges form a tree.
Output
If there is a configuration of real numbers written on edges of the tree that we can't achieve by performing the operations, output "NO".
Otherwise, output "YES".
You can print each letter in any case (upper or lower).
Examples
Input
2
1 2
Output
YES
Input
3
1 2
2 3
Output
NO
Input
5
1 2
1 3
1 4
2 5
Output
NO
Input
6
1 2
1 3
1 4
2 5
2 6
Output
YES
Note
In the first example, we can add any real x to the value written on the only edge (1, 2).
<image>
In the second example, one of configurations that we can't reach is 0 written on (1, 2) and 1 written on (2, 3).
<image>
Below you can see graphs from examples 3, 4:
<image> <image>
|
{
"input": [
"2\n1 2\n",
"3\n1 2\n2 3\n",
"5\n1 2\n1 3\n1 4\n2 5\n",
"6\n1 2\n1 3\n1 4\n2 5\n2 6\n"
],
"output": [
"YES",
"NO",
"NO",
"YES"
]
}
|
{
"input": [
"50\n16 4\n17 9\n31 19\n22 10\n8 1\n40 30\n3 31\n20 29\n47 27\n22 25\n32 34\n12 15\n40 32\n10 33\n47 12\n6 24\n46 41\n14 23\n12 35\n31 42\n46 28\n31 20\n46 37\n1 39\n29 49\n37 47\n40 6\n42 36\n47 2\n24 46\n2 13\n8 45\n41 3\n32 17\n4 7\n47 26\n28 8\n41 50\n34 44\n33 21\n25 5\n16 40\n3 14\n8 18\n28 11\n32 22\n2 38\n3 48\n44 43\n",
"10\n8 1\n1 2\n8 9\n8 5\n1 3\n1 10\n1 6\n1 7\n8 4\n",
"5\n5 1\n5 4\n4 3\n1 2\n",
"7\n1 2\n2 3\n1 4\n1 5\n3 6\n3 7\n",
"3\n1 3\n2 3\n",
"60\n26 6\n59 30\n31 12\n31 3\n38 23\n59 29\n53 9\n38 56\n53 54\n29 21\n17 55\n59 38\n26 16\n24 59\n24 25\n17 35\n24 41\n30 15\n31 27\n8 44\n26 5\n26 48\n8 32\n53 17\n3 34\n3 51\n30 28\n47 10\n53 60\n36 42\n24 53\n59 22\n53 40\n26 52\n36 4\n59 8\n29 37\n36 20\n17 47\n53 18\n3 50\n30 2\n17 7\n8 58\n59 1\n31 11\n24 26\n24 43\n53 57\n59 45\n47 13\n26 46\n17 33\n30 31\n26 39\n26 19\n24 36\n8 49\n38 14\n",
"7\n1 2\n2 3\n3 4\n3 5\n1 6\n1 7\n",
"20\n19 16\n19 18\n20 7\n9 4\n6 17\n14 2\n9 15\n2 13\n5 11\n19 12\n12 20\n16 9\n11 8\n19 5\n3 1\n19 14\n5 3\n12 10\n19 6\n",
"7\n1 2\n1 3\n2 4\n2 5\n3 6\n3 7\n",
"10\n9 5\n7 1\n9 10\n7 2\n5 4\n9 6\n2 9\n10 8\n1 3\n",
"4\n2 4\n2 3\n2 1\n",
"4\n1 4\n3 2\n1 3\n",
"3\n1 2\n1 3\n",
"5\n1 2\n1 5\n1 3\n1 4\n",
"20\n14 9\n12 13\n10 15\n2 1\n20 19\n16 6\n16 3\n17 14\n3 5\n2 11\n3 10\n15 8\n14 2\n6 4\n3 20\n5 18\n1 7\n1 16\n4 12\n",
"20\n7 5\n14 13\n17 6\n3 8\n16 12\n18 9\n3 18\n14 1\n17 3\n15 2\n17 4\n9 11\n2 7\n15 17\n3 20\n16 10\n17 14\n2 16\n1 19\n",
"8\n1 2\n2 3\n3 4\n1 7\n1 8\n4 5\n4 6\n",
"5\n5 1\n5 2\n5 3\n5 4\n",
"50\n49 6\n43 7\n1 27\n19 35\n15 37\n16 12\n19 21\n16 28\n49 9\n48 39\n13 1\n2 48\n9 50\n44 3\n41 32\n48 31\n49 33\n6 11\n13 20\n49 22\n13 41\n48 29\n13 46\n15 47\n34 2\n49 13\n48 14\n34 24\n16 36\n13 40\n49 34\n49 17\n43 25\n11 23\n10 15\n19 26\n34 44\n16 42\n19 18\n46 8\n29 38\n1 45\n12 43\n13 16\n46 30\n15 5\n49 10\n11 19\n32 4\n",
"20\n13 1\n18 2\n3 7\n18 5\n20 16\n3 12\n18 9\n3 10\n18 11\n13 6\n3 18\n20 15\n20 17\n3 13\n3 4\n13 14\n3 20\n18 8\n3 19\n",
"10\n8 2\n5 6\n1 8\n2 9\n1 4\n8 10\n10 5\n2 7\n2 3\n"
],
"output": [
"NO",
"YES",
"NO",
"NO",
"NO",
"YES",
"NO",
"NO",
"NO",
"NO",
"YES",
"NO",
"NO",
"YES",
"NO",
"NO",
"NO",
"YES",
"NO",
"YES",
"NO"
]
}
|
IN-CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
vector<long long> tree[500500];
long long num[500500], fine = 0, n;
bool visted[500500];
long long dfs(int v) {
visted[v] = true;
if (tree[v].size() == 1) num[v] = 1;
for (auto u : tree[v]) {
if (!visted[u]) {
num[v] += dfs(u);
}
}
return num[v];
}
long long solve(int v) {
visted[v] = true;
if (tree[v].size() == 1) {
fine++;
if (!visted[tree[v][0]]) solve(tree[v][0]);
} else {
for (auto u : tree[v]) {
if (!visted[u]) {
if (num[u] > 1 && num[1] - num[u] > 1) fine++;
solve(u);
}
}
}
}
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
cout.tie(NULL);
cin >> n;
for (int i = 1; i < n; i++) {
long long u, v;
cin >> u >> v;
tree[u].push_back(v);
tree[v].push_back(u);
}
if (n <= 2) return cout << "YES", 0;
dfs(1);
if (num[1] == 2) return cout << "NO", 0;
for (int i = 1; i <= n; i++) visted[i] = false;
solve(1);
if (fine == n - 1) return cout << "YES", 0;
cout << "NO";
}
|
1189_D1. Add on a Tree
|
Note that this is the first problem of the two similar problems. You can hack this problem only if you solve both problems.
You are given a tree with n nodes. In the beginning, 0 is written on all edges. In one operation, you can choose any 2 distinct leaves u, v and any real number x and add x to values written on all edges on the simple path between u and v.
For example, on the picture below you can see the result of applying two operations to the graph: adding 2 on the path from 7 to 6, and then adding -0.5 on the path from 4 to 5.
<image>
Is it true that for any configuration of real numbers written on edges, we can achieve it with a finite number of operations?
Leaf is a node of a tree of degree 1. Simple path is a path that doesn't contain any node twice.
Input
The first line contains a single integer n (2 ≤ n ≤ 10^5) — the number of nodes.
Each of the next n-1 lines contains two integers u and v (1 ≤ u, v ≤ n, u ≠ v), meaning that there is an edge between nodes u and v. It is guaranteed that these edges form a tree.
Output
If there is a configuration of real numbers written on edges of the tree that we can't achieve by performing the operations, output "NO".
Otherwise, output "YES".
You can print each letter in any case (upper or lower).
Examples
Input
2
1 2
Output
YES
Input
3
1 2
2 3
Output
NO
Input
5
1 2
1 3
1 4
2 5
Output
NO
Input
6
1 2
1 3
1 4
2 5
2 6
Output
YES
Note
In the first example, we can add any real x to the value written on the only edge (1, 2).
<image>
In the second example, one of configurations that we can't reach is 0 written on (1, 2) and 1 written on (2, 3).
<image>
Below you can see graphs from examples 3, 4:
<image> <image>
|
{
"input": [
"2\n1 2\n",
"3\n1 2\n2 3\n",
"5\n1 2\n1 3\n1 4\n2 5\n",
"6\n1 2\n1 3\n1 4\n2 5\n2 6\n"
],
"output": [
"YES",
"NO",
"NO",
"YES"
]
}
|
{
"input": [
"50\n16 4\n17 9\n31 19\n22 10\n8 1\n40 30\n3 31\n20 29\n47 27\n22 25\n32 34\n12 15\n40 32\n10 33\n47 12\n6 24\n46 41\n14 23\n12 35\n31 42\n46 28\n31 20\n46 37\n1 39\n29 49\n37 47\n40 6\n42 36\n47 2\n24 46\n2 13\n8 45\n41 3\n32 17\n4 7\n47 26\n28 8\n41 50\n34 44\n33 21\n25 5\n16 40\n3 14\n8 18\n28 11\n32 22\n2 38\n3 48\n44 43\n",
"10\n8 1\n1 2\n8 9\n8 5\n1 3\n1 10\n1 6\n1 7\n8 4\n",
"5\n5 1\n5 4\n4 3\n1 2\n",
"7\n1 2\n2 3\n1 4\n1 5\n3 6\n3 7\n",
"3\n1 3\n2 3\n",
"60\n26 6\n59 30\n31 12\n31 3\n38 23\n59 29\n53 9\n38 56\n53 54\n29 21\n17 55\n59 38\n26 16\n24 59\n24 25\n17 35\n24 41\n30 15\n31 27\n8 44\n26 5\n26 48\n8 32\n53 17\n3 34\n3 51\n30 28\n47 10\n53 60\n36 42\n24 53\n59 22\n53 40\n26 52\n36 4\n59 8\n29 37\n36 20\n17 47\n53 18\n3 50\n30 2\n17 7\n8 58\n59 1\n31 11\n24 26\n24 43\n53 57\n59 45\n47 13\n26 46\n17 33\n30 31\n26 39\n26 19\n24 36\n8 49\n38 14\n",
"7\n1 2\n2 3\n3 4\n3 5\n1 6\n1 7\n",
"20\n19 16\n19 18\n20 7\n9 4\n6 17\n14 2\n9 15\n2 13\n5 11\n19 12\n12 20\n16 9\n11 8\n19 5\n3 1\n19 14\n5 3\n12 10\n19 6\n",
"7\n1 2\n1 3\n2 4\n2 5\n3 6\n3 7\n",
"10\n9 5\n7 1\n9 10\n7 2\n5 4\n9 6\n2 9\n10 8\n1 3\n",
"4\n2 4\n2 3\n2 1\n",
"4\n1 4\n3 2\n1 3\n",
"3\n1 2\n1 3\n",
"5\n1 2\n1 5\n1 3\n1 4\n",
"20\n14 9\n12 13\n10 15\n2 1\n20 19\n16 6\n16 3\n17 14\n3 5\n2 11\n3 10\n15 8\n14 2\n6 4\n3 20\n5 18\n1 7\n1 16\n4 12\n",
"20\n7 5\n14 13\n17 6\n3 8\n16 12\n18 9\n3 18\n14 1\n17 3\n15 2\n17 4\n9 11\n2 7\n15 17\n3 20\n16 10\n17 14\n2 16\n1 19\n",
"8\n1 2\n2 3\n3 4\n1 7\n1 8\n4 5\n4 6\n",
"5\n5 1\n5 2\n5 3\n5 4\n",
"50\n49 6\n43 7\n1 27\n19 35\n15 37\n16 12\n19 21\n16 28\n49 9\n48 39\n13 1\n2 48\n9 50\n44 3\n41 32\n48 31\n49 33\n6 11\n13 20\n49 22\n13 41\n48 29\n13 46\n15 47\n34 2\n49 13\n48 14\n34 24\n16 36\n13 40\n49 34\n49 17\n43 25\n11 23\n10 15\n19 26\n34 44\n16 42\n19 18\n46 8\n29 38\n1 45\n12 43\n13 16\n46 30\n15 5\n49 10\n11 19\n32 4\n",
"20\n13 1\n18 2\n3 7\n18 5\n20 16\n3 12\n18 9\n3 10\n18 11\n13 6\n3 18\n20 15\n20 17\n3 13\n3 4\n13 14\n3 20\n18 8\n3 19\n",
"10\n8 2\n5 6\n1 8\n2 9\n1 4\n8 10\n10 5\n2 7\n2 3\n"
],
"output": [
"NO",
"YES",
"NO",
"NO",
"NO",
"YES",
"NO",
"NO",
"NO",
"NO",
"YES",
"NO",
"NO",
"YES",
"NO",
"NO",
"NO",
"YES",
"NO",
"YES",
"NO"
]
}
|
IN-CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
vector<int> AL[100005];
int main() {
int n;
scanf("%d", &n);
for (int i = 0; i < n - 1; i++) {
int u, v;
scanf("%d", &u);
scanf("%d", &v);
AL[u].push_back(v);
AL[v].push_back(u);
}
long long int l = 0;
for (int i = 1; i <= n; i++)
if (AL[i].size() == 1) l++;
if ((l * (l - 1)) / 2 >= n - 1) {
printf("YES\n");
} else {
printf("NO\n");
}
return 0;
}
|
1189_D1. Add on a Tree
|
Note that this is the first problem of the two similar problems. You can hack this problem only if you solve both problems.
You are given a tree with n nodes. In the beginning, 0 is written on all edges. In one operation, you can choose any 2 distinct leaves u, v and any real number x and add x to values written on all edges on the simple path between u and v.
For example, on the picture below you can see the result of applying two operations to the graph: adding 2 on the path from 7 to 6, and then adding -0.5 on the path from 4 to 5.
<image>
Is it true that for any configuration of real numbers written on edges, we can achieve it with a finite number of operations?
Leaf is a node of a tree of degree 1. Simple path is a path that doesn't contain any node twice.
Input
The first line contains a single integer n (2 ≤ n ≤ 10^5) — the number of nodes.
Each of the next n-1 lines contains two integers u and v (1 ≤ u, v ≤ n, u ≠ v), meaning that there is an edge between nodes u and v. It is guaranteed that these edges form a tree.
Output
If there is a configuration of real numbers written on edges of the tree that we can't achieve by performing the operations, output "NO".
Otherwise, output "YES".
You can print each letter in any case (upper or lower).
Examples
Input
2
1 2
Output
YES
Input
3
1 2
2 3
Output
NO
Input
5
1 2
1 3
1 4
2 5
Output
NO
Input
6
1 2
1 3
1 4
2 5
2 6
Output
YES
Note
In the first example, we can add any real x to the value written on the only edge (1, 2).
<image>
In the second example, one of configurations that we can't reach is 0 written on (1, 2) and 1 written on (2, 3).
