Search is not available for this dataset
name
stringlengths 2
88
| description
stringlengths 31
8.62k
| public_tests
dict | private_tests
dict | solution_type
stringclasses 2
values | programming_language
stringclasses 5
values | solution
stringlengths 1
983k
|
|---|---|---|---|---|---|---|
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 β€ n β€ 10^5) β the number of nodes.
Each of the next n-1 lines contains two integers u and v (1 β€ u, v β€ n, u β 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"
]
}
|
CORRECT
|
java
|
import java.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.HashSet;
import java.util.Iterator;
import java.util.Set;
import java.util.HashMap;
import java.util.InputMismatchException;
import java.io.IOException;
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;
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 out) {
int n = s.nextInt();
HashMap<Integer, HashSet<Integer>> map = new HashMap<>();
for (int i = 0; i < n - 1; i++) {
int src = s.nextInt() - 1;
int dest = s.nextInt() - 1;
HashSet<Integer> set = map.getOrDefault(src, new HashSet<>());
set.add(dest);
map.put(src, set);
set = map.getOrDefault(dest, new HashSet<>());
set.add(src);
map.put(dest, set);
}
Iterator<Integer> iter = map.keySet().iterator();
boolean ok = true;
while (iter.hasNext()) {
int curr = iter.next();
if (map.get(curr).size() == 2) {
ok = false;
break;
}
}
out.println(ok ? "YES" : "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 isWhitespace(c);
}
public static boolean isWhitespace(int 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 β€ n β€ 10^5) β the number of nodes.
Each of the next n-1 lines contains two integers u and v (1 β€ u, v β€ n, u β 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"
]
}
|
CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
const long long inff = 0x3f3f3f3f3f3f3f3f;
int n, x, y, du[100008];
int main() {
cin.tie(0);
cout.tie(0);
cin >> n;
for (int i(1); i <= (n - 1); ++i) {
scanf("%d", &x), scanf("%d", &y);
du[x]++, du[y]++;
}
for (int i(1); i <= (n); ++i)
if (du[i] == 2) return puts("NO"), 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 β€ n β€ 10^5) β the number of nodes.
Each of the next n-1 lines contains two integers u and v (1 β€ u, v β€ n, u β 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"
]
}
|
CORRECT
|
java
|
/*
Roses are red
Memes are neat
All my test cases time out
Lmao yeet
*/
import java.util.*;
import java.io.*;
public class D
{
public static void main(String args[]) throws Exception
{
BufferedReader infile = new BufferedReader(new InputStreamReader(System.in));
StringTokenizer st = new StringTokenizer(infile.readLine());
int N = Integer.parseInt(st.nextToken());
LinkedList<Integer>[] edges = new LinkedList[N+1];
for(int i=1; i <= N; i++)
edges[i] = new LinkedList<Integer>();
for(int i=0; i < N-1; i++)
{
st = new StringTokenizer(infile.readLine());
int a = Integer.parseInt(st.nextToken());
int b = Integer.parseInt(st.nextToken());
edges[a].add(b);
edges[b].add(a);
}
boolean b = true;
for(int i=1; i <= N; i++)
if(edges[i].size() == 2)
b = false;
if(b)
System.out.println("YES");
else
System.out.println("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 β€ n β€ 10^5) β the number of nodes.
Each of the next n-1 lines contains two integers u and v (1 β€ u, v β€ n, u β 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"
]
}
|
CORRECT
|
java
|
import java.io.*;
import java.util.*;
public class Solution {
public static void main(String[] args) {
FastReader in = new FastReader();
System.out.println(solve(in) ? "YES" : "NO");
}
static boolean solve(FastReader in) {
int n = in.nextInt();
int[] deg = new int[n + 1];
for (int i = 0; i < 2 * n - 2; i++) {
deg[in.nextInt()]++;
}
for (int i : deg) if (i == 2) return false;
return true;
}
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 β€ n β€ 10^5) β the number of nodes.
Each of the next n-1 lines contains two integers u and v (1 β€ u, v β€ n, u β 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"
]
}
|
CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
int main() {
int n, f = 0;
cin >> n;
vector<int> V[100008];
for (int i = 0; i < n - 1; i++) {
int u, v;
cin >> u >> v;
V[u].push_back(v);
V[v].push_back(u);
}
for (int i = 0; i <= n; i++) {
if (V[i].size() == 2) {
f = 1;
break;
}
}
if (f)
cout << "NO" << endl;
else
cout << "YES" << endl;
}
|
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 β€ n β€ 10^5) β the number of nodes.
Each of the next n-1 lines contains two integers u and v (1 β€ u, v β€ n, u β 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"
]
}
|
CORRECT
|
java
|
import java.util.*;
import java.io.*;
import java.text.*;
import java.math.*;
import static java.lang.Integer.*;
import static java.lang.Double.*;
import java.lang.Math.*;
public class add_on_a_tree {
public static void main(String[] args) throws Exception {
new add_on_a_tree().run();
}
public void run() throws Exception {
FastIO file = new FastIO();
int n = file.nextInt();
ArrayList<ArrayList<Integer>> al = new ArrayList<>();
for (int i = 0; i < n; i++) {
al.add(new ArrayList<>());
}
int[] vis = new int[n];
for (int i = 0; i < n; i++) {
vis[i] = 0;
}
for (int i = 0; i < n - 1; i++) {
int x = file.nextInt() - 1;
int y = file.nextInt() - 1;
al.get(x).add(y);
al.get(y).add(x);
}
boolean works = false;
for (int i = 0; i < n; i++) {
if (al.get(i).size() == 2) works = true;
}
if (works) System.out.println("NO");
else System.out.println("YES");
file.out.flush();
}
public static class FastIO {
BufferedReader br;
StringTokenizer st;
PrintWriter out;
public FastIO() {
br = new BufferedReader(new InputStreamReader(System.in));
out = new PrintWriter(System.out);
}
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 print(Object o) {
out.print(o);
}
void println(Object o) {
out.println(o);
}
void printf(String s, Object... o) {
out.printf(s, o);
}
}
public static long pow(long n, long p, long mod) {
if (p == 0)
return 1;
if (p == 1)
return n % mod;
if (p % 2 == 0) {
long temp = pow(n, p / 2, mod);
return (temp * temp) % mod;
} else {
long temp = pow(n, (p - 1) / 2, mod);
temp = (temp * temp) % mod;
return (temp * n) % mod;
}
}
public static long pow(long n, long p) {
if (p == 0)
return 1;
if (p == 1)
return n;
if (p % 2 == 0) {
long temp = pow(n, p / 2);
return (temp * temp);
} else {
long temp = pow(n, (p - 1) / 2);
temp = (temp * temp);
return (temp * n);
}
}
public static long gcd(long x, long y) {
if (x == 0)
return y;
else
return gcd(y % x, x);
}
public static boolean isPrime(int n) {
if (n <= 1)
return false;
if (n <= 3)
return true;
if (n % 2 == 0 || n % 3 == 0)
return false;
for (int i = 5; i * i <= n; i = i + 6)
if (n % i == 0 || n % (i + 2) == 0)
return false;
return true;
}
}
|
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 β€ n β€ 10^5) β the number of nodes.
Each of the next n-1 lines contains two integers u and v (1 β€ u, v β€ n, u β 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"
]
}
|
CORRECT
|
java
|
"use strict";
// readline().split(' ').map(value => +value);
var n = +readline();
var A = new Array(Math.pow(10, 5) + 100).fill(0);
for (var i = 0; i < n - 1; ++i) {
var input = readline().split(' ').map(value => +value);
var u = input[0];
var v = input[1];
++A[u];
++A[v];
}
var result = 'YES';
for (var i = 1; i <= n; ++i) {
if (A[i] === 2) {
result = 'NO';
break;
}
}
write(result);
|
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 β€ n β€ 10^5) β the number of nodes.
Each of the next n-1 lines contains two integers u and v (1 β€ u, v β€ n, u β 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"
]
}
|
CORRECT
|
java
|
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.io.IOException;
import java.util.StringTokenizer;
import java.util.Arrays;
public class c{
public static void main(String[] args)throws IOException{
br = new BufferedReader(new InputStreamReader(System.in));
PrintWriter out = new PrintWriter(System.out);
int n = nextInt();
int[] v = new int[n];
int a = 0, b = 0, ans = 0;
for(int i = 0; i<n-1;i++){
a = nextInt();
b = nextInt();
a--;b--;
v[a]+=1;
v[b]+=1;
if(v[a]==2){
ans++;
}else if(v[a]==3){
ans--;
}
if(v[b]==2){
ans++;
}else if(v[b]==3){
ans--;
}
}
if(ans==0){
out.println("YES");
}else{
out.println("NO");
}
out.close();
}
static BufferedReader br;
static StringTokenizer st = new StringTokenizer("");
static String next()throws IOException{
while(!st.hasMoreTokens()){
st = new StringTokenizer(br.readLine());
}
return st.nextToken();
}
static int nextInt()throws IOException{
return Integer.parseInt(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 β€ n β€ 10^5) β the number of nodes.
Each of the next n-1 lines contains two integers u and v (1 β€ u, v β€ n, u β 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"
]
}
|
CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
int cnt[100005];
int main() {
int n;
scanf("%d", &n);
if (n == 2) {
printf("YES\n");
return 0;
}
if (n == 3) {
printf("NO\n");
return 0;
}
for (int i = 0; i < n - 1; i++) {
int u, v;
scanf("%d%d", &u, &v);
cnt[u]++;
cnt[v]++;
}
for (int i = 1; i <= n; i++) {
if (cnt[i] == 2) {
printf("NO\n");
return 0;
}
}
printf("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 β€ n β€ 10^5) β the number of nodes.