<image>
Below you can see graphs from examples 3, 4:
<image> <image>
|
{
"input": [
"2\n1 2\n",
"3\n1 2\n2 3\n",
"5\n1 2\n1 3\n1 4\n2 5\n",
"6\n1 2\n1 3\n1 4\n2 5\n2 6\n"
],
"output": [
"YES",
"NO",
"NO",
"YES"
]
}
|
{
"input": [
"50\n16 4\n17 9\n31 19\n22 10\n8 1\n40 30\n3 31\n20 29\n47 27\n22 25\n32 34\n12 15\n40 32\n10 33\n47 12\n6 24\n46 41\n14 23\n12 35\n31 42\n46 28\n31 20\n46 37\n1 39\n29 49\n37 47\n40 6\n42 36\n47 2\n24 46\n2 13\n8 45\n41 3\n32 17\n4 7\n47 26\n28 8\n41 50\n34 44\n33 21\n25 5\n16 40\n3 14\n8 18\n28 11\n32 22\n2 38\n3 48\n44 43\n",
"10\n8 1\n1 2\n8 9\n8 5\n1 3\n1 10\n1 6\n1 7\n8 4\n",
"5\n5 1\n5 4\n4 3\n1 2\n",
"7\n1 2\n2 3\n1 4\n1 5\n3 6\n3 7\n",
"3\n1 3\n2 3\n",
"60\n26 6\n59 30\n31 12\n31 3\n38 23\n59 29\n53 9\n38 56\n53 54\n29 21\n17 55\n59 38\n26 16\n24 59\n24 25\n17 35\n24 41\n30 15\n31 27\n8 44\n26 5\n26 48\n8 32\n53 17\n3 34\n3 51\n30 28\n47 10\n53 60\n36 42\n24 53\n59 22\n53 40\n26 52\n36 4\n59 8\n29 37\n36 20\n17 47\n53 18\n3 50\n30 2\n17 7\n8 58\n59 1\n31 11\n24 26\n24 43\n53 57\n59 45\n47 13\n26 46\n17 33\n30 31\n26 39\n26 19\n24 36\n8 49\n38 14\n",
"7\n1 2\n2 3\n3 4\n3 5\n1 6\n1 7\n",
"20\n19 16\n19 18\n20 7\n9 4\n6 17\n14 2\n9 15\n2 13\n5 11\n19 12\n12 20\n16 9\n11 8\n19 5\n3 1\n19 14\n5 3\n12 10\n19 6\n",
"7\n1 2\n1 3\n2 4\n2 5\n3 6\n3 7\n",
"10\n9 5\n7 1\n9 10\n7 2\n5 4\n9 6\n2 9\n10 8\n1 3\n",
"4\n2 4\n2 3\n2 1\n",
"4\n1 4\n3 2\n1 3\n",
"3\n1 2\n1 3\n",
"5\n1 2\n1 5\n1 3\n1 4\n",
"20\n14 9\n12 13\n10 15\n2 1\n20 19\n16 6\n16 3\n17 14\n3 5\n2 11\n3 10\n15 8\n14 2\n6 4\n3 20\n5 18\n1 7\n1 16\n4 12\n",
"20\n7 5\n14 13\n17 6\n3 8\n16 12\n18 9\n3 18\n14 1\n17 3\n15 2\n17 4\n9 11\n2 7\n15 17\n3 20\n16 10\n17 14\n2 16\n1 19\n",
"8\n1 2\n2 3\n3 4\n1 7\n1 8\n4 5\n4 6\n",
"5\n5 1\n5 2\n5 3\n5 4\n",
"50\n49 6\n43 7\n1 27\n19 35\n15 37\n16 12\n19 21\n16 28\n49 9\n48 39\n13 1\n2 48\n9 50\n44 3\n41 32\n48 31\n49 33\n6 11\n13 20\n49 22\n13 41\n48 29\n13 46\n15 47\n34 2\n49 13\n48 14\n34 24\n16 36\n13 40\n49 34\n49 17\n43 25\n11 23\n10 15\n19 26\n34 44\n16 42\n19 18\n46 8\n29 38\n1 45\n12 43\n13 16\n46 30\n15 5\n49 10\n11 19\n32 4\n",
"20\n13 1\n18 2\n3 7\n18 5\n20 16\n3 12\n18 9\n3 10\n18 11\n13 6\n3 18\n20 15\n20 17\n3 13\n3 4\n13 14\n3 20\n18 8\n3 19\n",
"10\n8 2\n5 6\n1 8\n2 9\n1 4\n8 10\n10 5\n2 7\n2 3\n"
],
"output": [
"NO",
"YES",
"NO",
"NO",
"NO",
"YES",
"NO",
"NO",
"NO",
"NO",
"YES",
"NO",
"NO",
"YES",
"NO",
"NO",
"NO",
"YES",
"NO",
"YES",
"NO"
]
}
|
IN-CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 1000, mod = 1e9 + 7;
vector<int> vec[N];
long long bin_pow(int x, int y) {
if (y == 0) return 1;
if (y == 1) return x;
if (y % 2 == 0) {
long long z = bin_pow(x, y / 2);
return (z * z) % mod;
}
return (bin_pow(x, y - 1) * x) % mod;
}
bool check(int x) {}
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
int n;
cin >> n;
for (int i = 1; i < n; i++) {
int x, y;
cin >> x >> y;
vec[x].push_back(y);
vec[y].push_back(x);
}
int k = 0;
for (int i = 1; i <= n; i++) {
if (vec[i].size() == 1) {
k++;
}
}
if (n == 2) {
cout << "YES";
return 0;
}
if (k == 2) {
cout << "NO";
return 0;
}
if (k % 2 == 0) {
cout << "YES";
} else
cout << "NO";
return 0;
}
|
1189_D1. Add on a Tree
|
Note that this is the first problem of the two similar problems. You can hack this problem only if you solve both problems.
You are given a tree with n nodes. In the beginning, 0 is written on all edges. In one operation, you can choose any 2 distinct leaves u, v and any real number x and add x to values written on all edges on the simple path between u and v.
For example, on the picture below you can see the result of applying two operations to the graph: adding 2 on the path from 7 to 6, and then adding -0.5 on the path from 4 to 5.
<image>
Is it true that for any configuration of real numbers written on edges, we can achieve it with a finite number of operations?
Leaf is a node of a tree of degree 1. Simple path is a path that doesn't contain any node twice.
Input
The first line contains a single integer n (2 ≤ n ≤ 10^5) — the number of nodes.
Each of the next n-1 lines contains two integers u and v (1 ≤ u, v ≤ n, u ≠ v), meaning that there is an edge between nodes u and v. It is guaranteed that these edges form a tree.
Output
If there is a configuration of real numbers written on edges of the tree that we can't achieve by performing the operations, output "NO".
Otherwise, output "YES".
You can print each letter in any case (upper or lower).
Examples
Input
2
1 2
Output
YES
Input
3
1 2
2 3
Output
NO
Input
5
1 2
1 3
1 4
2 5
Output
NO
Input
6
1 2
1 3
1 4
2 5
2 6
Output
YES
Note
In the first example, we can add any real x to the value written on the only edge (1, 2).
<image>
In the second example, one of configurations that we can't reach is 0 written on (1, 2) and 1 written on (2, 3).
<image>
Below you can see graphs from examples 3, 4:
<image> <image>
|
{
"input": [
"2\n1 2\n",
"3\n1 2\n2 3\n",
"5\n1 2\n1 3\n1 4\n2 5\n",
"6\n1 2\n1 3\n1 4\n2 5\n2 6\n"
],
"output": [
"YES",
"NO",
"NO",
"YES"
]
}
|
{
"input": [
"50\n16 4\n17 9\n31 19\n22 10\n8 1\n40 30\n3 31\n20 29\n47 27\n22 25\n32 34\n12 15\n40 32\n10 33\n47 12\n6 24\n46 41\n14 23\n12 35\n31 42\n46 28\n31 20\n46 37\n1 39\n29 49\n37 47\n40 6\n42 36\n47 2\n24 46\n2 13\n8 45\n41 3\n32 17\n4 7\n47 26\n28 8\n41 50\n34 44\n33 21\n25 5\n16 40\n3 14\n8 18\n28 11\n32 22\n2 38\n3 48\n44 43\n",
"10\n8 1\n1 2\n8 9\n8 5\n1 3\n1 10\n1 6\n1 7\n8 4\n",
"5\n5 1\n5 4\n4 3\n1 2\n",
"7\n1 2\n2 3\n1 4\n1 5\n3 6\n3 7\n",
"3\n1 3\n2 3\n",
"60\n26 6\n59 30\n31 12\n31 3\n38 23\n59 29\n53 9\n38 56\n53 54\n29 21\n17 55\n59 38\n26 16\n24 59\n24 25\n17 35\n24 41\n30 15\n31 27\n8 44\n26 5\n26 48\n8 32\n53 17\n3 34\n3 51\n30 28\n47 10\n53 60\n36 42\n24 53\n59 22\n53 40\n26 52\n36 4\n59 8\n29 37\n36 20\n17 47\n53 18\n3 50\n30 2\n17 7\n8 58\n59 1\n31 11\n24 26\n24 43\n53 57\n59 45\n47 13\n26 46\n17 33\n30 31\n26 39\n26 19\n24 36\n8 49\n38 14\n",
"7\n1 2\n2 3\n3 4\n3 5\n1 6\n1 7\n",
"20\n19 16\n19 18\n20 7\n9 4\n6 17\n14 2\n9 15\n2 13\n5 11\n19 12\n12 20\n16 9\n11 8\n19 5\n3 1\n19 14\n5 3\n12 10\n19 6\n",
"7\n1 2\n1 3\n2 4\n2 5\n3 6\n3 7\n",
"10\n9 5\n7 1\n9 10\n7 2\n5 4\n9 6\n2 9\n10 8\n1 3\n",
"4\n2 4\n2 3\n2 1\n",
"4\n1 4\n3 2\n1 3\n",
"3\n1 2\n1 3\n",
"5\n1 2\n1 5\n1 3\n1 4\n",
"20\n14 9\n12 13\n10 15\n2 1\n20 19\n16 6\n16 3\n17 14\n3 5\n2 11\n3 10\n15 8\n14 2\n6 4\n3 20\n5 18\n1 7\n1 16\n4 12\n",
"20\n7 5\n14 13\n17 6\n3 8\n16 12\n18 9\n3 18\n14 1\n17 3\n15 2\n17 4\n9 11\n2 7\n15 17\n3 20\n16 10\n17 14\n2 16\n1 19\n",
"8\n1 2\n2 3\n3 4\n1 7\n1 8\n4 5\n4 6\n",
"5\n5 1\n5 2\n5 3\n5 4\n",
"50\n49 6\n43 7\n1 27\n19 35\n15 37\n16 12\n19 21\n16 28\n49 9\n48 39\n13 1\n2 48\n9 50\n44 3\n41 32\n48 31\n49 33\n6 11\n13 20\n49 22\n13 41\n48 29\n13 46\n15 47\n34 2\n49 13\n48 14\n34 24\n16 36\n13 40\n49 34\n49 17\n43 25\n11 23\n10 15\n19 26\n34 44\n16 42\n19 18\n46 8\n29 38\n1 45\n12 43\n13 16\n46 30\n15 5\n49 10\n11 19\n32 4\n",
"20\n13 1\n18 2\n3 7\n18 5\n20 16\n3 12\n18 9\n3 10\n18 11\n13 6\n3 18\n20 15\n20 17\n3 13\n3 4\n13 14\n3 20\n18 8\n3 19\n",
"10\n8 2\n5 6\n1 8\n2 9\n1 4\n8 10\n10 5\n2 7\n2 3\n"
],
"output": [
"NO",
"YES",
"NO",
"NO",
"NO",
"YES",
"NO",
"NO",
"NO",
"NO",
"YES",
"NO",
"NO",
"YES",
"NO",
"NO",
"NO",
"YES",
"NO",
"YES",
"NO"
]
}
|
IN-CORRECT
|
java
|
// Working program using Reader Class
import java.io.*;
import java.io.FileInputStream;
import java.io.IOException;
import java.io.InputStreamReader;
import java.math.BigInteger;
import java.util.*;
public class Main
{
static class Reader
{
final private int BUFFER_SIZE = 1 << 16;
private DataInputStream din;
private byte[] buffer;
private int bufferPointer, bytesRead;
public Reader()
{
din = new DataInputStream(System.in);
buffer = new byte[BUFFER_SIZE];
bufferPointer = bytesRead = 0;
}
public Reader(String file_name) throws IOException
{
din = new DataInputStream(new FileInputStream(file_name));
buffer = new byte[BUFFER_SIZE];
bufferPointer = bytesRead = 0;
}
public String readLine() throws IOException
{
byte[] buf = new byte[64]; // line length
int cnt = 0, c;
while ((c = read()) != -1)
{
if (c == '\n')
break;
buf[cnt++] = (byte) c;
}
return new String(buf, 0, cnt);
}
public int nextInt() throws IOException
{
int ret = 0;
byte c = read();
while (c <= ' ')
c = read();
boolean neg = (c == '-');
if (neg)
c = read();
do
{
ret = ret * 10 + c - '0';
} while ((c = read()) >= '0' && c <= '9');
if (neg)
return -ret;
return ret;
}
public long nextLong() throws IOException
{
long ret = 0;
byte c = read();
while (c <= ' ')
c = read();
boolean neg = (c == '-');
if (neg)
c = read();
do {
ret = ret * 10 + c - '0';
}
while ((c = read()) >= '0' && c <= '9');
if (neg)
return -ret;
return ret;
}
public double nextDouble() throws IOException
{
double ret = 0, div = 1;
byte c = read();
while (c <= ' ')
c = read();
boolean neg = (c == '-');
if (neg)
c = read();
do {
ret = ret * 10 + c - '0';
}
while ((c = read()) >= '0' && c <= '9');
if (c == '.')
{
while ((c = read()) >= '0' && c <= '9')
{
ret += (c - '0') / (div *= 10);
}
}
if (neg)
return -ret;
return ret;
}
private void fillBuffer() throws IOException
{
bytesRead = din.read(buffer, bufferPointer = 0, BUFFER_SIZE);
if (bytesRead == -1)
buffer[0] = -1;
}
private byte read() throws IOException
{
if (bufferPointer == bytesRead)
fillBuffer();
return buffer[bufferPointer++];
}
public void close() throws IOException
{
if (din == null)
return;
din.close();
}
}
public static int gcd(int a, int b)
{
// Everything divides 0
if (a == 0 || b == 0)
return 0;
// base case
if (a == b)
return a;
// a is greater
if (a > b)
return gcd(a-b, b);
return gcd(a, b-a);
}
public static class Node{
int val;
ArrayList<Integer> al;
Node(int v){
al=new ArrayList<Integer>();
}
}
public static void main(String[] args) throws IOException
{
//Reader scan=new Reader();
Reader scan=new Reader();
PrintWriter out=new PrintWriter(System.out);
int n=scan.nextInt();
Node[] node=new Node[n];
for(int i=0;i<n;i++) {
node[i]=new Node(i);
}
for(int i=0;i<n-1;i++) {
int x=scan.nextInt();
int y=scan.nextInt();
x--;
y--;
node[x].al.add(y);
node[y].al.add(x);
}
for(int i=0;i<n;i++) {
if(node[i].al.size()%2==0) {
out.println("NO");
out.close();
}
}
out.println("YES");
out.close();
}
}
|
1189_D1. Add on a Tree
|
Note that this is the first problem of the two similar problems. You can hack this problem only if you solve both problems.
You are given a tree with n nodes. In the beginning, 0 is written on all edges. In one operation, you can choose any 2 distinct leaves u, v and any real number x and add x to values written on all edges on the simple path between u and v.
For example, on the picture below you can see the result of applying two operations to the graph: adding 2 on the path from 7 to 6, and then adding -0.5 on the path from 4 to 5.
<image>
Is it true that for any configuration of real numbers written on edges, we can achieve it with a finite number of operations?
Leaf is a node of a tree of degree 1. Simple path is a path that doesn't contain any node twice.
Input
The first line contains a single integer n (2 ≤ n ≤ 10^5) — the number of nodes.
Each of the next n-1 lines contains two integers u and v (1 ≤ u, v ≤ n, u ≠ v), meaning that there is an edge between nodes u and v. It is guaranteed that these edges form a tree.
Output
If there is a configuration of real numbers written on edges of the tree that we can't achieve by performing the operations, output "NO".
Otherwise, output "YES".
You can print each letter in any case (upper or lower).
Examples
Input
2
1 2
Output
YES
Input
3
1 2
2 3
Output
NO
Input
5
1 2
1 3
1 4
2 5
Output
NO
Input
6
1 2
1 3
1 4
2 5
2 6
Output
YES
Note
In the first example, we can add any real x to the value written on the only edge (1, 2).
<image>
In the second example, one of configurations that we can't reach is 0 written on (1, 2) and 1 written on (2, 3).
<image>
Below you can see graphs from examples 3, 4:
<image> <image>
|
{
"input": [
"2\n1 2\n",
"3\n1 2\n2 3\n",
"5\n1 2\n1 3\n1 4\n2 5\n",
"6\n1 2\n1 3\n1 4\n2 5\n2 6\n"
],
"output": [
"YES",
"NO",
"NO",
"YES"
]
}
|
{
"input": [
"50\n16 4\n17 9\n31 19\n22 10\n8 1\n40 30\n3 31\n20 29\n47 27\n22 25\n32 34\n12 15\n40 32\n10 33\n47 12\n6 24\n46 41\n14 23\n12 35\n31 42\n46 28\n31 20\n46 37\n1 39\n29 49\n37 47\n40 6\n42 36\n47 2\n24 46\n2 13\n8 45\n41 3\n32 17\n4 7\n47 26\n28 8\n41 50\n34 44\n33 21\n25 5\n16 40\n3 14\n8 18\n28 11\n32 22\n2 38\n3 48\n44 43\n",
"10\n8 1\n1 2\n8 9\n8 5\n1 3\n1 10\n1 6\n1 7\n8 4\n",
"5\n5 1\n5 4\n4 3\n1 2\n",
"7\n1 2\n2 3\n1 4\n1 5\n3 6\n3 7\n",
"3\n1 3\n2 3\n",
"60\n26 6\n59 30\n31 12\n31 3\n38 23\n59 29\n53 9\n38 56\n53 54\n29 21\n17 55\n59 38\n26 16\n24 59\n24 25\n17 35\n24 41\n30 15\n31 27\n8 44\n26 5\n26 48\n8 32\n53 17\n3 34\n3 51\n30 28\n47 10\n53 60\n36 42\n24 53\n59 22\n53 40\n26 52\n36 4\n59 8\n29 37\n36 20\n17 47\n53 18\n3 50\n30 2\n17 7\n8 58\n59 1\n31 11\n24 26\n24 43\n53 57\n59 45\n47 13\n26 46\n17 33\n30 31\n26 39\n26 19\n24 36\n8 49\n38 14\n",
"7\n1 2\n2 3\n3 4\n3 5\n1 6\n1 7\n",
"20\n19 16\n19 18\n20 7\n9 4\n6 17\n14 2\n9 15\n2 13\n5 11\n19 12\n12 20\n16 9\n11 8\n19 5\n3 1\n19 14\n5 3\n12 10\n19 6\n",
"7\n1 2\n1 3\n2 4\n2 5\n3 6\n3 7\n",
"10\n9 5\n7 1\n9 10\n7 2\n5 4\n9 6\n2 9\n10 8\n1 3\n",
"4\n2 4\n2 3\n2 1\n",
"4\n1 4\n3 2\n1 3\n",
"3\n1 2\n1 3\n",
"5\n1 2\n1 5\n1 3\n1 4\n",
"20\n14 9\n12 13\n10 15\n2 1\n20 19\n16 6\n16 3\n17 14\n3 5\n2 11\n3 10\n15 8\n14 2\n6 4\n3 20\n5 18\n1 7\n1 16\n4 12\n",
"20\n7 5\n14 13\n17 6\n3 8\n16 12\n18 9\n3 18\n14 1\n17 3\n15 2\n17 4\n9 11\n2 7\n15 17\n3 20\n16 10\n17 14\n2 16\n1 19\n",
"8\n1 2\n2 3\n3 4\n1 7\n1 8\n4 5\n4 6\n",
"5\n5 1\n5 2\n5 3\n5 4\n",
"50\n49 6\n43 7\n1 27\n19 35\n15 37\n16 12\n19 21\n16 28\n49 9\n48 39\n13 1\n2 48\n9 50\n44 3\n41 32\n48 31\n49 33\n6 11\n13 20\n49 22\n13 41\n48 29\n13 46\n15 47\n34 2\n49 13\n48 14\n34 24\n16 36\n13 40\n49 34\n49 17\n43 25\n11 23\n10 15\n19 26\n34 44\n16 42\n19 18\n46 8\n29 38\n1 45\n12 43\n13 16\n46 30\n15 5\n49 10\n11 19\n32 4\n",
"20\n13 1\n18 2\n3 7\n18 5\n20 16\n3 12\n18 9\n3 10\n18 11\n13 6\n3 18\n20 15\n20 17\n3 13\n3 4\n13 14\n3 20\n18 8\n3 19\n",
"10\n8 2\n5 6\n1 8\n2 9\n1 4\n8 10\n10 5\n2 7\n2 3\n"
],
"output": [
"NO",
"YES",
"NO",
"NO",
"NO",
"YES",
"NO",
"NO",
"NO",
"NO",
"YES",
"NO",
"NO",
"YES",
"NO",
"NO",
"NO",
"YES",
"NO",
"YES",
"NO"
]
}
|
IN-CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
int n, cnt[100009], x, y, vis[100009], cnt2[100009];
vector<int> v[100009];
void solve(int node, int last) {
vis[node]++;
int yes = 0;
for (auto i : v[node])
if (!vis[i]) solve(i, node), yes++;
if (yes == 0) cnt2[last]++;
}
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
;
cin >> n;
if (n == 2) return cout << "YES", 0;
int node = 0;
for (int i = 1; i < n; i++) {
cin >> x >> y;
cnt[x]++, cnt[y]++;
v[x].push_back(y);
v[y].push_back(x);
if (cnt[x] > 1) node = x;
if (cnt[y] > 1) node = y;
}
solve(node, 0);
int cntt = 0;
for (int i = 1; i <= n; i++)
if (cnt2[i] >= 2)
cntt++;
else if (cnt2[i] == 1)
return cout << "NO", 0;
if (cntt > 1)
cout << "YES";
else
cout << "NO";
}
|
1189_D1. Add on a Tree
|
Note that this is the first problem of the two similar problems. You can hack this problem only if you solve both problems.
You are given a tree with n nodes. In the beginning, 0 is written on all edges. In one operation, you can choose any 2 distinct leaves u, v and any real number x and add x to values written on all edges on the simple path between u and v.
For example, on the picture below you can see the result of applying two operations to the graph: adding 2 on the path from 7 to 6, and then adding -0.5 on the path from 4 to 5.
<image>
Is it true that for any configuration of real numbers written on edges, we can achieve it with a finite number of operations?
Leaf is a node of a tree of degree 1. Simple path is a path that doesn't contain any node twice.
Input
The first line contains a single integer n (2 ≤ n ≤ 10^5) — the number of nodes.
Each of the next n-1 lines contains two integers u and v (1 ≤ u, v ≤ n, u ≠ v), meaning that there is an edge between nodes u and v. It is guaranteed that these edges form a tree.
Output
If there is a configuration of real numbers written on edges of the tree that we can't achieve by performing the operations, output "NO".
Otherwise, output "YES".
You can print each letter in any case (upper or lower).
Examples
Input
2
1 2
Output
YES
Input
3
1 2
2 3
Output
NO
Input
5
1 2
1 3
1 4
2 5
Output
NO
Input
6
1 2
1 3
1 4
2 5
2 6
Output
YES
Note
In the first example, we can add any real x to the value written on the only edge (1, 2).
<image>
In the second example, one of configurations that we can't reach is 0 written on (1, 2) and 1 written on (2, 3).
<image>
Below you can see graphs from examples 3, 4:
<image> <image>
|
{
"input": [
"2\n1 2\n",
"3\n1 2\n2 3\n",
"5\n1 2\n1 3\n1 4\n2 5\n",
"6\n1 2\n1 3\n1 4\n2 5\n2 6\n"
],
"output": [
"YES",
"NO",
"NO",
"YES"
]
}
|
{
"input": [
"50\n16 4\n17 9\n31 19\n22 10\n8 1\n40 30\n3 31\n20 29\n47 27\n22 25\n32 34\n12 15\n40 32\n10 33\n47 12\n6 24\n46 41\n14 23\n12 35\n31 42\n46 28\n31 20\n46 37\n1 39\n29 49\n37 47\n40 6\n42 36\n47 2\n24 46\n2 13\n8 45\n41 3\n32 17\n4 7\n47 26\n28 8\n41 50\n34 44\n33 21\n25 5\n16 40\n3 14\n8 18\n28 11\n32 22\n2 38\n3 48\n44 43\n",
"10\n8 1\n1 2\n8 9\n8 5\n1 3\n1 10\n1 6\n1 7\n8 4\n",
"5\n5 1\n5 4\n4 3\n1 2\n",
"7\n1 2\n2 3\n1 4\n1 5\n3 6\n3 7\n",
"3\n1 3\n2 3\n",
"60\n26 6\n59 30\n31 12\n31 3\n38 23\n59 29\n53 9\n38 56\n53 54\n29 21\n17 55\n59 38\n26 16\n24 59\n24 25\n17 35\n24 41\n30 15\n31 27\n8 44\n26 5\n26 48\n8 32\n53 17\n3 34\n3 51\n30 28\n47 10\n53 60\n36 42\n24 53\n59 22\n53 40\n26 52\n36 4\n59 8\n29 37\n36 20\n17 47\n53 18\n3 50\n30 2\n17 7\n8 58\n59 1\n31 11\n24 26\n24 43\n53 57\n59 45\n47 13\n26 46\n17 33\n30 31\n26 39\n26 19\n24 36\n8 49\n38 14\n",
"7\n1 2\n2 3\n3 4\n3 5\n1 6\n1 7\n",
"20\n19 16\n19 18\n20 7\n9 4\n6 17\n14 2\n9 15\n2 13\n5 11\n19 12\n12 20\n16 9\n11 8\n19 5\n3 1\n19 14\n5 3\n12 10\n19 6\n",
"7\n1 2\n1 3\n2 4\n2 5\n3 6\n3 7\n",
"10\n9 5\n7 1\n9 10\n7 2\n5 4\n9 6\n2 9\n10 8\n1 3\n",
"4\n2 4\n2 3\n2 1\n",
"4\n1 4\n3 2\n1 3\n",
"3\n1 2\n1 3\n",
"5\n1 2\n1 5\n1 3\n1 4\n",
"20\n14 9\n12 13\n10 15\n2 1\n20 19\n16 6\n16 3\n17 14\n3 5\n2 11\n3 10\n15 8\n14 2\n6 4\n3 20\n5 18\n1 7\n1 16\n4 12\n",
"20\n7 5\n14 13\n17 6\n3 8\n16 12\n18 9\n3 18\n14 1\n17 3\n15 2\n17 4\n9 11\n2 7\n15 17\n3 20\n16 10\n17 14\n2 16\n1 19\n",
"8\n1 2\n2 3\n3 4\n1 7\n1 8\n4 5\n4 6\n",
"5\n5 1\n5 2\n5 3\n5 4\n",
"50\n49 6\n43 7\n1 27\n19 35\n15 37\n16 12\n19 21\n16 28\n49 9\n48 39\n13 1\n2 48\n9 50\n44 3\n41 32\n48 31\n49 33\n6 11\n13 20\n49 22\n13 41\n48 29\n13 46\n15 47\n34 2\n49 13\n48 14\n34 24\n16 36\n13 40\n49 34\n49 17\n43 25\n11 23\n10 15\n19 26\n34 44\n16 42\n19 18\n46 8\n29 38\n1 45\n12 43\n13 16\n46 30\n15 5\n49 10\n11 19\n32 4\n",
"20\n13 1\n18 2\n3 7\n18 5\n20 16\n3 12\n18 9\n3 10\n18 11\n13 6\n3 18\n20 15\n20 17\n3 13\n3 4\n13 14\n3 20\n18 8\n3 19\n",
"10\n8 2\n5 6\n1 8\n2 9\n1 4\n8 10\n10 5\n2 7\n2 3\n"
],
"output": [
"NO",
"YES",
"NO",
"NO",
"NO",
"YES",
"NO",
"NO",
"NO",
"NO",
"YES",
"NO",
"NO",
"YES",
"NO",
"NO",
"NO",
"YES",
"NO",
"YES",
"NO"
]
}
|
IN-CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
int n, num, head[100010], f[100010][2];
struct node {
int v, pre;
} e[100010 * 2];
vector<int> G;
void Add(int from, int to) {
num++;
e[num].v = to;
e[num].pre = head[from];
head[from] = num;
}
void Dfs(int now, int from) {
f[now][0] = 0;
int falg = 0;
for (int i = head[now]; i; i = e[i].pre) {
int v = e[i].v;
if (v == from) continue;
falg = 1;
Dfs(v, now);
f[now][0] += f[v][0];
}
if (falg == 0) {
f[now][0] = 1;
G.push_back(now);
}
}
void dfs(int now, int from) {
f[now][1] = f[from][0] - f[now][0] + f[from][1];
if (now == 1) f[now][1] = 0;
for (int i = head[now]; i; i = e[i].pre) {
int v = e[i].v;
if (v == from) continue;
dfs(v, now);
}
}
bool check(int now, int from) {
for (int i = head[now]; i; i = e[i].pre) {
int v = e[i].v;
if (v == from) continue;
if (f[v][0] < 2 || f[v][1] < 2 || check(v, now) == 0) return 0;
}
return 1;
}
int main() {
scanf("%d", &n);
int u, v;
for (int i = 1; i < n; i++) {
scanf("%d%d", &u, &v);
Add(u, v);
Add(v, u);
}
if (n == 2) {
printf("YES\n");
return 0;
}
Dfs(1, 0);
dfs(1, 0);
for (int i = 0; i < G.size(); i++) f[G[i]][0] = 100;
if (check(1, 0))
printf("YES\n");
else
printf("NO\n");
}
|
1189_D1. Add on a Tree
|
Note that this is the first problem of the two similar problems. You can hack this problem only if you solve both problems.
You are given a tree with n nodes. In the beginning, 0 is written on all edges. In one operation, you can choose any 2 distinct leaves u, v and any real number x and add x to values written on all edges on the simple path between u and v.
For example, on the picture below you can see the result of applying two operations to the graph: adding 2 on the path from 7 to 6, and then adding -0.5 on the path from 4 to 5.
<image>
Is it true that for any configuration of real numbers written on edges, we can achieve it with a finite number of operations?
Leaf is a node of a tree of degree 1. Simple path is a path that doesn't contain any node twice.
Input
The first line contains a single integer n (2 ≤ n ≤ 10^5) — the number of nodes.
Each of the next n-1 lines contains two integers u and v (1 ≤ u, v ≤ n, u ≠ v), meaning that there is an edge between nodes u and v. It is guaranteed that these edges form a tree.
Output
If there is a configuration of real numbers written on edges of the tree that we can't achieve by performing the operations, output "NO".
Otherwise, output "YES".
You can print each letter in any case (upper or lower).
Examples
Input
2
1 2
Output
YES
Input
3
1 2
2 3
Output
NO
Input
5
1 2
1 3
1 4
2 5
Output
NO
Input
6
1 2
1 3
1 4
2 5
2 6
Output
YES
Note
In the first example, we can add any real x to the value written on the only edge (1, 2).
<image>
In the second example, one of configurations that we can't reach is 0 written on (1, 2) and 1 written on (2, 3).