Each of the next n-1 lines contains two integers u and v (1 β€ u, v β€ n, u β 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"
]
}
|
CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
int n;
vector<int> v[101010];
int main() {
scanf("%d", &n);
for (int i = 1; i < n; i++) {
int a, b;
scanf("%d %d", &a, &b);
v[a].push_back(b);
v[b].push_back(a);
}
for (int i = 1; i <= n; i++)
if (v[i].size() == 2) return !printf("NO");
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 β€ n β€ 10^5) β the number of nodes.
Each of the next n-1 lines contains two integers u and v (1 β€ u, v β€ n, u β 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"
]
}
|
CORRECT
|
java
|
import java.util.*;
public class Main {
public static void main(String[] args)
{
Scanner stdin = new Scanner(System.in);
/*int n = stdin.nextInt();
for(int i = 0; i < n; i++)
{
test(stdin);
}*/
test(stdin);
stdin.close();
}
public static void test(Scanner stdin)
{
int n = stdin.nextInt();
int dp[] = new int[n];
for(int i = 0; i < n - 1; i++)
{
dp[stdin.nextInt() - 1]++;
dp[stdin.nextInt() - 1]++;
}
boolean good = true;
for(int i = 0; i < n; i++)
{
if(dp[i] == 2)
{
good = false;
}
}
if(good)
{
System.out.println("YES");
}
else {
System.out.println("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 β€ n β€ 10^5) β the number of nodes.
Each of the next n-1 lines contains two integers u and v (1 β€ u, v β€ n, u β 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"
]
}
|
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] + b[i] == 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 β€ n β€ 10^5) β the number of nodes.
Each of the next n-1 lines contains two integers u and v (1 β€ u, v β€ n, u β 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"
]
}
|
CORRECT
|
java
|
import java.util.*;
import java.io.*;
import java.text.*;
public class D1_1189 {
public static void main(String[] args) throws IOException {
Scanner sc = new Scanner(System.in);
PrintWriter pw = new PrintWriter(System.out);
int count [] = new int [sc.nextInt()];
for(int i = 0;i<2*count.length-2;i++)
count [sc.nextInt()-1]++;
String x = "YES";
for (int i = 0; i < count.length; i++) {
if(count [i]==2)
x="NO";
}
pw.println(x);
pw.flush();
pw.close();
}
static class Scanner {
StringTokenizer st;
BufferedReader br;
public Scanner(InputStream s) {
br = new BufferedReader(new InputStreamReader(s));
}
public String next() throws IOException {
while (st == null || !st.hasMoreTokens())
st = new StringTokenizer(br.readLine());
return st.nextToken();
}
public int nextInt() throws IOException {
return Integer.parseInt(next());
}
public long nextLong() throws IOException {
return Long.parseLong(next());
}
public String nextLine() throws IOException {
return br.readLine();
}
public double nextDouble() throws IOException {
String x = next();
StringBuilder sb = new StringBuilder("0");
double res = 0, f = 1;
boolean dec = false, neg = false;
int start = 0;
if (x.charAt(0) == '-') {
neg = true;
start++;
}
for (int i = start; i < x.length(); i++)
if (x.charAt(i) == '.') {
res = Long.parseLong(sb.toString());
sb = new StringBuilder("0");
dec = true;
} else {
sb.append(x.charAt(i));
if (dec)
f *= 10;
}
res += Long.parseLong(sb.toString()) / f;
return res * (neg ? -1 : 1);
}
public boolean ready() throws IOException {
return br.ready();
}
}
}
|
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 β€ n β€ 10^5) β the number of nodes.
Each of the next n-1 lines contains two integers u and v (1 β€ u, v β€ n, u β 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"
]
}
|
CORRECT
|
python3
|
n=int(input())
a=[0]*n
for i in range(n-1):
c,d=map(int,input().split())
a[c-1]+=1
a[d-1]+=1
stat=2 in a
if(stat):
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 β€ n β€ 10^5) β the number of nodes.
Each of the next n-1 lines contains two integers u and v (1 β€ u, v β€ n, u β 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"
]
}
|
CORRECT
|
java
|
import java.util.*;
import java.io.*;
public class div572{
static class Reader
{
final private int BUFFER_SIZE = 1 << 16;
private DataInputStream din;
private byte[] buffer;
private int bufferPointer, bytesRead;
BufferedReader br;
public Reader()
{
din = new DataInputStream(System.in);
buffer = new byte[BUFFER_SIZE];
bufferPointer = bytesRead = 0;
br = new BufferedReader(new InputStreamReader(System.in));
}
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++];
}
StringTokenizer st;
String next() {
while (st == null || !st.hasMoreTokens()) {
try {
st = new StringTokenizer(br.readLine());
} catch (Exception r) {
r.printStackTrace();
}
}
return st.nextToken();
}
public void close() throws IOException
{
if (din == null)
return;
din.close();
}
}
public static void main(String s[]) throws IOException{
Reader sc = new Reader();
int n = sc.nextInt();
ArrayList<Integer> l1 = new ArrayList<>();
ArrayList<Integer> l2 = new ArrayList<>();
for(int i=0;i<n-1;i++){
l1.add(sc.nextInt());
l2.add(sc.nextInt());
}
HashMap<Integer,ArrayList<Integer>> tree = create(l1,l2);
ArrayList<Integer> keys = new ArrayList<>(tree.keySet());
for(int i=0;i<keys.size();i++){
if(tree.get(keys.get(i)).size()==2){
System.out.println("NO");
return;
}
}
System.out.println("YES");
}
static int count=0;
public static void bfs(int n, HashMap<Integer,ArrayList<Integer>> hm, int node){
HashMap<Integer,Integer> visited = new HashMap<>();
visited.put(node,0);
LinkedList<Integer> ll = new LinkedList<>();
ll.addLast(node);
while(ll.size()!=0){
int val = ll.removeFirst();
ArrayList<Integer> keys = hm.get(val);
for(int i=0;i<keys.size();i++){
if(!visited.containsKey(keys.get(i))){
ll.addLast(keys.get(i));
System.out.println(val+" "+keys.get(i));
visited.put(keys.get(i),0);
count++;
if(count==n-1)
return;
}
}
}
}
public static HashMap<Integer,ArrayList<Integer>> create(ArrayList<Integer> l1, ArrayList<Integer> l2){
HashMap<Integer,ArrayList<Integer>> hm = new HashMap<>();
for(int i=0;i<l1.size();i++){
int val1 = l1.get(i);
int val2 = l2.get(i);
if(!hm.containsKey(val1)){
ArrayList<Integer> temp = new ArrayList<>();
temp.add(val2);
hm.put(val1,temp);
}else if(hm.containsKey(val1)){
ArrayList<Integer> temp = hm.get(val1);
temp.add(val2);
hm.put(val1,temp);
}
if(!hm.containsKey(val2)){
ArrayList<Integer> temp = new ArrayList<>();
temp.add(val1);
hm.put(val2,temp);
}else if(hm.containsKey(val2)){
ArrayList<Integer> temp = hm.get(val2);
temp.add(val1);
hm.put(val2,temp);
}
}
return hm;
}
}
|
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 β€ n β€ 10^5) β the number of nodes.
Each of the next n-1 lines contains two integers u and v (1 β€ u, v β€ n, u β 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"
]
}
|
CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
vector<int> g[100005];
int find(int v, int* parent) {
if (parent[v] == v) {
return v;
}
return find(parent[v], parent);
}
int vis[100005];
int dp[100005];
int dfs(int u) {
vis[u] = 1;
for (int i = 0; i < g[u].size(); i++) {
int v = g[u][i];
if (!vis[v]) dfs(v);
}
}
int main() {
int n;
cin >> n;
for (int i = 0; i < n - 1; i++) {
int u, v;
cin >> u >> v;
g[u].push_back(v);
g[v].push_back(u);
}
for (int i = 1; i <= n; i++) {
if (g[i].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 β€ n β€ 10^5) β the number of nodes.
Each of the next n-1 lines contains two integers u and v (1 β€ u, v β€ n, u β 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"
]
}
|
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;
int flag = 1;
for (i = 1; i <= n; i++) {
if (in[i] == 1) num++;
if (in[i] == 2) flag = 0;
}
num--;
ans = (1 + num) * num / 2;
if (ans >= n - 1 && flag)
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 β€ n β€ 10^5) β the number of nodes.
Each of the next n-1 lines contains two integers u and v (1 β€ u, v β€ n, u β 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"
]
}
|
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 β€ n β€ 10^5) β the number of nodes.
Each of the next n-1 lines contains two integers u and v (1 β€ u, v β€ n, u β 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"
]
}
|
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!=root:
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 β€ n β€ 10^5) β the number of nodes.
Each of the next n-1 lines contains two integers u and v (1 β€ u, v β€ n, u β 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"
]
}
|
CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
int degree[100005];
int main() {
ios::sync_with_stdio(false);
cin.tie(0);
int n;
cin >> n;
for (int i = 1; i < n; ++i) {
int u, v;
cin >> u >> v;
++degree[u];
++degree[v];
}
for (int i = 1; i <= n; ++i) {
if (degree[i] == 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 β€ n β€ 10^5) β the number of nodes.
Each of the next n-1 lines contains two integers u and v (1 β€ u, v β€ n, u β 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"
]
}
|
CORRECT
|
java
|
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.util.List;
import java.util.StringTokenizer;
public class ProblemD1 {
static class FastScanner {
private BufferedReader reader;
private StringTokenizer tokenizer;
public FastScanner() {
reader = new BufferedReader(new InputStreamReader(System.in));
tokenizer = null;
}
public String next() throws IOException {
while (tokenizer == null || !tokenizer.hasMoreTokens()) {
tokenizer = new StringTokenizer(reader.readLine());
}
return tokenizer.nextToken();
}
public int nextInt() throws IOException {
return Integer.parseInt(next());
}
}
static class Solution {
int n;
List<List<Integer>> tree;
boolean dfs(int v, int p) {
if (tree.get(v).size() == 2) {
return false;
}
for (int u : tree.get(v)) {
if (u == p)
continue;
boolean result = dfs(u, v);
if (!result) {
return false;
}
}
return true;
}
void run() throws IOException {
FastScanner sc = new FastScanner();
n = sc.nextInt();
tree = new ArrayList<>();
for (int i = 0; i < n; i++) {
tree.add(new ArrayList<>());
}
for (int i = 0; i < n - 1; i++) {
int u = sc.nextInt() - 1;
int v = sc.nextInt() - 1;
tree.get(u).add(v);
tree.get(v).add(u);
}
if (dfs(0, -1)) {
System.out.println("YES");
} else {
System.out.println("NO");
}
}
}
public static void main(String[] args) throws IOException {
new Solution().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 β€ n β€ 10^5) β the number of nodes.