<image>
Below you can see graphs from examples 3, 4:
<image> <image>
|
{
"input": [
"2\n1 2\n",
"3\n1 2\n2 3\n",
"5\n1 2\n1 3\n1 4\n2 5\n",
"6\n1 2\n1 3\n1 4\n2 5\n2 6\n"
],
"output": [
"YES",
"NO",
"NO",
"YES"
]
}
|
{
"input": [
"50\n16 4\n17 9\n31 19\n22 10\n8 1\n40 30\n3 31\n20 29\n47 27\n22 25\n32 34\n12 15\n40 32\n10 33\n47 12\n6 24\n46 41\n14 23\n12 35\n31 42\n46 28\n31 20\n46 37\n1 39\n29 49\n37 47\n40 6\n42 36\n47 2\n24 46\n2 13\n8 45\n41 3\n32 17\n4 7\n47 26\n28 8\n41 50\n34 44\n33 21\n25 5\n16 40\n3 14\n8 18\n28 11\n32 22\n2 38\n3 48\n44 43\n",
"10\n8 1\n1 2\n8 9\n8 5\n1 3\n1 10\n1 6\n1 7\n8 4\n",
"5\n5 1\n5 4\n4 3\n1 2\n",
"7\n1 2\n2 3\n1 4\n1 5\n3 6\n3 7\n",
"3\n1 3\n2 3\n",
"60\n26 6\n59 30\n31 12\n31 3\n38 23\n59 29\n53 9\n38 56\n53 54\n29 21\n17 55\n59 38\n26 16\n24 59\n24 25\n17 35\n24 41\n30 15\n31 27\n8 44\n26 5\n26 48\n8 32\n53 17\n3 34\n3 51\n30 28\n47 10\n53 60\n36 42\n24 53\n59 22\n53 40\n26 52\n36 4\n59 8\n29 37\n36 20\n17 47\n53 18\n3 50\n30 2\n17 7\n8 58\n59 1\n31 11\n24 26\n24 43\n53 57\n59 45\n47 13\n26 46\n17 33\n30 31\n26 39\n26 19\n24 36\n8 49\n38 14\n",
"7\n1 2\n2 3\n3 4\n3 5\n1 6\n1 7\n",
"20\n19 16\n19 18\n20 7\n9 4\n6 17\n14 2\n9 15\n2 13\n5 11\n19 12\n12 20\n16 9\n11 8\n19 5\n3 1\n19 14\n5 3\n12 10\n19 6\n",
"7\n1 2\n1 3\n2 4\n2 5\n3 6\n3 7\n",
"10\n9 5\n7 1\n9 10\n7 2\n5 4\n9 6\n2 9\n10 8\n1 3\n",
"4\n2 4\n2 3\n2 1\n",
"4\n1 4\n3 2\n1 3\n",
"3\n1 2\n1 3\n",
"5\n1 2\n1 5\n1 3\n1 4\n",
"20\n14 9\n12 13\n10 15\n2 1\n20 19\n16 6\n16 3\n17 14\n3 5\n2 11\n3 10\n15 8\n14 2\n6 4\n3 20\n5 18\n1 7\n1 16\n4 12\n",
"20\n7 5\n14 13\n17 6\n3 8\n16 12\n18 9\n3 18\n14 1\n17 3\n15 2\n17 4\n9 11\n2 7\n15 17\n3 20\n16 10\n17 14\n2 16\n1 19\n",
"8\n1 2\n2 3\n3 4\n1 7\n1 8\n4 5\n4 6\n",
"5\n5 1\n5 2\n5 3\n5 4\n",
"50\n49 6\n43 7\n1 27\n19 35\n15 37\n16 12\n19 21\n16 28\n49 9\n48 39\n13 1\n2 48\n9 50\n44 3\n41 32\n48 31\n49 33\n6 11\n13 20\n49 22\n13 41\n48 29\n13 46\n15 47\n34 2\n49 13\n48 14\n34 24\n16 36\n13 40\n49 34\n49 17\n43 25\n11 23\n10 15\n19 26\n34 44\n16 42\n19 18\n46 8\n29 38\n1 45\n12 43\n13 16\n46 30\n15 5\n49 10\n11 19\n32 4\n",
"20\n13 1\n18 2\n3 7\n18 5\n20 16\n3 12\n18 9\n3 10\n18 11\n13 6\n3 18\n20 15\n20 17\n3 13\n3 4\n13 14\n3 20\n18 8\n3 19\n",
"10\n8 2\n5 6\n1 8\n2 9\n1 4\n8 10\n10 5\n2 7\n2 3\n"
],
"output": [
"NO",
"YES",
"NO",
"NO",
"NO",
"YES",
"NO",
"NO",
"NO",
"NO",
"YES",
"NO",
"NO",
"YES",
"NO",
"NO",
"NO",
"YES",
"NO",
"YES",
"NO"
]
}
|
IN-CORRECT
|
java
|
import java.util.*;
import java.lang.*;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.*;
public class Main {
int flag=1;
public void dfs(ArrayList<Integer>[] tree,int[] visited,int s) {
visited[s]=1;
int count=0;
for(int i:tree[s]) {
if(visited[i]==0) {
count++;
this.dfs(tree, visited, i);
}
}
if(count==1) {
flag=0;
}
}
public static void main(String[] args) throws Exception{
FastReader sc=new FastReader();
OutputStream outputStream = System.out;
PrintWriter out = new PrintWriter(outputStream);
Main mm=new Main();
int n=sc.nextInt();
ArrayList<Integer>[] tree=new ArrayList[n];
for(int i=0;i<n;i++) {
tree[i]=new ArrayList<Integer>();
}
for(int i=0;i<n-1;i++) {
int a=sc.nextInt()-1;
int b=sc.nextInt()-1;
tree[a].add(b);
tree[b].add(a);
}
mm.dfs(tree,new int[n], 0);
if(n==2) {
System.out.println("YES");
}
else {
if(mm.flag==1) {
System.out.println("YES");
}
else {
System.out.println("NO");
}
}
}
}
class FastReader
{
BufferedReader br;
StringTokenizer st;
public FastReader()
{
br = new BufferedReader(new
InputStreamReader(System.in));
}
String next()
{
while (st == null || !st.hasMoreElements())
{
try
{
st = new StringTokenizer(br.readLine());
}
catch (IOException e)
{
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt()
{
return Integer.parseInt(next());
}
long nextLong()
{
return Long.parseLong(next());
}
double nextDouble()
{
return Double.parseDouble(next());
}
String nextLine()
{
String str = "";
try
{
str = br.readLine();
}
catch (IOException e)
{
e.printStackTrace();
}
return str;
}
}
|
1189_D1. Add on a Tree
|
Note that this is the first problem of the two similar problems. You can hack this problem only if you solve both problems.
You are given a tree with n nodes. In the beginning, 0 is written on all edges. In one operation, you can choose any 2 distinct leaves u, v and any real number x and add x to values written on all edges on the simple path between u and v.
For example, on the picture below you can see the result of applying two operations to the graph: adding 2 on the path from 7 to 6, and then adding -0.5 on the path from 4 to 5.
<image>
Is it true that for any configuration of real numbers written on edges, we can achieve it with a finite number of operations?
Leaf is a node of a tree of degree 1. Simple path is a path that doesn't contain any node twice.
Input
The first line contains a single integer n (2 ≤ n ≤ 10^5) — the number of nodes.
Each of the next n-1 lines contains two integers u and v (1 ≤ u, v ≤ n, u ≠ v), meaning that there is an edge between nodes u and v. It is guaranteed that these edges form a tree.
Output
If there is a configuration of real numbers written on edges of the tree that we can't achieve by performing the operations, output "NO".
Otherwise, output "YES".
You can print each letter in any case (upper or lower).
Examples
Input
2
1 2
Output
YES
Input
3
1 2
2 3
Output
NO
Input
5
1 2
1 3
1 4
2 5
Output
NO
Input
6
1 2
1 3
1 4
2 5
2 6
Output
YES
Note
In the first example, we can add any real x to the value written on the only edge (1, 2).
<image>
In the second example, one of configurations that we can't reach is 0 written on (1, 2) and 1 written on (2, 3).
<image>
Below you can see graphs from examples 3, 4:
<image> <image>
|
{
"input": [
"2\n1 2\n",
"3\n1 2\n2 3\n",
"5\n1 2\n1 3\n1 4\n2 5\n",
"6\n1 2\n1 3\n1 4\n2 5\n2 6\n"
],
"output": [
"YES",
"NO",
"NO",
"YES"
]
}
|
{
"input": [
"50\n16 4\n17 9\n31 19\n22 10\n8 1\n40 30\n3 31\n20 29\n47 27\n22 25\n32 34\n12 15\n40 32\n10 33\n47 12\n6 24\n46 41\n14 23\n12 35\n31 42\n46 28\n31 20\n46 37\n1 39\n29 49\n37 47\n40 6\n42 36\n47 2\n24 46\n2 13\n8 45\n41 3\n32 17\n4 7\n47 26\n28 8\n41 50\n34 44\n33 21\n25 5\n16 40\n3 14\n8 18\n28 11\n32 22\n2 38\n3 48\n44 43\n",
"10\n8 1\n1 2\n8 9\n8 5\n1 3\n1 10\n1 6\n1 7\n8 4\n",
"5\n5 1\n5 4\n4 3\n1 2\n",
"7\n1 2\n2 3\n1 4\n1 5\n3 6\n3 7\n",
"3\n1 3\n2 3\n",
"60\n26 6\n59 30\n31 12\n31 3\n38 23\n59 29\n53 9\n38 56\n53 54\n29 21\n17 55\n59 38\n26 16\n24 59\n24 25\n17 35\n24 41\n30 15\n31 27\n8 44\n26 5\n26 48\n8 32\n53 17\n3 34\n3 51\n30 28\n47 10\n53 60\n36 42\n24 53\n59 22\n53 40\n26 52\n36 4\n59 8\n29 37\n36 20\n17 47\n53 18\n3 50\n30 2\n17 7\n8 58\n59 1\n31 11\n24 26\n24 43\n53 57\n59 45\n47 13\n26 46\n17 33\n30 31\n26 39\n26 19\n24 36\n8 49\n38 14\n",
"7\n1 2\n2 3\n3 4\n3 5\n1 6\n1 7\n",
"20\n19 16\n19 18\n20 7\n9 4\n6 17\n14 2\n9 15\n2 13\n5 11\n19 12\n12 20\n16 9\n11 8\n19 5\n3 1\n19 14\n5 3\n12 10\n19 6\n",
"7\n1 2\n1 3\n2 4\n2 5\n3 6\n3 7\n",
"10\n9 5\n7 1\n9 10\n7 2\n5 4\n9 6\n2 9\n10 8\n1 3\n",
"4\n2 4\n2 3\n2 1\n",
"4\n1 4\n3 2\n1 3\n",
"3\n1 2\n1 3\n",
"5\n1 2\n1 5\n1 3\n1 4\n",
"20\n14 9\n12 13\n10 15\n2 1\n20 19\n16 6\n16 3\n17 14\n3 5\n2 11\n3 10\n15 8\n14 2\n6 4\n3 20\n5 18\n1 7\n1 16\n4 12\n",
"20\n7 5\n14 13\n17 6\n3 8\n16 12\n18 9\n3 18\n14 1\n17 3\n15 2\n17 4\n9 11\n2 7\n15 17\n3 20\n16 10\n17 14\n2 16\n1 19\n",
"8\n1 2\n2 3\n3 4\n1 7\n1 8\n4 5\n4 6\n",
"5\n5 1\n5 2\n5 3\n5 4\n",
"50\n49 6\n43 7\n1 27\n19 35\n15 37\n16 12\n19 21\n16 28\n49 9\n48 39\n13 1\n2 48\n9 50\n44 3\n41 32\n48 31\n49 33\n6 11\n13 20\n49 22\n13 41\n48 29\n13 46\n15 47\n34 2\n49 13\n48 14\n34 24\n16 36\n13 40\n49 34\n49 17\n43 25\n11 23\n10 15\n19 26\n34 44\n16 42\n19 18\n46 8\n29 38\n1 45\n12 43\n13 16\n46 30\n15 5\n49 10\n11 19\n32 4\n",
"20\n13 1\n18 2\n3 7\n18 5\n20 16\n3 12\n18 9\n3 10\n18 11\n13 6\n3 18\n20 15\n20 17\n3 13\n3 4\n13 14\n3 20\n18 8\n3 19\n",
"10\n8 2\n5 6\n1 8\n2 9\n1 4\n8 10\n10 5\n2 7\n2 3\n"
],
"output": [
"NO",
"YES",
"NO",
"NO",
"NO",
"YES",
"NO",
"NO",
"NO",
"NO",
"YES",
"NO",
"NO",
"YES",
"NO",
"NO",
"NO",
"YES",
"NO",
"YES",
"NO"
]
}
|
IN-CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
const long long inf = LONG_MAX;
const long long arr = 1000000;
bool comp(pair<long long, long long> a, pair<long long, long long> b) {
if (a.first == b.first)
return a.second < b.second;
else
return a.first < b.first;
}
vector<vector<long long> > g(arr);
int main(void) {
long long n;
cin >> n;
for (int i = 0; i < n - 1; i++) {
long long l, r;
cin >> l >> r;
g[l].push_back(r);
g[r].push_back(l);
}
for (long long i = 2; i <= n; i++) {
if (g[i].size() == 2) return cout << "NO" << '\n', 0;
}
cout << "YES" << '\n';
return 0;
}
|
1189_D1. Add on a Tree
|
Note that this is the first problem of the two similar problems. You can hack this problem only if you solve both problems.
You are given a tree with n nodes. In the beginning, 0 is written on all edges. In one operation, you can choose any 2 distinct leaves u, v and any real number x and add x to values written on all edges on the simple path between u and v.
For example, on the picture below you can see the result of applying two operations to the graph: adding 2 on the path from 7 to 6, and then adding -0.5 on the path from 4 to 5.
<image>
Is it true that for any configuration of real numbers written on edges, we can achieve it with a finite number of operations?
Leaf is a node of a tree of degree 1. Simple path is a path that doesn't contain any node twice.
Input
The first line contains a single integer n (2 ≤ n ≤ 10^5) — the number of nodes.
Each of the next n-1 lines contains two integers u and v (1 ≤ u, v ≤ n, u ≠ v), meaning that there is an edge between nodes u and v. It is guaranteed that these edges form a tree.
Output
If there is a configuration of real numbers written on edges of the tree that we can't achieve by performing the operations, output "NO".
Otherwise, output "YES".
You can print each letter in any case (upper or lower).
Examples
Input
2
1 2
Output
YES
Input
3
1 2
2 3
Output
NO
Input
5
1 2
1 3
1 4
2 5
Output
NO
Input
6
1 2
1 3
1 4
2 5
2 6
Output
YES
Note
In the first example, we can add any real x to the value written on the only edge (1, 2).
<image>
In the second example, one of configurations that we can't reach is 0 written on (1, 2) and 1 written on (2, 3).
<image>
Below you can see graphs from examples 3, 4:
<image> <image>
|
{
"input": [
"2\n1 2\n",
"3\n1 2\n2 3\n",
"5\n1 2\n1 3\n1 4\n2 5\n",
"6\n1 2\n1 3\n1 4\n2 5\n2 6\n"
],
"output": [
"YES",
"NO",
"NO",
"YES"
]
}
|
{
"input": [
"50\n16 4\n17 9\n31 19\n22 10\n8 1\n40 30\n3 31\n20 29\n47 27\n22 25\n32 34\n12 15\n40 32\n10 33\n47 12\n6 24\n46 41\n14 23\n12 35\n31 42\n46 28\n31 20\n46 37\n1 39\n29 49\n37 47\n40 6\n42 36\n47 2\n24 46\n2 13\n8 45\n41 3\n32 17\n4 7\n47 26\n28 8\n41 50\n34 44\n33 21\n25 5\n16 40\n3 14\n8 18\n28 11\n32 22\n2 38\n3 48\n44 43\n",
"10\n8 1\n1 2\n8 9\n8 5\n1 3\n1 10\n1 6\n1 7\n8 4\n",
"5\n5 1\n5 4\n4 3\n1 2\n",
"7\n1 2\n2 3\n1 4\n1 5\n3 6\n3 7\n",
"3\n1 3\n2 3\n",
"60\n26 6\n59 30\n31 12\n31 3\n38 23\n59 29\n53 9\n38 56\n53 54\n29 21\n17 55\n59 38\n26 16\n24 59\n24 25\n17 35\n24 41\n30 15\n31 27\n8 44\n26 5\n26 48\n8 32\n53 17\n3 34\n3 51\n30 28\n47 10\n53 60\n36 42\n24 53\n59 22\n53 40\n26 52\n36 4\n59 8\n29 37\n36 20\n17 47\n53 18\n3 50\n30 2\n17 7\n8 58\n59 1\n31 11\n24 26\n24 43\n53 57\n59 45\n47 13\n26 46\n17 33\n30 31\n26 39\n26 19\n24 36\n8 49\n38 14\n",
"7\n1 2\n2 3\n3 4\n3 5\n1 6\n1 7\n",
"20\n19 16\n19 18\n20 7\n9 4\n6 17\n14 2\n9 15\n2 13\n5 11\n19 12\n12 20\n16 9\n11 8\n19 5\n3 1\n19 14\n5 3\n12 10\n19 6\n",
"7\n1 2\n1 3\n2 4\n2 5\n3 6\n3 7\n",
"10\n9 5\n7 1\n9 10\n7 2\n5 4\n9 6\n2 9\n10 8\n1 3\n",
"4\n2 4\n2 3\n2 1\n",
"4\n1 4\n3 2\n1 3\n",
"3\n1 2\n1 3\n",
"5\n1 2\n1 5\n1 3\n1 4\n",
"20\n14 9\n12 13\n10 15\n2 1\n20 19\n16 6\n16 3\n17 14\n3 5\n2 11\n3 10\n15 8\n14 2\n6 4\n3 20\n5 18\n1 7\n1 16\n4 12\n",
"20\n7 5\n14 13\n17 6\n3 8\n16 12\n18 9\n3 18\n14 1\n17 3\n15 2\n17 4\n9 11\n2 7\n15 17\n3 20\n16 10\n17 14\n2 16\n1 19\n",
"8\n1 2\n2 3\n3 4\n1 7\n1 8\n4 5\n4 6\n",
"5\n5 1\n5 2\n5 3\n5 4\n",
"50\n49 6\n43 7\n1 27\n19 35\n15 37\n16 12\n19 21\n16 28\n49 9\n48 39\n13 1\n2 48\n9 50\n44 3\n41 32\n48 31\n49 33\n6 11\n13 20\n49 22\n13 41\n48 29\n13 46\n15 47\n34 2\n49 13\n48 14\n34 24\n16 36\n13 40\n49 34\n49 17\n43 25\n11 23\n10 15\n19 26\n34 44\n16 42\n19 18\n46 8\n29 38\n1 45\n12 43\n13 16\n46 30\n15 5\n49 10\n11 19\n32 4\n",
"20\n13 1\n18 2\n3 7\n18 5\n20 16\n3 12\n18 9\n3 10\n18 11\n13 6\n3 18\n20 15\n20 17\n3 13\n3 4\n13 14\n3 20\n18 8\n3 19\n",
"10\n8 2\n5 6\n1 8\n2 9\n1 4\n8 10\n10 5\n2 7\n2 3\n"
],
"output": [
"NO",
"YES",
"NO",
"NO",
"NO",
"YES",
"NO",
"NO",
"NO",
"NO",
"YES",
"NO",
"NO",
"YES",
"NO",
"NO",
"NO",
"YES",
"NO",
"YES",
"NO"
]
}
|
IN-CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
vector<vector<long long> > g;
void endProg() {
cout << "NO";
exit(0);
}
void dfs(long long v, long long p) {
long long l = 0;
for (auto i : g[v]) {
if (i == p) continue;
if (g[i].size() == 1) l++;
dfs(i, v);
}
if (l == 0 && g[v].size() == 1) return;
if (v == 0 && l == 1 && g[v].size() == 1) return;
if (v == 0 && l <= 2 && g[v].size() <= 2)
endProg();
else if (v != 0 && l <= 1)
endProg();
}
int main() {
long long n;
cin >> n;
g.resize(n);
for (long long i = 0; i < n - 1; i++) {
long long a, b;
cin >> a >> b;
g[a - 1].push_back(b - 1);
g[b - 1].push_back(a - 1);
}
dfs(0, 0);
cout << "YES";
return 0;
}
|
1189_D1. Add on a Tree
|
Note that this is the first problem of the two similar problems. You can hack this problem only if you solve both problems.
You are given a tree with n nodes. In the beginning, 0 is written on all edges. In one operation, you can choose any 2 distinct leaves u, v and any real number x and add x to values written on all edges on the simple path between u and v.
For example, on the picture below you can see the result of applying two operations to the graph: adding 2 on the path from 7 to 6, and then adding -0.5 on the path from 4 to 5.
<image>
Is it true that for any configuration of real numbers written on edges, we can achieve it with a finite number of operations?
Leaf is a node of a tree of degree 1. Simple path is a path that doesn't contain any node twice.
Input
The first line contains a single integer n (2 ≤ n ≤ 10^5) — the number of nodes.
Each of the next n-1 lines contains two integers u and v (1 ≤ u, v ≤ n, u ≠ v), meaning that there is an edge between nodes u and v. It is guaranteed that these edges form a tree.
Output
If there is a configuration of real numbers written on edges of the tree that we can't achieve by performing the operations, output "NO".
Otherwise, output "YES".
You can print each letter in any case (upper or lower).
Examples
Input
2
1 2
Output
YES
Input
3
1 2
2 3
Output
NO
Input
5
1 2
1 3
1 4
2 5
Output
NO
Input
6
1 2
1 3
1 4
2 5
2 6
Output
YES
Note
In the first example, we can add any real x to the value written on the only edge (1, 2).
<image>
In the second example, one of configurations that we can't reach is 0 written on (1, 2) and 1 written on (2, 3).
<image>
Below you can see graphs from examples 3, 4:
<image> <image>
|
{
"input": [
"2\n1 2\n",
"3\n1 2\n2 3\n",
"5\n1 2\n1 3\n1 4\n2 5\n",
"6\n1 2\n1 3\n1 4\n2 5\n2 6\n"
],
"output": [
"YES",
"NO",
"NO",
"YES"
]
}
|
{
"input": [
"50\n16 4\n17 9\n31 19\n22 10\n8 1\n40 30\n3 31\n20 29\n47 27\n22 25\n32 34\n12 15\n40 32\n10 33\n47 12\n6 24\n46 41\n14 23\n12 35\n31 42\n46 28\n31 20\n46 37\n1 39\n29 49\n37 47\n40 6\n42 36\n47 2\n24 46\n2 13\n8 45\n41 3\n32 17\n4 7\n47 26\n28 8\n41 50\n34 44\n33 21\n25 5\n16 40\n3 14\n8 18\n28 11\n32 22\n2 38\n3 48\n44 43\n",
"10\n8 1\n1 2\n8 9\n8 5\n1 3\n1 10\n1 6\n1 7\n8 4\n",
"5\n5 1\n5 4\n4 3\n1 2\n",
"7\n1 2\n2 3\n1 4\n1 5\n3 6\n3 7\n",
"3\n1 3\n2 3\n",
"60\n26 6\n59 30\n31 12\n31 3\n38 23\n59 29\n53 9\n38 56\n53 54\n29 21\n17 55\n59 38\n26 16\n24 59\n24 25\n17 35\n24 41\n30 15\n31 27\n8 44\n26 5\n26 48\n8 32\n53 17\n3 34\n3 51\n30 28\n47 10\n53 60\n36 42\n24 53\n59 22\n53 40\n26 52\n36 4\n59 8\n29 37\n36 20\n17 47\n53 18\n3 50\n30 2\n17 7\n8 58\n59 1\n31 11\n24 26\n24 43\n53 57\n59 45\n47 13\n26 46\n17 33\n30 31\n26 39\n26 19\n24 36\n8 49\n38 14\n",
"7\n1 2\n2 3\n3 4\n3 5\n1 6\n1 7\n",
"20\n19 16\n19 18\n20 7\n9 4\n6 17\n14 2\n9 15\n2 13\n5 11\n19 12\n12 20\n16 9\n11 8\n19 5\n3 1\n19 14\n5 3\n12 10\n19 6\n",
"7\n1 2\n1 3\n2 4\n2 5\n3 6\n3 7\n",
"10\n9 5\n7 1\n9 10\n7 2\n5 4\n9 6\n2 9\n10 8\n1 3\n",
"4\n2 4\n2 3\n2 1\n",
"4\n1 4\n3 2\n1 3\n",
"3\n1 2\n1 3\n",
"5\n1 2\n1 5\n1 3\n1 4\n",
"20\n14 9\n12 13\n10 15\n2 1\n20 19\n16 6\n16 3\n17 14\n3 5\n2 11\n3 10\n15 8\n14 2\n6 4\n3 20\n5 18\n1 7\n1 16\n4 12\n",
"20\n7 5\n14 13\n17 6\n3 8\n16 12\n18 9\n3 18\n14 1\n17 3\n15 2\n17 4\n9 11\n2 7\n15 17\n3 20\n16 10\n17 14\n2 16\n1 19\n",
"8\n1 2\n2 3\n3 4\n1 7\n1 8\n4 5\n4 6\n",
"5\n5 1\n5 2\n5 3\n5 4\n",
"50\n49 6\n43 7\n1 27\n19 35\n15 37\n16 12\n19 21\n16 28\n49 9\n48 39\n13 1\n2 48\n9 50\n44 3\n41 32\n48 31\n49 33\n6 11\n13 20\n49 22\n13 41\n48 29\n13 46\n15 47\n34 2\n49 13\n48 14\n34 24\n16 36\n13 40\n49 34\n49 17\n43 25\n11 23\n10 15\n19 26\n34 44\n16 42\n19 18\n46 8\n29 38\n1 45\n12 43\n13 16\n46 30\n15 5\n49 10\n11 19\n32 4\n",
"20\n13 1\n18 2\n3 7\n18 5\n20 16\n3 12\n18 9\n3 10\n18 11\n13 6\n3 18\n20 15\n20 17\n3 13\n3 4\n13 14\n3 20\n18 8\n3 19\n",
"10\n8 2\n5 6\n1 8\n2 9\n1 4\n8 10\n10 5\n2 7\n2 3\n"
],
"output": [
"NO",
"YES",
"NO",
"NO",
"NO",
"YES",
"NO",
"NO",
"NO",
"NO",
"YES",
"NO",
"NO",
"YES",
"NO",
"NO",
"NO",
"YES",
"NO",
"YES",
"NO"
]
}
|
IN-CORRECT
|
cpp
|
#include <bits/stdc++.h>
#pragma GCC optimize("Ofast")
#pragma GCC target("sse4")
const long double EPS = 1e-9;
const long long INF = 1e9;
const long long mod = 1e9 + 7;
using namespace std;
mt19937 rnd(chrono::high_resolution_clock::now().time_since_epoch().count());
long long n, m, k;
long long q;
vector<vector<long long>> vec;
vector<bool> used;
vector<long long> cnt;
vector<long long> ans;
vector<long long> pref;
int main() {
ios::sync_with_stdio(false);
cin.tie(nullptr);
cin >> n;
vec.resize(n);
cnt.resize(n, 0);
for (int i = 0; i < n - 1; ++i) {
int v, to;
cin >> v >> to;
--v;
--to;
vec[v].push_back(to);
vec[to].push_back(v);
cnt[v]++;
cnt[to]++;
}
int cntl = 0, cntnl = 0;
for (int i = 0; i < n; ++i) {
if (cnt[i] == 1)
cntl++;
else {
int cntal = 0;
for (int j = 0; j < vec[i].size(); ++j) {
if (cnt[vec[i][j]] == 1) {
cntal++;
}
}
if (cntal <= 1) {
cntnl++;
}
}
}
if ((cntl == 2 && n > 2) || cntnl) {
cout << "NO";
} else {
cout << "YES";
}
return 0;
}
|
1189_D1. Add on a Tree
|
Note that this is the first problem of the two similar problems. You can hack this problem only if you solve both problems.
You are given a tree with n nodes. In the beginning, 0 is written on all edges. In one operation, you can choose any 2 distinct leaves u, v and any real number x and add x to values written on all edges on the simple path between u and v.
For example, on the picture below you can see the result of applying two operations to the graph: adding 2 on the path from 7 to 6, and then adding -0.5 on the path from 4 to 5.
<image>
Is it true that for any configuration of real numbers written on edges, we can achieve it with a finite number of operations?
Leaf is a node of a tree of degree 1. Simple path is a path that doesn't contain any node twice.
Input
The first line contains a single integer n (2 ≤ n ≤ 10^5) — the number of nodes.
Each of the next n-1 lines contains two integers u and v (1 ≤ u, v ≤ n, u ≠ v), meaning that there is an edge between nodes u and v. It is guaranteed that these edges form a tree.
Output
If there is a configuration of real numbers written on edges of the tree that we can't achieve by performing the operations, output "NO".
Otherwise, output "YES".
You can print each letter in any case (upper or lower).
Examples
Input
2
1 2
Output
YES
Input
3
1 2
2 3
Output
NO
Input
5
1 2
1 3
1 4
2 5
Output
NO
Input
6
1 2
1 3
1 4
2 5
2 6
Output
YES
Note
In the first example, we can add any real x to the value written on the only edge (1, 2).
<image>
In the second example, one of configurations that we can't reach is 0 written on (1, 2) and 1 written on (2, 3).