Each of the next n-1 lines contains two integers u and v (1 β€ u, v β€ n, u β 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"
]
}
|
CORRECT
|
java
|
import java.util.Scanner;
public class Q4 {
public static void main(String[] args) {
// TODO Auto-generated method stub
Scanner read = new Scanner(System.in);
int n = read.nextInt();
int[] all = new int[n];
for(int i=0;i<n-1;i++) {
int a = read.nextInt()-1;
int b = read.nextInt()-1;
all[a]++;
all[b]++;
}
int flag = 1;
for(int i=0;i<n;i++) {
if(all[i]==2) {
flag = 0;
break;
}
}
if(flag==1) {
System.out.println("YES");
}
else {
System.out.println("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 β€ n β€ 10^5) β the number of nodes.
Each of the next n-1 lines contains two integers u and v (1 β€ u, v β€ n, u β 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"
]
}
|
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 (adj[v].size() == 2) 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);
adj[u].push_back(v);
}
if (n == 2) {
cout << "YES";
return 0;
}
for (long long i = 0; i < n; i++) {
if (adj[i].size() == 1 && !mark[i]) 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 β€ n β€ 10^5) β the number of nodes.
Each of the next n-1 lines contains two integers u and v (1 β€ u, v β€ n, u β 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"
]
}
|
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()==2)
{
out1.println("NO");
return;
}
}
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 β€ n β€ 10^5) β the number of nodes.
Each of the next n-1 lines contains two integers u and v (1 β€ u, v β€ n, u β 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"
]
}
|
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]++;
}
long long 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 β€ n β€ 10^5) β the number of nodes.
Each of the next n-1 lines contains two integers u and v (1 β€ u, v β€ n, u β 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"
]
}
|
CORRECT
|
python3
|
n = int(input())
degree = [0 for i in range(n)]
for i in range(n-1):
a = [int(i) for i in input().split()]
x = a[0]-1
y = a[1]-1
degree[x]+=1
degree[y]+=1
flag = 0
for i in range(n):
if(degree[i] == 2):
flag=1
break
if(flag==0):
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 β€ n β€ 10^5) β the number of nodes.
Each of the next n-1 lines contains two integers u and v (1 β€ u, v β€ n, u β 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"
]
}
|
CORRECT
|
java
|
import java.io.*;
import java.util.*;
public class tr0 {
static PrintWriter out;
static StringBuilder sb;
static final double EPS = 1e-9;
static int mod = 1000000007;
static int inf = (int) 1e9 + 2;
static long[] fac;
static int[] si;
static ArrayList<Integer> primes;
static ArrayList<Integer>[] ad;
static int[] ans;
static int m, k;
static HashMap<Integer, Integer> t;
static boolean f;
public static void main(String[] args) throws IOException {
Scanner sc = new Scanner(System.in);
out = new PrintWriter(System.out);
int n=sc.nextInt();
ad=new ArrayList[n];
for(int i=0;i<n;i++)
ad[i]=new ArrayList<>();
for(int i=0;i<n-1;i++) {
int a=sc.nextInt()-1;
int b=sc.nextInt()-1;
ad[a].add(b);
ad[b].add(a);
}
for(int i=0;i<n;i++)
if(ad[i].size()==2) {
System.out.println("NO");
return;
}
out.println("YES");
out.close();
}
static class seg implements Comparable<seg> {
int s;
int e;
int num;
int id;
seg(int s, int e) {
this.s = s;
this.e = e;
}
public String toString() {
return s + " " + e+" "+num+" "+id;
}
@Override
public int compareTo(seg o) {
if(o.s==s)
return o.e-e;
return s - o.s;
}
}
static public class SegmentTree { // 1-based DS, OOP
int N; // the number of elements in the array as a power of 2 (i.e. after padding)
int[] array, sTree, lazy,candies;
SegmentTree(int[] in) {
array = in.clone();
N = in.length - 1;
sTree = new int[N << 1]; // no. of nodes = 2*N - 1, we add one to cross out index zero
lazy = new int[N << 1];
candies = new int[N << 1];
build(1, 1, N);
build2(1, 1, N);
}
public String toString() {
return Arrays.toString(sTree)+"\n"+Arrays.toString(candies);
}
void build(int node, int b, int e) // O(n)
{
if (b == e)
sTree[node] = array[b];
else {
int mid = b + e >> 1;
build(node << 1, b, mid);
build(node << 1 | 1, mid + 1, e);
sTree[node] = (sTree[node << 1] + sTree[node << 1 | 1])%10;
if(sTree[node<<1]+sTree[node << 1 | 1]>=10)
candies[node]++;
}
}
void build2(int node, int b, int e) // O(n)
{
if (b == e)
b=e;
else {
int mid = b + e >> 1;
build(node << 1, b, mid);
build(node << 1 | 1, mid + 1, e);
candies[node] = (candies[node << 1] + candies[node << 1 | 1]);
}
}
void update_point(int index, int val) // O(log n)
{
index += N - 1;
sTree[index] += val;
while (index > 1) {
index >>= 1;
sTree[index] = sTree[index << 1] + sTree[index << 1 | 1];
}
}
void update_range(int i, int j, int val) // O(log n)
{
update_range(1, 1, N, i, j, val);
}
void update_range(int node, int b, int e, int i, int j, int val) {
if (i > e || j < b)
return;
if (b >= i && e <= j) {
sTree[node] += (e - b + 1) * val;
lazy[node] += val;
} else {
int mid = b + e >> 1;
propagate(node, b, mid, e);
update_range(node << 1, b, mid, i, j, val);
update_range(node << 1 | 1, mid + 1, e, i, j, val);
sTree[node] = sTree[node << 1] + sTree[node << 1 | 1];
}
}
void propagate(int node, int b, int mid, int e) {
lazy[node << 1] += lazy[node];
lazy[node << 1 | 1] += lazy[node];
sTree[node << 1] += (mid - b + 1) * lazy[node];
sTree[node << 1 | 1] += (e - mid) * lazy[node];
lazy[node] = 0;
}
// static int c=0;
int query(int i, int j) {
// c=0;
return query(1, 1, N, i, j);
}
int query(int node, int b, int e, int i, int j) // O(log n)
{
if (i > e || j < b)
return 0;
if (b >= i && e <= j)
return candies[node];
int mid = b + e >> 1;
//propagate(node, b, mid, e);
int q1 = query(node << 1, b, mid, i, j);
int q2 = query(node << 1 | 1, mid + 1, e, i, j);
return q1+q2;
}
}
static int x, y, d;
static long power(int i) {
// if(i==0)
// return 1;
long a = 1;
for (int k = 0; k < i; k++)
a *= i;
return a;
}
static void extendedEuclid(int a, int b) {
if (b == 0) {
x = 1;
y = 0;
d = a;
return;
}
extendedEuclid(b, a % b);
int x1 = y;
int y1 = x - a / b * y;
x = x1;
y = y1;
}
static void dfs(int u) {
int counter = 1;
for (int v : ad[u])
if (ans[v] == 0) {
if (counter == ans[u]) {
counter++;
ans[v] = counter;
} else
ans[v] = counter++;
if (counter > k)
f = false;
dfs(v);
}
}
static void bfs(int s) {
Queue<Integer> q = new LinkedList();
q.add(s);
ans[s] = 1;
while (!q.isEmpty()) {
int u = q.remove();
int c = 1;
for (int v : ad[u]) {
if (ans[v] == 0) {
// System.out.println(u);
if (c == ans[u]) {
c++;
ans[v] = c++;
} else
ans[v] = c++;
// System.out.println(v+" "+c);
if (ans[v] > k)
f = false;
q.add(v);
}
}
}
}
static void seive() {
si = new int[1000001];
primes = new ArrayList<>();
int N = 1000001;
si[1] = 1;
for (int i = 2; i < N; i++) {
if (si[i] == 0) {
si[i] = i;
primes.add(i);
}
for (int j = 0; j < primes.size() && primes.get(j) <= si[i] && (i * primes.get(j)) < N; j++)
si[primes.get(j) * i] = primes.get(j);
}
}
static long inver(long x) {
int a = (int) x;
long e = (mod - 2);
long res = 1;
while (e > 0) {
if ((e & 1) == 1) {
// System.out.println(res*a);
res = (int) ((1l * res * a) % mod);
}
a = (int) ((1l * a * a) % mod);
e >>= 1;
}
// out.println(res+" "+x);
return res % mod;
}
static long fac(int n) {
if (n == 0)
return fac[n] = 1;
if (n == 1)
return fac[n] = 1;
long ans = 1;
for (int i = 1; i <= n; i++)
fac[i] = ans = (i % mod * ans % mod) % mod;
return ans % mod;
}
static int gcd(int a, int b) {
if (b == 0)
return a;
return gcd(b, a % b);
}
static class unionfind {
int[] p;
int[] size;
unionfind(int n) {
p = new int[n];
size = new int[n];
for (int i = 0; i < n; i++) {
p[i] = i;
}
Arrays.fill(size, 1);
}
int findSet(int v) {
if (v == p[v])
return v;
return p[v] = findSet(p[v]);
}
boolean sameSet(int a, int b) {
a = findSet(a);
b = findSet(b);
if (a == b)
return true;
return false;
}
int max() {
int max = 0;
for (int i = 0; i < size.length; i++)
if (size[i] > max)
max = size[i];
return max;
}
void combine(int a, int b) {
a = findSet(a);
b = findSet(b);
if (a == b)
return;
if (size[a] > size[b]) {
p[b] = a;
size[a] += size[b];
} else {
p[a] = b;
size[b] += size[a];
}
}
}
static class Scanner {
StringTokenizer st;
BufferedReader br;
public Scanner(InputStream system) {
br = new BufferedReader(new InputStreamReader(system));
}
public Scanner(String file) throws Exception {
br = new BufferedReader(new FileReader(file));
}
public String next() throws IOException {
while (st == null || !st.hasMoreTokens())
st = new StringTokenizer(br.readLine());
return st.nextToken();
}
public String nextLine() throws IOException {
return br.readLine();
}
public int nextInt() throws IOException {
return Integer.parseInt(next());
}
public double nextDouble() throws IOException {
return Double.parseDouble(next());
}
public char nextChar() throws IOException {
return next().charAt(0);
}
public Long nextLong() throws IOException {
return Long.parseLong(next());
}
public boolean ready() throws IOException {
return br.ready();
}
public void waitForInput() throws InterruptedException {
Thread.sleep(3000);
}
}
}
|
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 β€ n β€ 10^5) β the number of nodes.
Each of the next n-1 lines contains two integers u and v (1 β€ u, v β€ n, u β 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"
]
}
|
CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
int n, cnt[100009];
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
;
cin >> n;
for (int i = 1; i < n; i++) {
int x, y;
cin >> x >> y;
cnt[x]++, cnt[y]++;
}
for (int i = 1; i <= n; i++)
if (cnt[i] == 2) 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 β€ n β€ 10^5) β the number of nodes.
Each of the next n-1 lines contains two integers u and v (1 β€ u, v β€ n, u β 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"
]
}
|
CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
const long long maxn = 1e5 + 10;
long long n, deg[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("");
}
signed main() {
n = read();
bool flag = 0;
for (long long i = 2; i <= n; i++) deg[read()]++, deg[read()]++;
for (long long i = 1; i <= n; i++)
if (deg[i] == 2) {
puts("NO");
flag = 1;
break;
}
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 β€ n β€ 10^5) β the number of nodes.
Each of the next n-1 lines contains two integers u and v (1 β€ u, v β€ n, u β 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"
]
}
|
CORRECT
|
python3
|
n=int(input())
ar=[0]*n
for i in range(n-1):
a,b=map(int,input().split())
ar[a-1]+=1
ar[b-1]+=1
ans="YES"
for i in range(n):
if(ar[i]==2):
ans="NO"
break
print(ans)
|
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 β€ n β€ 10^5) β the number of nodes.
Each of the next n-1 lines contains two integers u and v (1 β€ u, v β€ n, u β 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"
]
}
|
CORRECT
|
java
|
import java.io.*;
import java.util.*;
public class A {
public static void main(String[] args) throws IOException {
Scanner sc = new Scanner();
PrintWriter out = new PrintWriter(System.out);
int n=sc.nextInt();
int []deg=new int [n];
for(int i=1;i<n;i++) {
int u=sc.nextInt()-1,v=sc.nextInt()-1;
deg[u]++;
deg[v]++;
}
boolean no=false;
for(int x:deg)
if(x==2)
no=true;
out.println(no?"NO":"YES");
out.close();
}
static class Scanner {
BufferedReader br;
StringTokenizer st;
Scanner() {
br = new BufferedReader(new InputStreamReader(System.in));
}
Scanner(String fileName) throws FileNotFoundException {
br = new BufferedReader(new FileReader(fileName));
}
String next() throws IOException {
while (st == null || !st.hasMoreTokens())
st = new StringTokenizer(br.readLine());
return st.nextToken();
}
String nextLine() throws IOException {
return br.readLine();
}
int nextInt() throws IOException {
return Integer.parseInt(next());
}
long nextLong() throws NumberFormatException, IOException {
return Long.parseLong(next());
}
double nextDouble() throws NumberFormatException, IOException {
return Double.parseDouble(next());
}
boolean ready() throws IOException {
return br.ready();
}
}
}
|
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 β€ n β€ 10^5) β the number of nodes.
Each of the next n-1 lines contains two integers u and v (1 β€ u, v β€ n, u β 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"
]
}
|
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);
}
int flag=1;
for(int i=0;i<n;i++) {
if(tree[i].size()==2) {
flag=0;
break;
}
}
if(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 β€ n β€ 10^5) β the number of nodes.
Each of the next n-1 lines contains two integers u and v (1 β€ u, v β€ n, u β 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"
]
}
|
CORRECT
|
java
|
//package Contest572;
//package ... ctrl+alt+l //Π²ΡΡΠ°Π²Π½ΡΡΡ ΠΊΠΎΠ΄
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++) {
int a=g.adj[mass1[i]].size();
int b=g.adj[mass2[i]].size();
if(a==2 || b==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 β€ n β€ 10^5) β the number of nodes.
Each of the next n-1 lines contains two integers u and v (1 β€ u, v β€ n, u β 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"
]
}
|
CORRECT
|
python3
|
import sys
input = sys.stdin.readline
def getN():
return int(input())
def getList():
return list(map(int, input().split()))
from sys import exit
n = getN()
vertex = [[] for i in range(n)]
for i in range(n-1):
a, b = getList()
vertex[a-1].append(b-1)
vertex[b - 1].append(a - 1)
for v in vertex:
if len(v) == 2:
print("NO")
exit()
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 β€ n β€ 10^5) β the number of nodes.
Each of the next n-1 lines contains two integers u and v (1 β€ u, v β€ n, u β 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"
]
}
|
CORRECT
|
java
|
import java.io.*;
import java.util.*;
import java.util.stream.*;
import static java.lang.Integer.parseInt;
public class m implements Runnable {
private static final boolean ONLINE_JUDGE = true; //System.getProperty("ONLINE_JUDGE") != null;
public static void main(String[] args) {
new Thread(new m()).start();
}
@Override
public void run() {
try (BufferedReader reader = ONLINE_JUDGE ? new BufferedReader(new InputStreamReader(System.in))
: new BufferedReader(new FileReader("input"));
Writer writer = ONLINE_JUDGE ? new OutputStreamWriter(System.out) : new FileWriter("output");
PrintWriter printWriter = new PrintWriter(writer);) {
Solver solver = new Solver(reader, printWriter);
while (solver.solve());
} catch (IOException e) {
e.printStackTrace(System.err);
}
}
static final class Solver {
final BufferedReader in;
final PrintWriter out;
public Solver(BufferedReader reader, PrintWriter writer) {
in = reader;
out = writer;
}
boolean solve() throws IOException {
String line = in.readLine();
if (line == null) {
return false;
}
if (line.isEmpty()) {
return true;
}
StringTokenizer tok = new StringTokenizer(line);
int n = parseInt(tok.nextToken());
List<Integer> [] ar = new List[n];
int[] deg = new int[n];
for (int i = 1; i < n; ++i) {
tok = new StringTokenizer(in.readLine());
int u = parseInt(tok.nextToken()) - 1;
int v = parseInt(tok.nextToken()) - 1;
deg[u]++;
deg[v]++;
if (ar[u] == null) ar[u] = new ArrayList<Integer>();
if (ar[v] == null) ar[v] = new ArrayList<Integer>();
ar[u].add(v);
ar[v].add(u);
}
for (int i = 0; i < n; ++i) {
if (deg[i] == 2) {
out.println("NO");
return true;
}
}
out.println("YES");
return true;
}
}
static void debug(String s) {
System.out.println(s);
}
}
|
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 β€ n β€ 10^5) β the number of nodes.
Each of the next n-1 lines contains two integers u and v (1 β€ u, v β€ n, u β 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"
]
}
|
CORRECT
|
java
|
import java.io.InputStream;
import java.io.OutputStream;
import java.io.PrintWriter;
import java.util.*;
public class MAIN {
public static void main(String[] args) {
InputStream inputStream = System.in;
OutputStream outputStream = System.out;
Scanner in = new Scanner(inputStream);
PrintWriter out = new PrintWriter(outputStream);
int n = in.nextInt();
List<ArrayList<Integer>> tree = new ArrayList<>(n+1);
for(int i=0;i<=n;i++) tree.add(new ArrayList<>());
for(int i=1;i<n;i++){
int a = in.nextInt();
int b = in.nextInt();
tree.get(a).add(b);
tree.get(b).add(a);
}
for(int i=1;i<=n;i++){
if(tree.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 β€ n β€ 10^5) β the number of nodes.
Each of the next n-1 lines contains two integers u and v (1 β€ u, v β€ n, u β 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"
]
}
|
CORRECT
|
python3
|
n = int(input())
# n, k = map(int, input().split())
# a = [int(i) for i in input().split()]
d = [0 for i in range(n + 1)]
for i in range(n - 1):
a, b = map(int, input().split())
d[a] += 1; d[b] += 1
for val in d:
if val == 2:
print("NO")
exit()
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 β€ n β€ 10^5) β the number of nodes.
Each of the next n-1 lines contains two integers u and v (1 β€ u, v β€ n, u β 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"
]
}
|
CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
vector<long long> tree[500500];
long long num[500500], fine = 0, n;
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);
}
for (int i = 1; i <= n; i++)
if (tree[i].size() == 2) 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 β€ n β€ 10^5) β the number of nodes.