<image>
Below you can see graphs from examples 3, 4:
<image> <image>
|
{
"input": [
"2\n1 2\n",
"3\n1 2\n2 3\n",
"5\n1 2\n1 3\n1 4\n2 5\n",
"6\n1 2\n1 3\n1 4\n2 5\n2 6\n"
],
"output": [
"YES",
"NO",
"NO",
"YES"
]
}
|
{
"input": [
"50\n16 4\n17 9\n31 19\n22 10\n8 1\n40 30\n3 31\n20 29\n47 27\n22 25\n32 34\n12 15\n40 32\n10 33\n47 12\n6 24\n46 41\n14 23\n12 35\n31 42\n46 28\n31 20\n46 37\n1 39\n29 49\n37 47\n40 6\n42 36\n47 2\n24 46\n2 13\n8 45\n41 3\n32 17\n4 7\n47 26\n28 8\n41 50\n34 44\n33 21\n25 5\n16 40\n3 14\n8 18\n28 11\n32 22\n2 38\n3 48\n44 43\n",
"10\n8 1\n1 2\n8 9\n8 5\n1 3\n1 10\n1 6\n1 7\n8 4\n",
"5\n5 1\n5 4\n4 3\n1 2\n",
"7\n1 2\n2 3\n1 4\n1 5\n3 6\n3 7\n",
"3\n1 3\n2 3\n",
"60\n26 6\n59 30\n31 12\n31 3\n38 23\n59 29\n53 9\n38 56\n53 54\n29 21\n17 55\n59 38\n26 16\n24 59\n24 25\n17 35\n24 41\n30 15\n31 27\n8 44\n26 5\n26 48\n8 32\n53 17\n3 34\n3 51\n30 28\n47 10\n53 60\n36 42\n24 53\n59 22\n53 40\n26 52\n36 4\n59 8\n29 37\n36 20\n17 47\n53 18\n3 50\n30 2\n17 7\n8 58\n59 1\n31 11\n24 26\n24 43\n53 57\n59 45\n47 13\n26 46\n17 33\n30 31\n26 39\n26 19\n24 36\n8 49\n38 14\n",
"7\n1 2\n2 3\n3 4\n3 5\n1 6\n1 7\n",
"20\n19 16\n19 18\n20 7\n9 4\n6 17\n14 2\n9 15\n2 13\n5 11\n19 12\n12 20\n16 9\n11 8\n19 5\n3 1\n19 14\n5 3\n12 10\n19 6\n",
"7\n1 2\n1 3\n2 4\n2 5\n3 6\n3 7\n",
"10\n9 5\n7 1\n9 10\n7 2\n5 4\n9 6\n2 9\n10 8\n1 3\n",
"4\n2 4\n2 3\n2 1\n",
"4\n1 4\n3 2\n1 3\n",
"3\n1 2\n1 3\n",
"5\n1 2\n1 5\n1 3\n1 4\n",
"20\n14 9\n12 13\n10 15\n2 1\n20 19\n16 6\n16 3\n17 14\n3 5\n2 11\n3 10\n15 8\n14 2\n6 4\n3 20\n5 18\n1 7\n1 16\n4 12\n",
"20\n7 5\n14 13\n17 6\n3 8\n16 12\n18 9\n3 18\n14 1\n17 3\n15 2\n17 4\n9 11\n2 7\n15 17\n3 20\n16 10\n17 14\n2 16\n1 19\n",
"8\n1 2\n2 3\n3 4\n1 7\n1 8\n4 5\n4 6\n",
"5\n5 1\n5 2\n5 3\n5 4\n",
"50\n49 6\n43 7\n1 27\n19 35\n15 37\n16 12\n19 21\n16 28\n49 9\n48 39\n13 1\n2 48\n9 50\n44 3\n41 32\n48 31\n49 33\n6 11\n13 20\n49 22\n13 41\n48 29\n13 46\n15 47\n34 2\n49 13\n48 14\n34 24\n16 36\n13 40\n49 34\n49 17\n43 25\n11 23\n10 15\n19 26\n34 44\n16 42\n19 18\n46 8\n29 38\n1 45\n12 43\n13 16\n46 30\n15 5\n49 10\n11 19\n32 4\n",
"20\n13 1\n18 2\n3 7\n18 5\n20 16\n3 12\n18 9\n3 10\n18 11\n13 6\n3 18\n20 15\n20 17\n3 13\n3 4\n13 14\n3 20\n18 8\n3 19\n",
"10\n8 2\n5 6\n1 8\n2 9\n1 4\n8 10\n10 5\n2 7\n2 3\n"
],
"output": [
"NO",
"YES",
"NO",
"NO",
"NO",
"YES",
"NO",
"NO",
"NO",
"NO",
"YES",
"NO",
"NO",
"YES",
"NO",
"NO",
"NO",
"YES",
"NO",
"YES",
"NO"
]
}
|
IN-CORRECT
|
java
|
import java.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.StringTokenizer;
import java.io.IOException;
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.InputStream;
/**
* Built using CHelper plug-in
* Actual solution is at the top
*/
public class Main {
public static void main(String[] args) {
InputStream inputStream = System.in;
OutputStream outputStream = System.out;
InputReader sc = new InputReader(inputStream);
PrintWriter out = new PrintWriter(outputStream);
Task solver = new Task();
solver.solve(1, sc, out);
out.close();
}
static class Task {
public void solve(int testNumber, InputReader sc, PrintWriter out) {
int n=sc.nextInt();
ArrayList<Integer>[] map=new ArrayList[n+1];
for(int i=1;i<=n;i++)
map[i]=new ArrayList<Integer>();
for(int i=0;i<n-1;i++) {
int u=sc.nextInt();
int v=sc.nextInt();
map[u].add(v);
}
boolean[] jud=new boolean[n+1];
for(int i=1;i<=n;i++) {
if(map[i].size()==0)
jud[i]=true;
}
for(int i=1;i<=n;i++) {
if(!jud[i]&&map[i].size()<2) {
out.println("NO");
return ;
}
}
out.println("YES");
}
}
static class InputReader {
public BufferedReader reader;
public StringTokenizer tokenizer;
public InputReader(InputStream stream) {
reader = new BufferedReader(new InputStreamReader(stream), 32768);
tokenizer = null;
}
public String next() {
while (tokenizer == null || !tokenizer.hasMoreTokens()) {
try {
tokenizer = new StringTokenizer(reader.readLine());
} catch (IOException e) {
throw new RuntimeException(e);
}
}
return tokenizer.nextToken();
}
public int nextInt() {
return Integer.parseInt(next());
}
public long nextLong() {
return Long.parseLong(next());
}
}
}
|
1189_D1. Add on a Tree
|
Note that this is the first problem of the two similar problems. You can hack this problem only if you solve both problems.
You are given a tree with n nodes. In the beginning, 0 is written on all edges. In one operation, you can choose any 2 distinct leaves u, v and any real number x and add x to values written on all edges on the simple path between u and v.
For example, on the picture below you can see the result of applying two operations to the graph: adding 2 on the path from 7 to 6, and then adding -0.5 on the path from 4 to 5.
<image>
Is it true that for any configuration of real numbers written on edges, we can achieve it with a finite number of operations?
Leaf is a node of a tree of degree 1. Simple path is a path that doesn't contain any node twice.
Input
The first line contains a single integer n (2 ≤ n ≤ 10^5) — the number of nodes.
Each of the next n-1 lines contains two integers u and v (1 ≤ u, v ≤ n, u ≠ v), meaning that there is an edge between nodes u and v. It is guaranteed that these edges form a tree.
Output
If there is a configuration of real numbers written on edges of the tree that we can't achieve by performing the operations, output "NO".
Otherwise, output "YES".
You can print each letter in any case (upper or lower).
Examples
Input
2
1 2
Output
YES
Input
3
1 2
2 3
Output
NO
Input
5
1 2
1 3
1 4
2 5
Output
NO
Input
6
1 2
1 3
1 4
2 5
2 6
Output
YES
Note
In the first example, we can add any real x to the value written on the only edge (1, 2).
<image>
In the second example, one of configurations that we can't reach is 0 written on (1, 2) and 1 written on (2, 3).
<image>
Below you can see graphs from examples 3, 4:
<image> <image>
|
{
"input": [
"2\n1 2\n",
"3\n1 2\n2 3\n",
"5\n1 2\n1 3\n1 4\n2 5\n",
"6\n1 2\n1 3\n1 4\n2 5\n2 6\n"
],
"output": [
"YES",
"NO",
"NO",
"YES"
]
}
|
{
"input": [
"50\n16 4\n17 9\n31 19\n22 10\n8 1\n40 30\n3 31\n20 29\n47 27\n22 25\n32 34\n12 15\n40 32\n10 33\n47 12\n6 24\n46 41\n14 23\n12 35\n31 42\n46 28\n31 20\n46 37\n1 39\n29 49\n37 47\n40 6\n42 36\n47 2\n24 46\n2 13\n8 45\n41 3\n32 17\n4 7\n47 26\n28 8\n41 50\n34 44\n33 21\n25 5\n16 40\n3 14\n8 18\n28 11\n32 22\n2 38\n3 48\n44 43\n",
"10\n8 1\n1 2\n8 9\n8 5\n1 3\n1 10\n1 6\n1 7\n8 4\n",
"5\n5 1\n5 4\n4 3\n1 2\n",
"7\n1 2\n2 3\n1 4\n1 5\n3 6\n3 7\n",
"3\n1 3\n2 3\n",
"60\n26 6\n59 30\n31 12\n31 3\n38 23\n59 29\n53 9\n38 56\n53 54\n29 21\n17 55\n59 38\n26 16\n24 59\n24 25\n17 35\n24 41\n30 15\n31 27\n8 44\n26 5\n26 48\n8 32\n53 17\n3 34\n3 51\n30 28\n47 10\n53 60\n36 42\n24 53\n59 22\n53 40\n26 52\n36 4\n59 8\n29 37\n36 20\n17 47\n53 18\n3 50\n30 2\n17 7\n8 58\n59 1\n31 11\n24 26\n24 43\n53 57\n59 45\n47 13\n26 46\n17 33\n30 31\n26 39\n26 19\n24 36\n8 49\n38 14\n",
"7\n1 2\n2 3\n3 4\n3 5\n1 6\n1 7\n",
"20\n19 16\n19 18\n20 7\n9 4\n6 17\n14 2\n9 15\n2 13\n5 11\n19 12\n12 20\n16 9\n11 8\n19 5\n3 1\n19 14\n5 3\n12 10\n19 6\n",
"7\n1 2\n1 3\n2 4\n2 5\n3 6\n3 7\n",
"10\n9 5\n7 1\n9 10\n7 2\n5 4\n9 6\n2 9\n10 8\n1 3\n",
"4\n2 4\n2 3\n2 1\n",
"4\n1 4\n3 2\n1 3\n",
"3\n1 2\n1 3\n",
"5\n1 2\n1 5\n1 3\n1 4\n",
"20\n14 9\n12 13\n10 15\n2 1\n20 19\n16 6\n16 3\n17 14\n3 5\n2 11\n3 10\n15 8\n14 2\n6 4\n3 20\n5 18\n1 7\n1 16\n4 12\n",
"20\n7 5\n14 13\n17 6\n3 8\n16 12\n18 9\n3 18\n14 1\n17 3\n15 2\n17 4\n9 11\n2 7\n15 17\n3 20\n16 10\n17 14\n2 16\n1 19\n",
"8\n1 2\n2 3\n3 4\n1 7\n1 8\n4 5\n4 6\n",
"5\n5 1\n5 2\n5 3\n5 4\n",
"50\n49 6\n43 7\n1 27\n19 35\n15 37\n16 12\n19 21\n16 28\n49 9\n48 39\n13 1\n2 48\n9 50\n44 3\n41 32\n48 31\n49 33\n6 11\n13 20\n49 22\n13 41\n48 29\n13 46\n15 47\n34 2\n49 13\n48 14\n34 24\n16 36\n13 40\n49 34\n49 17\n43 25\n11 23\n10 15\n19 26\n34 44\n16 42\n19 18\n46 8\n29 38\n1 45\n12 43\n13 16\n46 30\n15 5\n49 10\n11 19\n32 4\n",
"20\n13 1\n18 2\n3 7\n18 5\n20 16\n3 12\n18 9\n3 10\n18 11\n13 6\n3 18\n20 15\n20 17\n3 13\n3 4\n13 14\n3 20\n18 8\n3 19\n",
"10\n8 2\n5 6\n1 8\n2 9\n1 4\n8 10\n10 5\n2 7\n2 3\n"
],
"output": [
"NO",
"YES",
"NO",
"NO",
"NO",
"YES",
"NO",
"NO",
"NO",
"NO",
"YES",
"NO",
"NO",
"YES",
"NO",
"NO",
"NO",
"YES",
"NO",
"YES",
"NO"
]
}
|
IN-CORRECT
|
python3
|
n = int(input())
tr = [0] * (n + 1)
for i in range(n - 1):
u, v = map(int, input().split())
tr[u] += 1
tr[v] += 1
if tr[u] == 2 or tr[v] == 2:
print('No')
break
else:
print('Yes')
|
1189_D1. Add on a Tree
|
Note that this is the first problem of the two similar problems. You can hack this problem only if you solve both problems.
You are given a tree with n nodes. In the beginning, 0 is written on all edges. In one operation, you can choose any 2 distinct leaves u, v and any real number x and add x to values written on all edges on the simple path between u and v.
For example, on the picture below you can see the result of applying two operations to the graph: adding 2 on the path from 7 to 6, and then adding -0.5 on the path from 4 to 5.
<image>
Is it true that for any configuration of real numbers written on edges, we can achieve it with a finite number of operations?
Leaf is a node of a tree of degree 1. Simple path is a path that doesn't contain any node twice.
Input
The first line contains a single integer n (2 ≤ n ≤ 10^5) — the number of nodes.
Each of the next n-1 lines contains two integers u and v (1 ≤ u, v ≤ n, u ≠ v), meaning that there is an edge between nodes u and v. It is guaranteed that these edges form a tree.
Output
If there is a configuration of real numbers written on edges of the tree that we can't achieve by performing the operations, output "NO".
Otherwise, output "YES".
You can print each letter in any case (upper or lower).
Examples
Input
2
1 2
Output
YES
Input
3
1 2
2 3
Output
NO
Input
5
1 2
1 3
1 4
2 5
Output
NO
Input
6
1 2
1 3
1 4
2 5
2 6
Output
YES
Note
In the first example, we can add any real x to the value written on the only edge (1, 2).
<image>
In the second example, one of configurations that we can't reach is 0 written on (1, 2) and 1 written on (2, 3).
<image>
Below you can see graphs from examples 3, 4:
<image> <image>
|
{
"input": [
"2\n1 2\n",
"3\n1 2\n2 3\n",
"5\n1 2\n1 3\n1 4\n2 5\n",
"6\n1 2\n1 3\n1 4\n2 5\n2 6\n"
],
"output": [
"YES",
"NO",
"NO",
"YES"
]
}
|
{
"input": [
"50\n16 4\n17 9\n31 19\n22 10\n8 1\n40 30\n3 31\n20 29\n47 27\n22 25\n32 34\n12 15\n40 32\n10 33\n47 12\n6 24\n46 41\n14 23\n12 35\n31 42\n46 28\n31 20\n46 37\n1 39\n29 49\n37 47\n40 6\n42 36\n47 2\n24 46\n2 13\n8 45\n41 3\n32 17\n4 7\n47 26\n28 8\n41 50\n34 44\n33 21\n25 5\n16 40\n3 14\n8 18\n28 11\n32 22\n2 38\n3 48\n44 43\n",
"10\n8 1\n1 2\n8 9\n8 5\n1 3\n1 10\n1 6\n1 7\n8 4\n",
"5\n5 1\n5 4\n4 3\n1 2\n",
"7\n1 2\n2 3\n1 4\n1 5\n3 6\n3 7\n",
"3\n1 3\n2 3\n",
"60\n26 6\n59 30\n31 12\n31 3\n38 23\n59 29\n53 9\n38 56\n53 54\n29 21\n17 55\n59 38\n26 16\n24 59\n24 25\n17 35\n24 41\n30 15\n31 27\n8 44\n26 5\n26 48\n8 32\n53 17\n3 34\n3 51\n30 28\n47 10\n53 60\n36 42\n24 53\n59 22\n53 40\n26 52\n36 4\n59 8\n29 37\n36 20\n17 47\n53 18\n3 50\n30 2\n17 7\n8 58\n59 1\n31 11\n24 26\n24 43\n53 57\n59 45\n47 13\n26 46\n17 33\n30 31\n26 39\n26 19\n24 36\n8 49\n38 14\n",
"7\n1 2\n2 3\n3 4\n3 5\n1 6\n1 7\n",
"20\n19 16\n19 18\n20 7\n9 4\n6 17\n14 2\n9 15\n2 13\n5 11\n19 12\n12 20\n16 9\n11 8\n19 5\n3 1\n19 14\n5 3\n12 10\n19 6\n",
"7\n1 2\n1 3\n2 4\n2 5\n3 6\n3 7\n",
"10\n9 5\n7 1\n9 10\n7 2\n5 4\n9 6\n2 9\n10 8\n1 3\n",
"4\n2 4\n2 3\n2 1\n",
"4\n1 4\n3 2\n1 3\n",
"3\n1 2\n1 3\n",
"5\n1 2\n1 5\n1 3\n1 4\n",
"20\n14 9\n12 13\n10 15\n2 1\n20 19\n16 6\n16 3\n17 14\n3 5\n2 11\n3 10\n15 8\n14 2\n6 4\n3 20\n5 18\n1 7\n1 16\n4 12\n",
"20\n7 5\n14 13\n17 6\n3 8\n16 12\n18 9\n3 18\n14 1\n17 3\n15 2\n17 4\n9 11\n2 7\n15 17\n3 20\n16 10\n17 14\n2 16\n1 19\n",
"8\n1 2\n2 3\n3 4\n1 7\n1 8\n4 5\n4 6\n",
"5\n5 1\n5 2\n5 3\n5 4\n",
"50\n49 6\n43 7\n1 27\n19 35\n15 37\n16 12\n19 21\n16 28\n49 9\n48 39\n13 1\n2 48\n9 50\n44 3\n41 32\n48 31\n49 33\n6 11\n13 20\n49 22\n13 41\n48 29\n13 46\n15 47\n34 2\n49 13\n48 14\n34 24\n16 36\n13 40\n49 34\n49 17\n43 25\n11 23\n10 15\n19 26\n34 44\n16 42\n19 18\n46 8\n29 38\n1 45\n12 43\n13 16\n46 30\n15 5\n49 10\n11 19\n32 4\n",
"20\n13 1\n18 2\n3 7\n18 5\n20 16\n3 12\n18 9\n3 10\n18 11\n13 6\n3 18\n20 15\n20 17\n3 13\n3 4\n13 14\n3 20\n18 8\n3 19\n",
"10\n8 2\n5 6\n1 8\n2 9\n1 4\n8 10\n10 5\n2 7\n2 3\n"
],
"output": [
"NO",
"YES",
"NO",
"NO",
"NO",
"YES",
"NO",
"NO",
"NO",
"NO",
"YES",
"NO",
"NO",
"YES",
"NO",
"NO",
"NO",
"YES",
"NO",
"YES",
"NO"
]
}
|
IN-CORRECT
|
java
|
import java.io.*;
import java.util.*;
public class Main {
public static void main(String[] args) throws IOException {
InputStream inputStream = System.in;
OutputStream outputStream = System.out;
InputReader in = new InputReader(inputStream);
PrintWriter out = new PrintWriter(outputStream);
// InputReader in = new InputReader(".in");
// PrintWriter out = new PrintWriter(new FileWriter(".out"));
Task solver = new Task();
solver.solve(1, in, out);
out.close();
}
static class Task {
public void solve(int testNumber, InputReader in, PrintWriter out) {
int n = in.nextInt();
int[] a = new int[n];
int[] b = new int[n];
for (int i = 0; i < n - 1; i++) {
a[in.nextInt() - 1]++;
b[in.nextInt() - 1]++;
}
for (int i = 0; i < n; i++) {
if (a[i] == 1 && b[i] == 1) {
out.println("NO");
return;
}
}
out.println("YES");
}
}
static class InputReader {
public BufferedReader reader;
public StringTokenizer tokenizer;
public InputReader(InputStream stream) {
reader = new BufferedReader(new InputStreamReader(stream), 32768);
tokenizer = null;
}
public InputReader(String s) {
try {
reader = new BufferedReader(new FileReader(s));
} catch (FileNotFoundException e) {
e.printStackTrace();
}
}
public String nextToken() {
while (tokenizer == null || !tokenizer.hasMoreTokens()) {
try {
tokenizer = new StringTokenizer(reader.readLine());
} catch (IOException e) {
throw new RuntimeException(e);
}
}
return tokenizer.nextToken();
}
public int nextInt() {
return Integer.parseInt(nextToken());
}
public long nextLong() {
return Long.parseLong(nextToken());
}
public double nextDouble() {
return Double.parseDouble(nextToken());
}
}
}
|
1189_D1. Add on a Tree
|
Note that this is the first problem of the two similar problems. You can hack this problem only if you solve both problems.