Each of the next n-1 lines contains two integers u and v (1 β€ u, v β€ n, u β 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"
]
}
|
CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
const long long mod = 1000000007;
void solve() {
int n, u, v;
cin >> n;
vector<int> Adj[n];
for (int i = 1; i < n; ++i) {
cin >> u >> v;
--u;
--v;
Adj[u].push_back(v);
Adj[v].push_back(u);
}
for (int i = 0; i < n; ++i) {
if (Adj[i].size() == 2) {
cout << "NO\n";
return;
}
}
cout << "YES\n";
}
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
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 β€ n β€ 10^5) β the number of nodes.
Each of the next n-1 lines contains two integers u and v (1 β€ u, v β€ n, u β 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"
]
}
|
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 - 1; 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 β€ n β€ 10^5) β the number of nodes.
Each of the next n-1 lines contains two integers u and v (1 β€ u, v β€ n, u β 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"
]
}
|
CORRECT
|
java
|
import java.util.*;
import java.io.*;
import java.math.*;
public class Solution{
public static void main(String[] args)throws IOException{
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
StringTokenizer st = new StringTokenizer(br.readLine());
int n = Integer.parseInt(st.nextToken());
int[] deg = new int[n];
for(int i=0;i<n-1;i++){
st = new StringTokenizer(br.readLine());
int l= Integer.parseInt(st.nextToken())-1;
int r= Integer.parseInt(st.nextToken())-1;
deg[l]++;
deg[r]++;
}
boolean ok = true;
for(int i=0;i<n;i++) if(deg[i]==2) ok = false;
if(ok) System.out.println("YES");
else System.out.println("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 β€ n β€ 10^5) β the number of nodes.
Each of the next n-1 lines contains two integers u and v (1 β€ u, v β€ n, u β 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"
]
}
|
CORRECT
|
java
|
import java.io.*;
import java.math.BigInteger;
import java.util.*;
public class GFG770 {
static int mod1 = (int) (1e9 + 7);
static class Reader {
final private int BUFFER_SIZE = 1 << 16;
Scanner scan = new Scanner(new BufferedReader(new InputStreamReader(System.in)));
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 nextString() throws IOException {
String str00 = scan.next();
return str00;
}
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);
}
String next() throws IOException {
int c;
for (c = read(); c <= 32; c = read());
StringBuilder sb = new StringBuilder();
for (; c > 32; c = read()) {
sb.append((char) c);
}
return sb.toString();
}
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 int[] nextArray(int n) throws IOException {
int[] a = new int[n];
for (int i = 0; i < n; i++) {
a[i] = nextInt();
}
return a;
}
}
static long GCD(long a, long b)
{
if (b == 0)
return a;
return GCD(b, a % b);
}
static long power(long x, long y, long p)
{
long res = 1; // Initialize result
x = x % p; // Update x if it is more than or
// equal to p
while (y > 0)
{
// If y is odd, multiply x with result
if (y%2!=0)
res = (res*x) % p;
// y must be even now
y = y>>1; // y = y/2
x = (x*x) % p;
}
return res;
}
static boolean primeCheck(long num0) {
boolean b1 = true;
if (num0 == 1) {
b1 = false;
} else {
int num01 = (int) (Math.sqrt(num0)) + 1;
me1:
for (int i = 2; i < num01; i++) {
if (num0 % i == 0) {
b1 = false;
break me1;
}
}
}
return b1;
}
public static int dev(long num1)
{
int count00=0;
while (num1%2==0)
{
count00++;
num1/=2;
}
HashMap<Long,Long> hashMap=new HashMap<>();
if(count00!=0)
{
hashMap.put(2L,(long)count00);
}
for (int i = 3; i <= (int)Math.sqrt(num1); i = i + 2)
{
// While i divides n, print i and divide n
if(num1%i==0) {
int count01 = 0;
while (num1 % i == 0) {
num1 /= i;
count01++;
}
hashMap.put((long)i,(long)count01);
}
}
if(num1>2)
{
hashMap.put((long)num1,1L);
}
long numOfDiv=1;
for(long num00:hashMap.keySet())
{
long cDiv0=hashMap.get(num00);
numOfDiv*=(cDiv0+1);
}
return (int)(numOfDiv);
}
public static void main(String[] args) throws IOException {
Reader r = new Reader();
//PrintWriter writer=new PrintWriter(System.out);
//Scanner r = new Scanner(new BufferedReader(new InputStreamReader(System.in)));
//Scanner r=new Scanner(System.in);
OutputWriter770 out77 = new OutputWriter770(System.out);
int num1=r.nextInt();
int[] arr1=new int[num1+1];
for(int i=0;i<num1-1;i++)
{
int u = r.nextInt();
int v = r.nextInt();
arr1[u]++;
arr1[v]++;
}
String ans1="YES";
me1: for(int i=1;i<num1+1;i++)
{
if(arr1[i]==2)
{
ans1="NO";
break me1;
}
}
out77.print(ans1+"");
r.close();
out77.close();
}
}
class OutputWriter770
{
BufferedWriter writer;
public OutputWriter770(OutputStream stream)
{
writer = new BufferedWriter(new OutputStreamWriter(stream));
}
public void print(int i) throws IOException
{
writer.write(i + "");
}
public void println(int i) throws IOException
{
writer.write(i + "\n");
}
public void print(String s) throws IOException
{
writer.write(s + "");
}
public void println(String s) throws IOException
{
writer.write(s + "\n");
}
public void print(char[] c) throws IOException
{
writer.write(c);
}
public void close() throws IOException
{
writer.flush();
writer.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 β€ n β€ 10^5) β the number of nodes.
Each of the next n-1 lines contains two integers u and v (1 β€ u, v β€ n, u β 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"
]
}
|
CORRECT
|
python3
|
n = int(input())
adj = [0 for i in range(n)]
for i in range(n-1):
iarr = list(map(int,input().split()))
u = iarr[0]-1
v = iarr[1]-1
adj[u]+=1
adj[v]+=1
flag = 0
for i in adj:
if i==2:
flag=1
break
if flag==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 β€ n β€ 10^5) β the number of nodes.
Each of the next n-1 lines contains two integers u and v (1 β€ u, v β€ n, u β 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"
]
}
|
CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
int main() {
int n;
cin >> n;
vector<vector<int>> g(n);
int f, t;
for (int i = 0; i < n - 1; ++i) {
cin >> f >> t;
f--;
t--;
g[f].push_back(t);
g[t].push_back(f);
}
for (int i = 0; i < n; ++i) {
if (g[i].size() == 2) {
cout << "NO\n";
return 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 β€ n β€ 10^5) β the number of nodes.
Each of the next n-1 lines contains two integers u and v (1 β€ u, v β€ n, u β 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"
]
}
|
CORRECT
|
python3
|
n=int(input())
d={}
o,t=0,0
for _ in range(n-1):
s,e=map(int,input().split())
if s in d:
d[s]+=1
else:
d[s]=1
if e in d:
d[e]+=1
else:
d[e]=1
for i in d:
if d[i]==1:
o+=1
elif d[i]==2:
t+=1
if t:
print('NO')
else:
if (o*(o-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 β€ n β€ 10^5) β the number of nodes.
Each of the next n-1 lines contains two integers u and v (1 β€ u, v β€ n, u β 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"
]
}
|
CORRECT
|
python3
|
from sys import stdin
input = stdin.readline
n = int(input())
degree = [0 for i in range(n+1)]
for _ in range(n-1):
i, j = [int(i) for i in input().split()]
degree[i] += 1
degree[j] += 1
res = False
for i in range(1, n+1):
if degree[i] == 2: res = True
if res: 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 β€ n β€ 10^5) β the number of nodes.
Each of the next n-1 lines contains two integers u and v (1 β€ u, v β€ n, u β 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"
]
}
|
CORRECT
|
java
|
import java.util.*;
public class Graph{
int V;
ArrayList<Integer>[] adj;
Graph(int v){
V=v;
adj=new ArrayList[V+1];
for(int i=1;i<V+1;i++) {
adj[i]=new ArrayList<>();
}
}
void addEdge(int a,int b) {
adj[a].add(b);
adj[b].add(a);
}
void check() {
int three=0;
int leaf=0;
int ans=1;
for(int i=1;i<adj.length;i++) {
if(adj[i].size()==2) {
ans=0;
break;
}
if(adj[i].size()==1) {
leaf++;
}
if(adj[i].size()>=3) {
three++;
}
}
if(ans==0) {
System.out.println("NO");
}else {
System.out.println("YES");
}
}
public static void main(String[] args) {
Scanner ip=new Scanner(System.in);
int n=ip.nextInt();
Graph g=new Graph(n);
for(int i=1;i<n;i++) {
int p=ip.nextInt();
int c=ip.nextInt();
g.addEdge(p,c);
}
if(n==2) {
System.out.println("YES");
}else {
g.check();
}
}
}
|
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 β€ n β€ 10^5) β the number of nodes.
Each of the next n-1 lines contains two integers u and v (1 β€ u, v β€ n, u β 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"
]
}
|
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()
#nγ4δ»₯δΈ
for i in range(n):
if len(tree[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 β€ n β€ 10^5) β the number of nodes.
Each of the next n-1 lines contains two integers u and v (1 β€ u, v β€ n, u β 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"
]
}
|
CORRECT
|
python3
|
if __name__ == '__main__':
n = int(input())
g = [[] for _ in range(n)]
for _ in range(n - 1):
u, v = map(int, input().split())
g[u - 1].append(v - 1)
g[v - 1].append(u - 1)
if list(filter(lambda x: len(x) == 2, g)):
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 β€ n β€ 10^5) β the number of nodes.
Each of the next n-1 lines contains two integers u and v (1 β€ u, v β€ n, u β 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"
]
}
|
CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 10;
int a[N];
int main() {
int n;
scanf("%d", &n);
for (int i = 1; i < n; i++) {
int x, y;
scanf("%d%d", &x, &y);
a[x]++;
a[y]++;
}
for (int i = 1; i <= n; i++)
if (a[i] == 2) {
printf("NO\n");
return 0;
}
printf("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 β€ n β€ 10^5) β the number of nodes.
Each of the next n-1 lines contains two integers u and v (1 β€ u, v β€ n, u β 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"
]
}
|
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 Vaibhav Pulastya
*/
public class Main {
public static void main(String[] args) {
InputStream inputStream = System.in;
OutputStream outputStream = System.out;
InputReader in = new InputReader(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, InputReader in, PrintWriter out) {
int n = in.nextInt();
int[] deg = new int[n];
for (int i = 0; i < n - 1; i++) {
int u = in.nextInt() - 1;
int v = in.nextInt() - 1;
deg[u]++;
deg[v]++;
}
for (int i = 0; i < n; i++) {
if (deg[i] == 2) {
out.println("NO");
return;
}
}
out.println("YES");
}
}
static class InputReader {
private InputStream stream;
private byte[] buf = new byte[1024];
private int curChar;
private int numChars;
public InputReader(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;
}
private boolean isSpaceChar(int c) {
return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
}
}
}
|
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 β€ n β€ 10^5) β the number of nodes.
Each of the next n-1 lines contains two integers u and v (1 β€ u, v β€ n, u β 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"
]
}
|
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 β€ n β€ 10^5) β the number of nodes.
Each of the next n-1 lines contains two integers u and v (1 β€ u, v β€ n, u β 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"
]
}
|
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(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 β€ n β€ 10^5) β the number of nodes.
Each of the next n-1 lines contains two integers u and v (1 β€ u, v β€ n, u β 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"
]
}
|
CORRECT
|
python3
|
from collections import defaultdict
g = defaultdict(set)
n = int(input())
for _ in range(n - 1):
u, v = map(int, input().split())
g[u].add(v)
g[v].add(u)
def solve(n, g):
for i in range(1, n + 1):
if len(g[i]) == 1 or len(g[i]) >= 3:
continue
else:
return False
return True
solution = solve(n, g)
print('YES' if solution 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 β€ n β€ 10^5) β the number of nodes.
Each of the next n-1 lines contains two integers u and v (1 β€ u, v β€ n, u β 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"
]
}
|
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.second].size() == 2 || 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 β€ n β€ 10^5) β the number of nodes.
Each of the next n-1 lines contains two integers u and v (1 β€ u, v β€ n, u β 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"
]
}
|
CORRECT
|
java
|
import java.io.*;
import java.util.*;
public class D {
void run() {
int n = in.nextInt();
int[] d = new int[n];
int i,u,v;
for(i=0;i<n-1;i++){
u = in.nextInt() - 1;
v = in.nextInt() - 1;
d[u]++;
d[v]++;
}
for(i=0;i<n;i++)
if(d[i] == 2)
break;
out.println(i < n ? "NO" : "YES");
}
static MyScanner in;
static PrintWriter out;
public static void main(String[] args) throws IOException {
boolean stdStream = true;
if(stdStream){
in = new MyScanner();
out = new PrintWriter(System.out);
}else{
String fileName = D.class.getSimpleName();
int n_test = 1;
String inputFileName = fileName + String.format(".%02d", n_test) + ".inp";
String outputFileName = fileName + String.format(".%02d", n_test) + ".out";
in = new MyScanner(new BufferedReader(new FileReader(inputFileName)));
out = new PrintWriter(outputFileName);
}
new D().run();
in.close();
out.close();
}
static class MyScanner {
BufferedReader br;
StringTokenizer st;
MyScanner(){
this.br = new BufferedReader(new InputStreamReader(System.in));
}
MyScanner(BufferedReader br) {
this.br = br;
}
void close() throws IOException {
br.close();
}
void findToken() {
while (st == null || !st.hasMoreTokens()) {
try {
st = new StringTokenizer(br.readLine());
} catch (IOException e) {
throw new RuntimeException(e);
}
}
}
String next() {
findToken();
return st.nextToken();
}
int nextInt() {
return Integer.parseInt(next());
}
long nextLong() {
return Long.parseLong(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 β€ n β€ 10^5) β the number of nodes.
Each of the next n-1 lines contains two integers u and v (1 β€ u, v β€ n, u β 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"
]
}
|
CORRECT
|
python2
|
#!/usr/bin/env python
from __future__ import division, print_function
import os
import sys
from io import BytesIO, IOBase
if sys.version_info[0] < 3:
from __builtin__ import xrange as range
from future_builtins import ascii, filter, hex, map, oct, zip
import random
import collections
import math
import itertools
import bisect
def real_main():
n = read_int()
deg = [0]*n
for _ in range(n-1):
a, b = read_int_array()
deg[a-1] += 1
deg[b-1] += 1
if any(x == 2 for x in deg):
print("NO")
else:
print("YES")
def solve():
pass
def main():
if False and 'PYCHARM_HOSTED' in os.environ:
main_pycharm()
else:
real_main()
from copy import deepcopy
def main_pycharm():
solution = solve
test_inputs = None
test_outputs = None
judge = None
slow_solution = None
if solution is not None:
if test_outputs is not None:
test_with_answers(solution, test_inputs, test_outputs)
if judge is not None:
test_with_judge(solution, test_inputs, judge)
if slow_solution is not None:
test_with_slow_solution(solution, test_inputs, slow_solution)
def test_with_answers(solution, inputs, answers):
total, wrong = 0, 0
for args, test_ans in zip(inputs, answers):
ans = solution(*deepcopy(args))
if ans != test_ans:
print('WRONG! ans=%s, test_ans=%s, args=%s' % (ans, test_ans, args))
wrong += 1
else:
print('GOOD')
total += 1
print('ALL %d TESTS PASSED' % total if not wrong else '%d out of %d tests are WRONG' % (wrong, total))
def test_with_judge(solution, inputs_gen, judge):
total, wrong = 0, 0
for args in inputs_gen:
ans = solution(*deepcopy(args))
if not judge(deepcopy(ans), *deepcopy(args)):
print('WRONG! ans=%s, args=%s' % (ans, args))
wrong += 1
total += 1
print('ALL %d TESTS PASSED' % total if not wrong else '%d out of %d tests are WRONG' % (wrong, total))
def test_with_slow_solution(solution, inputs_gen, solution_slow):
total, wrong = 0, 0
for args in inputs_gen:
ans = solution(*deepcopy(args))
slow = solution_slow(*deepcopy(args))
if ans != slow:
print('WRONG! ans=%s, slow=%s, args=%s' % (ans, slow, args))
wrong += 1
total += 1
print('ALL %d TESTS PASSED' % total if not wrong else '%d out of %d tests are WRONG' % (wrong, total))
def generate_nums(n, min, max, check_if_good=None):
while True:
nums = [random.randint(min, max) for _ in range(n)]
if check_if_good is None or check_if_good(nums):
return nums
# This mergesort can be like 7 times faster than build in sort
# (for stupid reasons)
def mergesort(A, key=lambda x: x, reverse=False):
C = A
A = list(range(len(A)))
B = list(A)
n = len(A)
for i in range(0, n - 1, 2):
if key(C[A[i]]) > key(C[A[i ^ 1]]):
A[i], A[i ^ 1] = A[i ^ 1], A[i]
width = 2
while width < n:
for i in range(0, n, 2 * width):
R1, R2 = min(i + width, n), min(i + 2 * width, n)
j, k = R1, i
while i < R1 and j < R2:
if key(C[A[i]]) > key(C[A[j]]):
B[k] = A[j]
j += 1
else:
B[k] = A[i]
i += 1
k += 1
while i < R1:
B[k] = A[i]
k += 1
i += 1
while k < R2:
B[k] = A[k]
k += 1
A, B = B, A
width *= 2
if reverse:
A.reverse()
return A
def mergesort_simple(A, reverse=False):
C = A
A = list(range(len(A)))
B = list(A)
n = len(A)
for i in range(0, n - 1, 2):
if C[A[i]] > C[A[i ^ 1]]:
A[i], A[i ^ 1] = A[i ^ 1], A[i]
width = 2
while width < n:
for i in range(0, n, 2 * width):
R1, R2 = min(i + width, n), min(i + 2 * width, n)
j, k = R1, i
while i < R1 and j < R2:
if C[A[i]] > C[A[j]]:
B[k] = A[j]
j += 1
else:
B[k] = A[i]
i += 1
k += 1
while i < R1:
B[k] = A[i]
k += 1
i += 1
while k < R2:
B[k] = A[k]
k += 1
A, B = B, A
width *= 2
if reverse:
A.reverse()
return A
# region fastio
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
def print(*args, **kwargs):
"""Prints the values to a stream, or to sys.stdout by default."""
sep, file = kwargs.pop("sep", " "), kwargs.pop("file", sys.stdout)
at_start = True
for x in args:
if not at_start:
file.write(sep)
file.write(str(x))
at_start = False
file.write(kwargs.pop("end", "\n"))
if kwargs.pop("flush", False):
file.flush()
if sys.version_info[0] < 3:
sys.stdin, sys.stdout = FastIO(sys.stdin), FastIO(sys.stdout)
else:
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
def read():
return input()
def read_int():
return int(input())
def read_array(sep=None, maxsplit=-1):
return input().split(sep, maxsplit)
def read_int_array():
return [int(a) for a in read_array()]
# endregion
if __name__ == "__main__":
main()
|
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 β€ n β€ 10^5) β the number of nodes.
Each of the next n-1 lines contains two integers u and v (1 β€ u, v β€ n, u β 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"
]
}
|
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(count==0&&c<3||count==1&&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 β€ n β€ 10^5) β the number of nodes.
Each of the next n-1 lines contains two integers u and v (1 β€ u, v β€ n, u β 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"
]
}
|
CORRECT
|
python2
|
n = input()
adj = [ [] for i in range(n+1) ]
for i in range(n-1):
u, v = map(int,raw_input().split())
adj[u].append(v)
adj[v].append(u)
for i in range(1,n+1):
if len(adj[i]) == 2:
print "NO"
exit()
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 β€ n β€ 10^5) β the number of nodes.