You are given a tree with n nodes. In the beginning, 0 is written on all edges. In one operation, you can choose any 2 distinct leaves u, v and any real number x and add x to values written on all edges on the simple path between u and v.
For example, on the picture below you can see the result of applying two operations to the graph: adding 2 on the path from 7 to 6, and then adding -0.5 on the path from 4 to 5.
<image>
Is it true that for any configuration of real numbers written on edges, we can achieve it with a finite number of operations?
Leaf is a node of a tree of degree 1. Simple path is a path that doesn't contain any node twice.
Input
The first line contains a single integer n (2 ≤ n ≤ 10^5) — the number of nodes.
Each of the next n-1 lines contains two integers u and v (1 ≤ u, v ≤ n, u ≠ v), meaning that there is an edge between nodes u and v. It is guaranteed that these edges form a tree.
Output
If there is a configuration of real numbers written on edges of the tree that we can't achieve by performing the operations, output "NO".
Otherwise, output "YES".
You can print each letter in any case (upper or lower).
Examples
Input
2
1 2
Output
YES
Input
3
1 2
2 3
Output
NO
Input
5
1 2
1 3
1 4
2 5
Output
NO
Input
6
1 2
1 3
1 4
2 5
2 6
Output
YES
Note
In the first example, we can add any real x to the value written on the only edge (1, 2).
<image>
In the second example, one of configurations that we can't reach is 0 written on (1, 2) and 1 written on (2, 3).
<image>
Below you can see graphs from examples 3, 4:
<image> <image>
|
{
"input": [
"2\n1 2\n",
"3\n1 2\n2 3\n",
"5\n1 2\n1 3\n1 4\n2 5\n",
"6\n1 2\n1 3\n1 4\n2 5\n2 6\n"
],
"output": [
"YES",
"NO",
"NO",
"YES"
]
}
|
{
"input": [
"50\n16 4\n17 9\n31 19\n22 10\n8 1\n40 30\n3 31\n20 29\n47 27\n22 25\n32 34\n12 15\n40 32\n10 33\n47 12\n6 24\n46 41\n14 23\n12 35\n31 42\n46 28\n31 20\n46 37\n1 39\n29 49\n37 47\n40 6\n42 36\n47 2\n24 46\n2 13\n8 45\n41 3\n32 17\n4 7\n47 26\n28 8\n41 50\n34 44\n33 21\n25 5\n16 40\n3 14\n8 18\n28 11\n32 22\n2 38\n3 48\n44 43\n",
"10\n8 1\n1 2\n8 9\n8 5\n1 3\n1 10\n1 6\n1 7\n8 4\n",
"5\n5 1\n5 4\n4 3\n1 2\n",
"7\n1 2\n2 3\n1 4\n1 5\n3 6\n3 7\n",
"3\n1 3\n2 3\n",
"60\n26 6\n59 30\n31 12\n31 3\n38 23\n59 29\n53 9\n38 56\n53 54\n29 21\n17 55\n59 38\n26 16\n24 59\n24 25\n17 35\n24 41\n30 15\n31 27\n8 44\n26 5\n26 48\n8 32\n53 17\n3 34\n3 51\n30 28\n47 10\n53 60\n36 42\n24 53\n59 22\n53 40\n26 52\n36 4\n59 8\n29 37\n36 20\n17 47\n53 18\n3 50\n30 2\n17 7\n8 58\n59 1\n31 11\n24 26\n24 43\n53 57\n59 45\n47 13\n26 46\n17 33\n30 31\n26 39\n26 19\n24 36\n8 49\n38 14\n",
"7\n1 2\n2 3\n3 4\n3 5\n1 6\n1 7\n",
"20\n19 16\n19 18\n20 7\n9 4\n6 17\n14 2\n9 15\n2 13\n5 11\n19 12\n12 20\n16 9\n11 8\n19 5\n3 1\n19 14\n5 3\n12 10\n19 6\n",
"7\n1 2\n1 3\n2 4\n2 5\n3 6\n3 7\n",
"10\n9 5\n7 1\n9 10\n7 2\n5 4\n9 6\n2 9\n10 8\n1 3\n",
"4\n2 4\n2 3\n2 1\n",
"4\n1 4\n3 2\n1 3\n",
"3\n1 2\n1 3\n",
"5\n1 2\n1 5\n1 3\n1 4\n",
"20\n14 9\n12 13\n10 15\n2 1\n20 19\n16 6\n16 3\n17 14\n3 5\n2 11\n3 10\n15 8\n14 2\n6 4\n3 20\n5 18\n1 7\n1 16\n4 12\n",
"20\n7 5\n14 13\n17 6\n3 8\n16 12\n18 9\n3 18\n14 1\n17 3\n15 2\n17 4\n9 11\n2 7\n15 17\n3 20\n16 10\n17 14\n2 16\n1 19\n",
"8\n1 2\n2 3\n3 4\n1 7\n1 8\n4 5\n4 6\n",
"5\n5 1\n5 2\n5 3\n5 4\n",
"50\n49 6\n43 7\n1 27\n19 35\n15 37\n16 12\n19 21\n16 28\n49 9\n48 39\n13 1\n2 48\n9 50\n44 3\n41 32\n48 31\n49 33\n6 11\n13 20\n49 22\n13 41\n48 29\n13 46\n15 47\n34 2\n49 13\n48 14\n34 24\n16 36\n13 40\n49 34\n49 17\n43 25\n11 23\n10 15\n19 26\n34 44\n16 42\n19 18\n46 8\n29 38\n1 45\n12 43\n13 16\n46 30\n15 5\n49 10\n11 19\n32 4\n",
"20\n13 1\n18 2\n3 7\n18 5\n20 16\n3 12\n18 9\n3 10\n18 11\n13 6\n3 18\n20 15\n20 17\n3 13\n3 4\n13 14\n3 20\n18 8\n3 19\n",
"10\n8 2\n5 6\n1 8\n2 9\n1 4\n8 10\n10 5\n2 7\n2 3\n"
],
"output": [
"NO",
"YES",
"NO",
"NO",
"NO",
"YES",
"NO",
"NO",
"NO",
"NO",
"YES",
"NO",
"NO",
"YES",
"NO",
"NO",
"NO",
"YES",
"NO",
"YES",
"NO"
]
}
|
IN-CORRECT
|
cpp
|
#include <bits/stdc++.h>
#pragma GCC optimize("Ofast")
using namespace std;
long int mod = 1e9 + 7;
vector<long int> v[100500];
bool vis[100500];
long int n, m, t, c, re, from[100500];
void bfs(long int s) {
queue<long int> q;
long int ans = 0;
q.push(s);
vis[s] = 1;
while (!q.empty()) {
ans = 0;
long int u = q.front();
q.pop();
for (auto i : v[u]) {
if (!vis[i]) {
from[i] = u;
vis[i] = 1;
q.push(i);
ans++;
}
}
if (!ans) re++;
}
}
int main() {
ios::sync_with_stdio(0);
cin.tie(0);
cin >> n;
for (long int i = 1; i < n; i++) {
long int x, y;
cin >> x >> y;
v[x].push_back(y);
v[y].push_back(x);
}
for (long int i = 1; i < n + 1; i++)
if (!vis[i]) bfs(i);
for (long int i = 2; i < n + 1; i++)
if (from[i] == 1) t++;
if (t == n - 1) {
if (n % 2 == 0)
cout << "YES";
else
cout << "NO";
return 0;
}
if (re % 2)
cout << "NO";
else
cout << "YES";
return 0;
}
|
1189_D1. Add on a Tree
|
Note that this is the first problem of the two similar problems. You can hack this problem only if you solve both problems.
You are given a tree with n nodes. In the beginning, 0 is written on all edges. In one operation, you can choose any 2 distinct leaves u, v and any real number x and add x to values written on all edges on the simple path between u and v.
For example, on the picture below you can see the result of applying two operations to the graph: adding 2 on the path from 7 to 6, and then adding -0.5 on the path from 4 to 5.
<image>
Is it true that for any configuration of real numbers written on edges, we can achieve it with a finite number of operations?
Leaf is a node of a tree of degree 1. Simple path is a path that doesn't contain any node twice.
Input
The first line contains a single integer n (2 ≤ n ≤ 10^5) — the number of nodes.
Each of the next n-1 lines contains two integers u and v (1 ≤ u, v ≤ n, u ≠ v), meaning that there is an edge between nodes u and v. It is guaranteed that these edges form a tree.
Output
If there is a configuration of real numbers written on edges of the tree that we can't achieve by performing the operations, output "NO".
Otherwise, output "YES".
You can print each letter in any case (upper or lower).
Examples
Input
2
1 2
Output
YES
Input
3
1 2
2 3
Output
NO
Input
5
1 2
1 3
1 4
2 5
Output
NO
Input
6
1 2
1 3
1 4
2 5
2 6
Output
YES
Note
In the first example, we can add any real x to the value written on the only edge (1, 2).
<image>
In the second example, one of configurations that we can't reach is 0 written on (1, 2) and 1 written on (2, 3).
<image>
Below you can see graphs from examples 3, 4:
<image> <image>
|
{
"input": [
"2\n1 2\n",
"3\n1 2\n2 3\n",
"5\n1 2\n1 3\n1 4\n2 5\n",
"6\n1 2\n1 3\n1 4\n2 5\n2 6\n"
],
"output": [
"YES",
"NO",
"NO",
"YES"
]
}
|
{
"input": [
"50\n16 4\n17 9\n31 19\n22 10\n8 1\n40 30\n3 31\n20 29\n47 27\n22 25\n32 34\n12 15\n40 32\n10 33\n47 12\n6 24\n46 41\n14 23\n12 35\n31 42\n46 28\n31 20\n46 37\n1 39\n29 49\n37 47\n40 6\n42 36\n47 2\n24 46\n2 13\n8 45\n41 3\n32 17\n4 7\n47 26\n28 8\n41 50\n34 44\n33 21\n25 5\n16 40\n3 14\n8 18\n28 11\n32 22\n2 38\n3 48\n44 43\n",
"10\n8 1\n1 2\n8 9\n8 5\n1 3\n1 10\n1 6\n1 7\n8 4\n",
"5\n5 1\n5 4\n4 3\n1 2\n",
"7\n1 2\n2 3\n1 4\n1 5\n3 6\n3 7\n",
"3\n1 3\n2 3\n",
"60\n26 6\n59 30\n31 12\n31 3\n38 23\n59 29\n53 9\n38 56\n53 54\n29 21\n17 55\n59 38\n26 16\n24 59\n24 25\n17 35\n24 41\n30 15\n31 27\n8 44\n26 5\n26 48\n8 32\n53 17\n3 34\n3 51\n30 28\n47 10\n53 60\n36 42\n24 53\n59 22\n53 40\n26 52\n36 4\n59 8\n29 37\n36 20\n17 47\n53 18\n3 50\n30 2\n17 7\n8 58\n59 1\n31 11\n24 26\n24 43\n53 57\n59 45\n47 13\n26 46\n17 33\n30 31\n26 39\n26 19\n24 36\n8 49\n38 14\n",
"7\n1 2\n2 3\n3 4\n3 5\n1 6\n1 7\n",
"20\n19 16\n19 18\n20 7\n9 4\n6 17\n14 2\n9 15\n2 13\n5 11\n19 12\n12 20\n16 9\n11 8\n19 5\n3 1\n19 14\n5 3\n12 10\n19 6\n",
"7\n1 2\n1 3\n2 4\n2 5\n3 6\n3 7\n",
"10\n9 5\n7 1\n9 10\n7 2\n5 4\n9 6\n2 9\n10 8\n1 3\n",
"4\n2 4\n2 3\n2 1\n",
"4\n1 4\n3 2\n1 3\n",
"3\n1 2\n1 3\n",
"5\n1 2\n1 5\n1 3\n1 4\n",
"20\n14 9\n12 13\n10 15\n2 1\n20 19\n16 6\n16 3\n17 14\n3 5\n2 11\n3 10\n15 8\n14 2\n6 4\n3 20\n5 18\n1 7\n1 16\n4 12\n",
"20\n7 5\n14 13\n17 6\n3 8\n16 12\n18 9\n3 18\n14 1\n17 3\n15 2\n17 4\n9 11\n2 7\n15 17\n3 20\n16 10\n17 14\n2 16\n1 19\n",
"8\n1 2\n2 3\n3 4\n1 7\n1 8\n4 5\n4 6\n",
"5\n5 1\n5 2\n5 3\n5 4\n",
"50\n49 6\n43 7\n1 27\n19 35\n15 37\n16 12\n19 21\n16 28\n49 9\n48 39\n13 1\n2 48\n9 50\n44 3\n41 32\n48 31\n49 33\n6 11\n13 20\n49 22\n13 41\n48 29\n13 46\n15 47\n34 2\n49 13\n48 14\n34 24\n16 36\n13 40\n49 34\n49 17\n43 25\n11 23\n10 15\n19 26\n34 44\n16 42\n19 18\n46 8\n29 38\n1 45\n12 43\n13 16\n46 30\n15 5\n49 10\n11 19\n32 4\n",
"20\n13 1\n18 2\n3 7\n18 5\n20 16\n3 12\n18 9\n3 10\n18 11\n13 6\n3 18\n20 15\n20 17\n3 13\n3 4\n13 14\n3 20\n18 8\n3 19\n",
"10\n8 2\n5 6\n1 8\n2 9\n1 4\n8 10\n10 5\n2 7\n2 3\n"
],
"output": [
"NO",
"YES",
"NO",
"NO",
"NO",
"YES",
"NO",
"NO",
"NO",
"NO",
"YES",
"NO",
"NO",
"YES",
"NO",
"NO",
"NO",
"YES",
"NO",
"YES",
"NO"
]
}
|
IN-CORRECT
|
java
|
import java.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.InputMismatchException;
import java.io.IOException;
import java.io.InputStream;
/**
* Built using CHelper plug-in
* Actual solution is at the top
*
* @author KharYusuf
*/
public class Main {
public static void main(String[] args) {
InputStream inputStream = System.in;
OutputStream outputStream = System.out;
FastReader in = new FastReader(inputStream);
PrintWriter out = new PrintWriter(outputStream);
D1AddOnATree solver = new D1AddOnATree();
solver.solve(1, in, out);
out.close();
}
static class D1AddOnATree {
public void solve(int testNumber, FastReader s, PrintWriter w) {
int n = s.nextInt(), f[] = new int[n];
for (int i = 1; i < n; i++) {
f[s.nextInt() - 1]++;
f[s.nextInt() - 1]++;
}
for (int i = 0; i < n; i++) {
if ((f[i] & 1) == 0) {
w.println("NO");
return;
}
}
w.println("YES");
}
}
static class FastReader {
private InputStream stream;
private byte[] buf = new byte[1024];
private int curChar;
private int numChars;
private FastReader.SpaceCharFilter filter;
public FastReader(InputStream stream) {
this.stream = stream;
}
public int read() {
if (numChars == -1)
throw new InputMismatchException();
if (curChar >= numChars) {
curChar = 0;
try {
numChars = stream.read(buf);
} catch (IOException e) {
throw new InputMismatchException();
}
if (numChars <= 0)
return -1;
}
return buf[curChar++];
}
public int nextInt() {
int c = read();
while (isSpaceChar(c))
c = read();
int sgn = 1;
if (c == '-') {
sgn = -1;
c = read();
}
int res = 0;
do {
if (c < '0' || c > '9')
throw new InputMismatchException();
res *= 10;
res += c - '0';
c = read();
}
while (!isSpaceChar(c));
return res * sgn;
}
public boolean isSpaceChar(int c) {
if (filter != null)
return filter.isSpaceChar(c);
return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
}
public interface SpaceCharFilter {
public boolean isSpaceChar(int ch);
}
}
}
|
1189_D1. Add on a Tree
|
Note that this is the first problem of the two similar problems. You can hack this problem only if you solve both problems.