Each of the next n-1 lines contains two integers u and v (1 β€ u, v β€ n, u β 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"
]
}
|
CORRECT
|
java
|
/*
9 2
1 1 1 1 1 0 0 0 0
1 2
1 5
5 6
5 7
2 3
2 4
3 8
3 9
*/
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.util.*;
public class Question {
static int[][] arr;
static LinkedList[] adj;
static boolean visited[];
static boolean ans = true;
public static void main (String[] args) throws IOException {
Reader.init(System.in);
int n = Reader.nextInt();
int[] arr = new int[n];
adj = new LinkedList[n+1];
visited = new boolean[n+1];
for (int i = 1 ; i < n+1 ; i++){
adj[i] = new LinkedList<>();
}
for (int i = 0 ; i < n-1 ; i++){
int n1= Reader.nextInt();
int n2 = Reader.nextInt();
adj[n1].addLast(n2);
adj[n2].addLast(n1);
}
int leaf = 0;
for (int i = 1 ; i <=n ; i++){
if (adj[i].size()==1){
leaf = i;
break;
}
}
DFSUtil(leaf,0);
if (ans){
System.out.println("YES");
}
else{
System.out.println("NO");
}
}
static void DFSUtil(int v , int num)
{
// Mark the current node as visited and print it
visited[v] = true;
//System.out.print(v+" ");
if (adj[v].size()==2 && num==1){
ans&=false;
}
// Recur for all the vertices adjacent to this vertex
Iterator<Integer> i = adj[v].listIterator();
while (i.hasNext())
{
int n = i.next();
if (!visited[n])
DFSUtil(n,1);
}
}
}
/*This is a function problem.You only need to complete the function given below*/
/*Complete the function below*/
/*
ArrayList<ArrayList<Integer>> list: to represent graph containing 'v'
vertices and edges between them
V: represent number of vertices
*/
class DetectCycle
{
static int ans = 0;
static boolean isCyclic(ArrayList<ArrayList<Integer>> list, int V)
{
ans = 0;
for(int i = 0 ; i < V ; i++){
dfs(i , new boolean[V],list,new boolean[V]);
if(ans==1){
return true;
}
}
return false;
}
static void dfs(int src , boolean[] visited,ArrayList<ArrayList<Integer>> list,boolean[] stack){
if(stack[src]==true){
ans = 1;
return;
}
if (visited[src]){
return;
}
stack[src] = true;
visited[src] = true;
for(int i = 0 ; i < list.get(src).size() ;i++){
//int ans = 0;
int n = list.get(src).get(i);
if(!visited[n]){
dfs(n,visited , list,stack);
}
else{
System.out.println(n);
ans = 1;
}
}
stack[src] = false;
}
static void colSwap(int[][] arr , int x , int y,int m , int n){
for (int i = 0 ; i < n ; i++){
int temp = arr[i][x];
arr[i][x] = arr[i][y];
arr[i][y] = temp;
}
}
public static void sortbyColumn(int arr[][], int col)
{
// Using built-in sort function Arrays.sort
Arrays.sort(arr, new Comparator<int[]>() {
@Override
// Compare values according to columns
public int compare(final int[] entry1,
final int[] entry2) {
// To sort in descending order revert
// the '>' Operator
if (entry1[col] > entry2[col])
return 1;
else
return -1;
}
}); // End of function call sort().
}
}
class Reader {
static BufferedReader reader;
static StringTokenizer tokenizer;
/** call this method to initialize reader for InputStream */
static void init(InputStream input) {
reader = new BufferedReader(
new InputStreamReader(input) );
tokenizer = new StringTokenizer("");
}
/** get next word */
static String next() throws IOException {
while ( ! tokenizer.hasMoreTokens() ) {
//TODO add check for eof if necessary
tokenizer = new StringTokenizer(
reader.readLine() );
}
return tokenizer.nextToken();
}
static int nextInt() throws IOException {
return Integer.parseInt( next() );
}
static long nextLong() throws IOException {
return Long.parseLong( next() );
}
static double nextDouble() throws IOException {
return Double.parseDouble( next() );
}
}
class MergeSort
{
// Merges two subarrays of arr[].
// First subarray is arr[l..m]
// Second subarray is arr[m+1..r]
void merge(int arr[], int l, int m, int r)
{
// Find sizes of two subarrays to be merged
int n1 = m - l + 1;
int n2 = r - m;
/* Create temp arrays */
int L[] = new int [n1];
int R[] = new int [n2];
/*Copy data to temp arrays*/
for (int i=0; i<n1; ++i)
L[i] = arr[l + i];
for (int j=0; j<n2; ++j)
R[j] = arr[m + 1+ j];
/* Merge the temp arrays */
// Initial indexes of first and second subarrays
int i = 0, j = 0;
// Initial index of merged subarry array
int k = l;
while (i < n1 && j < n2)
{
if (L[i] <= R[j])
{
arr[k] = L[i];
i++;
}
else
{
arr[k] = R[j];
j++;
}
k++;
}
/* Copy remaining elements of L[] if any */
while (i < n1)
{
arr[k] = L[i];
i++;
k++;
}
/* Copy remaining elements of R[] if any */
while (j < n2)
{
arr[k] = R[j];
j++;
k++;
}
}
// Main function that sorts arr[l..r] using
// merge()
void sort(int arr[], int l, int r)
{
if (l < r)
{
// Find the middle point
int m = (l+r)/2;
// Sort first and second halves
sort(arr, l, m);
sort(arr , m+1, r);
// Merge the sorted halves
merge(arr, l, m, r);
}
}
/* A utility function to print array of size n */
static void printArray(int arr[])
{
int n = arr.length;
for (int i=0; i<n; ++i)
System.out.print(arr[i] + " ");
System.out.println();
}
// Driver method
}
class Node{
int data;
int freq;
public Node(int data, int freq) {
this.data = data;
this.freq = freq;
}
}
class GFG {
// limit for array size
static int N = 1000000;
static int n; // array size
// Max size of tree
static int []tree = new int[2 * N];
// function to build the tree
static void build( char []arr,int n)
{
// insert leaf nodes in tree
for (int i = 0; i < n-1; i++) {
if (arr[i]!=arr[i+1]) {
System.out.println(i);
tree[n + i] = 1;
}
}
// build the tree by calculating
// parents
for (int i = n - 1; i > 0; --i)
tree[i] = tree[i << 1] +
tree[i << 1 | 1];
}
static int query(int l, int r)
{
int res = 0;
// loop to find the sum in the range
for (l += n, r += n; l < r;
l >>= 1, r >>= 1)
{
if ((l & 1) > 0)
res += tree[l++];
if ((r & 1) > 0)
res += tree[--r];
}
return res;
}
// driver program to test the
// above function
}
class SegmentTree{
int[] arr ;
int[] tree;
SegmentTree(int[] arr ,int n){
this.arr = arr;
tree = new int[2*n];
}
void updateRangeUtil(int si, int ss, int se, int us,
int ue, int val)
{
if (ss>se || ss>ue || se<us)
return ;
if (ss==se)
{
tree[si] |= val;
return;
}
int mid = (ss+se)/2;
updateRangeUtil(si*2+1, ss, mid, us, ue, val);
updateRangeUtil(si*2+2, mid+1, se, us, ue, val);
tree[si] = tree[si*2+1] + tree[si*2+2];
}
void constructSTUtil(int ss, int se, int si)
{
// out of range as ss can never be greater than se
if (ss > se)
return ;
// If there is one element in array, store it in
// current node of segment tree and return
if (ss == se)
{
//tree[si] = arr[ss];
System.out.println(tree[si]);
return;
}
// If there are more than one elements, then recur
// for left and right subtrees and store the sum
// of values in this node
int mid = (ss + se)/2;
constructSTUtil(ss, mid, si*2+1);
constructSTUtil(mid+1, se, si*2+2);
//tree[si] = tree[si*2 + 1] + tree[si*2 + 2];
}
}
class Heap{
int[] Heap;
int size;
int maxSize;
public Heap(int maxSize){
this.maxSize = maxSize;
this.size = 0;
Heap = new int[maxSize + 1];
Heap[0] = Integer.MIN_VALUE;
}
int parent(int pos){
return pos/2;
}
int left(int pos){
return 2*pos;
}
int right(int pos){
return 2*pos + 1;
}
boolean isLeaf(int pos){
if (pos>=size/2 && pos<=size){
return true;
}
return false;
}
void swap(int i , int j){
int temp = Heap[i];
Heap[i] = Heap[j];
Heap[j]= temp;
}
void minHeapify(int pos){
if (!isLeaf(pos)){
if (Heap[pos]>Heap[left(pos)] || Heap[pos] > Heap[right(pos)]){
if (Heap[left(pos)] < Heap[right(pos)]) {
swap(pos, left(pos));
minHeapify(left(pos));
}
// Swap with the right child and heapify
// the right child
else {
swap(pos, right(pos));
minHeapify(right(pos));
}
}
}
}
}
class Graph{
}
class DijkstraNode implements Comparator<DijkstraNode>{
public int node;
public int cost;
public int[] visited;
public boolean[] settled;
public DijkstraNode(int node, int cost,int vertices) {
this.node = node;
this.cost = cost;
visited = new int[vertices];
settled = new boolean[vertices];
}
@Override
public int compare(DijkstraNode node1 , DijkstraNode node2){
if (node1.cost < node2.cost){
return -1;
}
if (node1.cost > node2.cost){
return 1;
}
return 0;
}
ArrayList<Integer> list = new ArrayList<>(10000);
}
|
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 β€ n β€ 10^5) β the number of nodes.
Each of the next n-1 lines contains two integers u and v (1 β€ u, v β€ n, u β 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"
]
}
|
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] == 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 β€ n β€ 10^5) β the number of nodes.