You are given a tree with n nodes. In the beginning, 0 is written on all edges. In one operation, you can choose any 2 distinct leaves u, v and any real number x and add x to values written on all edges on the simple path between u and v.
For example, on the picture below you can see the result of applying two operations to the graph: adding 2 on the path from 7 to 6, and then adding -0.5 on the path from 4 to 5.
<image>
Is it true that for any configuration of real numbers written on edges, we can achieve it with a finite number of operations?
Leaf is a node of a tree of degree 1. Simple path is a path that doesn't contain any node twice.
Input
The first line contains a single integer n (2 ≤ n ≤ 10^5) — the number of nodes.
Each of the next n-1 lines contains two integers u and v (1 ≤ u, v ≤ n, u ≠ v), meaning that there is an edge between nodes u and v. It is guaranteed that these edges form a tree.
Output
If there is a configuration of real numbers written on edges of the tree that we can't achieve by performing the operations, output "NO".
Otherwise, output "YES".
You can print each letter in any case (upper or lower).
Examples
Input
2
1 2
Output
YES
Input
3
1 2
2 3
Output
NO
Input
5
1 2
1 3
1 4
2 5
Output
NO
Input
6
1 2
1 3
1 4
2 5
2 6
Output
YES
Note
In the first example, we can add any real x to the value written on the only edge (1, 2).
<image>
In the second example, one of configurations that we can't reach is 0 written on (1, 2) and 1 written on (2, 3).
<image>
Below you can see graphs from examples 3, 4:
<image> <image>
|
{
"input": [
"2\n1 2\n",
"3\n1 2\n2 3\n",
"5\n1 2\n1 3\n1 4\n2 5\n",
"6\n1 2\n1 3\n1 4\n2 5\n2 6\n"
],
"output": [
"YES",
"NO",
"NO",
"YES"
]
}
|
{
"input": [
"50\n16 4\n17 9\n31 19\n22 10\n8 1\n40 30\n3 31\n20 29\n47 27\n22 25\n32 34\n12 15\n40 32\n10 33\n47 12\n6 24\n46 41\n14 23\n12 35\n31 42\n46 28\n31 20\n46 37\n1 39\n29 49\n37 47\n40 6\n42 36\n47 2\n24 46\n2 13\n8 45\n41 3\n32 17\n4 7\n47 26\n28 8\n41 50\n34 44\n33 21\n25 5\n16 40\n3 14\n8 18\n28 11\n32 22\n2 38\n3 48\n44 43\n",
"10\n8 1\n1 2\n8 9\n8 5\n1 3\n1 10\n1 6\n1 7\n8 4\n",
"5\n5 1\n5 4\n4 3\n1 2\n",
"7\n1 2\n2 3\n1 4\n1 5\n3 6\n3 7\n",
"3\n1 3\n2 3\n",
"60\n26 6\n59 30\n31 12\n31 3\n38 23\n59 29\n53 9\n38 56\n53 54\n29 21\n17 55\n59 38\n26 16\n24 59\n24 25\n17 35\n24 41\n30 15\n31 27\n8 44\n26 5\n26 48\n8 32\n53 17\n3 34\n3 51\n30 28\n47 10\n53 60\n36 42\n24 53\n59 22\n53 40\n26 52\n36 4\n59 8\n29 37\n36 20\n17 47\n53 18\n3 50\n30 2\n17 7\n8 58\n59 1\n31 11\n24 26\n24 43\n53 57\n59 45\n47 13\n26 46\n17 33\n30 31\n26 39\n26 19\n24 36\n8 49\n38 14\n",
"7\n1 2\n2 3\n3 4\n3 5\n1 6\n1 7\n",
"20\n19 16\n19 18\n20 7\n9 4\n6 17\n14 2\n9 15\n2 13\n5 11\n19 12\n12 20\n16 9\n11 8\n19 5\n3 1\n19 14\n5 3\n12 10\n19 6\n",
"7\n1 2\n1 3\n2 4\n2 5\n3 6\n3 7\n",
"10\n9 5\n7 1\n9 10\n7 2\n5 4\n9 6\n2 9\n10 8\n1 3\n",
"4\n2 4\n2 3\n2 1\n",
"4\n1 4\n3 2\n1 3\n",
"3\n1 2\n1 3\n",
"5\n1 2\n1 5\n1 3\n1 4\n",
"20\n14 9\n12 13\n10 15\n2 1\n20 19\n16 6\n16 3\n17 14\n3 5\n2 11\n3 10\n15 8\n14 2\n6 4\n3 20\n5 18\n1 7\n1 16\n4 12\n",
"20\n7 5\n14 13\n17 6\n3 8\n16 12\n18 9\n3 18\n14 1\n17 3\n15 2\n17 4\n9 11\n2 7\n15 17\n3 20\n16 10\n17 14\n2 16\n1 19\n",
"8\n1 2\n2 3\n3 4\n1 7\n1 8\n4 5\n4 6\n",
"5\n5 1\n5 2\n5 3\n5 4\n",
"50\n49 6\n43 7\n1 27\n19 35\n15 37\n16 12\n19 21\n16 28\n49 9\n48 39\n13 1\n2 48\n9 50\n44 3\n41 32\n48 31\n49 33\n6 11\n13 20\n49 22\n13 41\n48 29\n13 46\n15 47\n34 2\n49 13\n48 14\n34 24\n16 36\n13 40\n49 34\n49 17\n43 25\n11 23\n10 15\n19 26\n34 44\n16 42\n19 18\n46 8\n29 38\n1 45\n12 43\n13 16\n46 30\n15 5\n49 10\n11 19\n32 4\n",
"20\n13 1\n18 2\n3 7\n18 5\n20 16\n3 12\n18 9\n3 10\n18 11\n13 6\n3 18\n20 15\n20 17\n3 13\n3 4\n13 14\n3 20\n18 8\n3 19\n",
"10\n8 2\n5 6\n1 8\n2 9\n1 4\n8 10\n10 5\n2 7\n2 3\n"
],
"output": [
"NO",
"YES",
"NO",
"NO",
"NO",
"YES",
"NO",
"NO",
"NO",
"NO",
"YES",
"NO",
"NO",
"YES",
"NO",
"NO",
"NO",
"YES",
"NO",
"YES",
"NO"
]
}
|
IN-CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
const long long int inf = 1e18, M = 1e9 + 7;
const long long int N = 1e5 + 5;
vector<long long int> v[N], dp(N, 0);
bool dfs(long long int x, long long int p) {
if (v[x].size() == 1ll && p != -1) {
dp[x] = 1ll;
return 0;
}
bool res = 0;
for (auto u : v[x]) {
if (u == p) continue;
res |= dfs(u, x);
dp[x] += dp[u];
if (v[u].size() > 1 && dp[u] == 1ll) {
return 1;
}
}
return res;
}
void solve() {
long long int n;
cin >> n;
long long int a, b;
for (long long int i = 1; i < n; ++i) {
cin >> a >> b;
v[a].push_back(b);
v[b].push_back(a);
}
if (dfs(1, -1))
cout << "NO";
else
cout << "YES";
}
int32_t main() {
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
;
long long int t = 1;
while (t--) {
solve();
}
return 0;
}
|
1189_D1. Add on a Tree
|
Note that this is the first problem of the two similar problems. You can hack this problem only if you solve both problems.
You are given a tree with n nodes. In the beginning, 0 is written on all edges. In one operation, you can choose any 2 distinct leaves u, v and any real number x and add x to values written on all edges on the simple path between u and v.
For example, on the picture below you can see the result of applying two operations to the graph: adding 2 on the path from 7 to 6, and then adding -0.5 on the path from 4 to 5.
<image>
Is it true that for any configuration of real numbers written on edges, we can achieve it with a finite number of operations?
Leaf is a node of a tree of degree 1. Simple path is a path that doesn't contain any node twice.
Input
The first line contains a single integer n (2 ≤ n ≤ 10^5) — the number of nodes.
Each of the next n-1 lines contains two integers u and v (1 ≤ u, v ≤ n, u ≠ v), meaning that there is an edge between nodes u and v. It is guaranteed that these edges form a tree.
Output
If there is a configuration of real numbers written on edges of the tree that we can't achieve by performing the operations, output "NO".
Otherwise, output "YES".
You can print each letter in any case (upper or lower).
Examples
Input
2
1 2
Output
YES
Input
3
1 2
2 3
Output
NO
Input
5
1 2
1 3
1 4
2 5
Output
NO
Input
6
1 2
1 3
1 4
2 5
2 6
Output
YES
Note
In the first example, we can add any real x to the value written on the only edge (1, 2).
<image>
In the second example, one of configurations that we can't reach is 0 written on (1, 2) and 1 written on (2, 3).
<image>
Below you can see graphs from examples 3, 4:
<image> <image>
|
{
"input": [
"2\n1 2\n",
"3\n1 2\n2 3\n",
"5\n1 2\n1 3\n1 4\n2 5\n",
"6\n1 2\n1 3\n1 4\n2 5\n2 6\n"
],
"output": [
"YES",
"NO",
"NO",
"YES"
]
}
|
{
"input": [
"50\n16 4\n17 9\n31 19\n22 10\n8 1\n40 30\n3 31\n20 29\n47 27\n22 25\n32 34\n12 15\n40 32\n10 33\n47 12\n6 24\n46 41\n14 23\n12 35\n31 42\n46 28\n31 20\n46 37\n1 39\n29 49\n37 47\n40 6\n42 36\n47 2\n24 46\n2 13\n8 45\n41 3\n32 17\n4 7\n47 26\n28 8\n41 50\n34 44\n33 21\n25 5\n16 40\n3 14\n8 18\n28 11\n32 22\n2 38\n3 48\n44 43\n",
"10\n8 1\n1 2\n8 9\n8 5\n1 3\n1 10\n1 6\n1 7\n8 4\n",
"5\n5 1\n5 4\n4 3\n1 2\n",
"7\n1 2\n2 3\n1 4\n1 5\n3 6\n3 7\n",
"3\n1 3\n2 3\n",
"60\n26 6\n59 30\n31 12\n31 3\n38 23\n59 29\n53 9\n38 56\n53 54\n29 21\n17 55\n59 38\n26 16\n24 59\n24 25\n17 35\n24 41\n30 15\n31 27\n8 44\n26 5\n26 48\n8 32\n53 17\n3 34\n3 51\n30 28\n47 10\n53 60\n36 42\n24 53\n59 22\n53 40\n26 52\n36 4\n59 8\n29 37\n36 20\n17 47\n53 18\n3 50\n30 2\n17 7\n8 58\n59 1\n31 11\n24 26\n24 43\n53 57\n59 45\n47 13\n26 46\n17 33\n30 31\n26 39\n26 19\n24 36\n8 49\n38 14\n",
"7\n1 2\n2 3\n3 4\n3 5\n1 6\n1 7\n",
"20\n19 16\n19 18\n20 7\n9 4\n6 17\n14 2\n9 15\n2 13\n5 11\n19 12\n12 20\n16 9\n11 8\n19 5\n3 1\n19 14\n5 3\n12 10\n19 6\n",
"7\n1 2\n1 3\n2 4\n2 5\n3 6\n3 7\n",
"10\n9 5\n7 1\n9 10\n7 2\n5 4\n9 6\n2 9\n10 8\n1 3\n",
"4\n2 4\n2 3\n2 1\n",
"4\n1 4\n3 2\n1 3\n",
"3\n1 2\n1 3\n",
"5\n1 2\n1 5\n1 3\n1 4\n",
"20\n14 9\n12 13\n10 15\n2 1\n20 19\n16 6\n16 3\n17 14\n3 5\n2 11\n3 10\n15 8\n14 2\n6 4\n3 20\n5 18\n1 7\n1 16\n4 12\n",
"20\n7 5\n14 13\n17 6\n3 8\n16 12\n18 9\n3 18\n14 1\n17 3\n15 2\n17 4\n9 11\n2 7\n15 17\n3 20\n16 10\n17 14\n2 16\n1 19\n",
"8\n1 2\n2 3\n3 4\n1 7\n1 8\n4 5\n4 6\n",
"5\n5 1\n5 2\n5 3\n5 4\n",
"50\n49 6\n43 7\n1 27\n19 35\n15 37\n16 12\n19 21\n16 28\n49 9\n48 39\n13 1\n2 48\n9 50\n44 3\n41 32\n48 31\n49 33\n6 11\n13 20\n49 22\n13 41\n48 29\n13 46\n15 47\n34 2\n49 13\n48 14\n34 24\n16 36\n13 40\n49 34\n49 17\n43 25\n11 23\n10 15\n19 26\n34 44\n16 42\n19 18\n46 8\n29 38\n1 45\n12 43\n13 16\n46 30\n15 5\n49 10\n11 19\n32 4\n",
"20\n13 1\n18 2\n3 7\n18 5\n20 16\n3 12\n18 9\n3 10\n18 11\n13 6\n3 18\n20 15\n20 17\n3 13\n3 4\n13 14\n3 20\n18 8\n3 19\n",
"10\n8 2\n5 6\n1 8\n2 9\n1 4\n8 10\n10 5\n2 7\n2 3\n"
],
"output": [
"NO",
"YES",
"NO",
"NO",
"NO",
"YES",
"NO",
"NO",
"NO",
"NO",
"YES",
"NO",
"NO",
"YES",
"NO",
"NO",
"NO",
"YES",
"NO",
"YES",
"NO"
]
}
|
IN-CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
const long long inf = LONG_MAX;
const long long arr = 1000000;
bool comp(pair<long long, long long> a, pair<long long, long long> b) {
if (a.first == b.first)
return a.second < b.second;
else
return a.first < b.first;
}
vector<vector<long long> > g(arr);
int main(void) {
long long n;
cin >> n;
for (int i = 0; i < n - 1; i++) {
long long l, r;
cin >> l >> r;
g[l].push_back(r);
g[r].push_back(l);
}
for (long long i = 2; i <= n; i++) {
if (g[i].size() == 2) return cout << "NO" << '\n', 0;
}
if (n % 2 == 1) return cout << "NO" << '\n', 0;
cout << "YES" << '\n';
return 0;
}
|
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