Each of the next n-1 lines contains two integers u and v (1 β€ u, v β€ n, u β 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"
]
}
|
CORRECT
|
python3
|
from collections import Counter
from sys import stdin
n=int(input())
lst=[]
for _ in range(n-1):
a,b=map(int,stdin.readline().split())
lst.append(a)
lst.append(b)
fg=0
for _,i in Counter(lst).items():
if i==2:
fg=1
break
if fg==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 β€ n β€ 10^5) β the number of nodes.
Each of the next n-1 lines contains two integers u and v (1 β€ u, v β€ n, u β 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"
]
}
|
CORRECT
|
cpp
|
#include <bits/stdc++.h>
#pragma GCC optimize("-O2")
using namespace std;
const int LIM = 1e5 + 5, MOD = 1e9 + 7;
const long double EPS = 1e-9;
vector<vector<int> > g;
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
int n;
cin >> n;
g.resize(n);
for (int i = 0; i < n - 1; ++i) {
int u, v;
cin >> u >> v;
u--, v--;
g[u].push_back(v);
g[v].push_back(u);
}
for (int i = 0; i < n; ++i) {
if (g[i].size() == 2) {
cout << "NO" << '\n';
exit(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 β€ n β€ 10^5) β the number of nodes.
Each of the next n-1 lines contains two integers u and v (1 β€ u, v β€ n, u β 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"
]
}
|
CORRECT
|
python3
|
n = int(input())
edges = [0]*(n+1)
isPossible = True
for i in range(n-1):
a,b = [int(i) for i in input().split()]
edges[a]+=1
edges[b]+=1
for d in edges:
if d == 2:
isPossible = False
break
print("YES" if isPossible 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 β€ n β€ 10^5) β the number of nodes.
Each of the next n-1 lines contains two integers u and v (1 β€ u, v β€ n, u β 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"
]
}
|
CORRECT
|
python3
|
import sys
N = int(input())
adj_list = [[] for _ in range(N)]
for _ in range(N-1):
(u,v) = list(map(int, input().split()))
u -= 1
v -= 1
adj_list[u].append(v)
adj_list[v].append(u)
num_leaf = 0
for node in range(N):
if len(adj_list[node]) == 2:
print("NO")
sys.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 β€ n β€ 10^5) β the number of nodes.
Each of the next n-1 lines contains two integers u and v (1 β€ u, v β€ n, u β 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"
]
}
|
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)==2:
exit(print('NO'))
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 β€ n β€ 10^5) β the number of nodes.
Each of the next n-1 lines contains two integers u and v (1 β€ u, v β€ n, u β 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"
]
}
|
CORRECT
|
python3
|
# |
# _` | __ \ _` | __| _ \ __ \ _` | _` |
# ( | | | ( | ( ( | | | ( | ( |
# \__,_| _| _| \__,_| \___| \___/ _| _| \__,_| \__,_|
import sys
import math
def read_line():
return sys.stdin.readline()[:-1]
def read_int():
return int(sys.stdin.readline())
def read_int_line():
return [int(v) for v in sys.stdin.readline().split()]
def read_float_line():
return [float(v) for v in sys.stdin.readline().split()]
n = read_int()
d = {}
for i in range(n-1):
u, v = read_int_line()
if u in d:
d[u] += 1
else:
d[u] = 1
if v in d:
d[v] += 1
else:
d[v] = 1
f = False
for i in d:
if d[i]==2:
f = True
if f:
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 β€ n β€ 10^5) β the number of nodes.
Each of the next n-1 lines contains two integers u and v (1 β€ u, v β€ n, u β 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"
]
}
|
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], c = 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] == 2) c++;
}
w.println(c > 0 ? "NO" : "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 β€ n β€ 10^5) β the number of nodes.
Each of the next n-1 lines contains two integers u and v (1 β€ u, v β€ n, u β 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"
]
}
|
CORRECT
|
java
|
import java.io.*;
import java.util.*;
public class R572D {
public static void main(String[] args) {
JS scan = new JS();
int n = scan.nextInt();
Integer[] arr =new Integer[n+1];
Arrays.fill(arr,0);
for(int i = 0;i<n-1;i++){
int a = scan.nextInt();
int b = scan.nextInt();
arr[a]++;
arr[b]++;
}
for(int i = 1;i<=n;i++){
if(arr[i]==2){
System.out.println("NO");
return;
}
}
System.out.println("YES");
}
static class JS{
public int BS = 1<<16;
public char NC = (char)0;
byte[] buf = new byte[BS];
int bId = 0, size = 0;
char c = NC;
double num = 1;
BufferedInputStream in;
public JS() {
in = new BufferedInputStream(System.in, BS);
}
public JS(String s) throws FileNotFoundException {
in = new BufferedInputStream(new FileInputStream(new File(s)), BS);
}
public char nextChar(){
while(bId==size) {
try {
size = in.read(buf);
}catch(Exception e) {
return NC;
}
if(size==-1)return NC;
bId=0;
}
return (char)buf[bId++];
}
public int nextInt() {
return (int)nextLong();
}
public long nextLong() {
num=1;
boolean neg = false;
if(c==NC)c=nextChar();
for(;(c<'0' || c>'9'); c = nextChar()) {
if(c=='-')neg=true;
}
long res = 0;
for(; c>='0' && c <='9'; c=nextChar()) {
res = (res<<3)+(res<<1)+c-'0';
num*=10;
}
return neg?-res:res;
}
public double nextDouble() {
double cur = nextLong();
return c!='.' ? cur:cur+nextLong()/num;
}
public String next() {
StringBuilder res = new StringBuilder();
while(c<=32)c=nextChar();
while(c>32) {
res.append(c);
c=nextChar();
}
return res.toString();
}
public String nextLine() {
StringBuilder res = new StringBuilder();
while(c<=32)c=nextChar();
while(c!='\n') {
res.append(c);
c=nextChar();
}
return res.toString();
}
public boolean hasNext() {
if(c>32)return true;
while(true) {
c=nextChar();
if(c==NC)return false;
else if(c>32)return true;
}
}
}
}
|
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 β€ n β€ 10^5) β the number of nodes.
Each of the next n-1 lines contains two integers u and v (1 β€ u, v β€ n, u β 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"
]
}
|
CORRECT
|
python3
|
def ii():
return int(input())
def ss():
return [x for x in input()]
def si():
return [int(x) for x in input().split()]
def mi():
return map(int, input().split())
a = ii()
s = [0 for i in range(a)]
for i in range(a - 1):
c, d = [int(x) - 1 for x in input().split()]
s[c] += 1
s[d] += 1
if 2 in s:
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 β€ n β€ 10^5) β the number of nodes.
Each of the next n-1 lines contains two integers u and v (1 β€ u, v β€ n, u β 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"
]
}
|
CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
void func() {
int n;
cin >> n;
vector<int> adj[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);
}
for (int i = 1; i <= n; i++) {
if (adj[i].size() == 1) continue;
if (adj[i].size() < 3) {
cout << "NO" << endl;
return;
}
}
cout << "YES" << endl;
return;
}
int main() {
std::ios::sync_with_stdio(false);
int t = 1;
while (t--) func();
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 β€ n β€ 10^5) β the number of nodes.
Each of the next n-1 lines contains two integers u and v (1 β€ u, v β€ n, u β 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"
]
}
|
CORRECT
|
python3
|
import sys
input=sys.stdin.readline
from collections import deque
n=int(input())
if n==2:
print('YES')
exit()
if n==3:
print('NO')
exit()
Edges=[[] for _ in range(n)]
for _ in range(n-1):
u,v=map(lambda x: int(x)-1,input().split())
Edges[u].append(v)
Edges[v].append(u)
for i,E in enumerate(Edges):
if len(E)>=3:
root=i
break
else:
print('NO')
exit()
Chi=[[] for _ in range(n)]
Par=[0]*n
q=deque()
q.append(root)
Used=[False]*n
Used[root]=True
while q:
v=q.popleft()
for c in Edges[v]:
if Used[c]:
continue
Chi[v].append(c)
Par[c]=v
Used[c]=True
q.append(c)
Leaf=[]
for v,l in enumerate(Chi):
if not l:
Leaf.append(v)
for l in Leaf:
while True:
p=Par[l]
if p==root:
break
if len(Chi[p])==1:
print('NO')
exit()
l=p
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 β€ n β€ 10^5) β the number of nodes.
Each of the next n-1 lines contains two integers u and v (1 β€ u, v β€ n, u β 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"
]
}
|
CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
const long long MAX = 1e5 + 10;
long long arr[MAX];
int main() {
ios_base::sync_with_stdio(false);
cin.tie(0);
int n;
cin >> n;
for (int i = 0; i < n - 1; i++) {
int u, v;
cin >> u >> v;
arr[u]++;
arr[v]++;
}
for (int i = 1; i <= n; i++)
if (arr[i] == 2) return puts("NO");
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 β€ n β€ 10^5) β the number of nodes.
Each of the next n-1 lines contains two integers u and v (1 β€ u, v β€ n, u β 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"
]
}
|
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];
int main() {
scanf("%d", &n);
for (int i = 1; i < n; ++i) {
scanf("%d %d", &u, &v);
++deg[u], ++deg[v];
}
for (int i = 1; i <= n; ++i) {
if (deg[i] == 2) {
printf("NO\n");
return 0;
}
}
printf("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 β€ n β€ 10^5) β the number of nodes.
Each of the next n-1 lines contains two integers u and v (1 β€ u, v β€ n, u β 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"
]
}
|
CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
int a[100010];
int main() {
int n, flag = 1;
scanf("%d", &n);
for (int i = 1, x, y; i < n; i++) {
scanf("%d %d", &x, &y);
a[x]++;
a[y]++;
}
for (int i = 1; i <= n; i++)
if (a[i] == 2) flag = 0;
flag ? printf("YES") : 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 β€ n β€ 10^5) β the number of nodes.
Each of the next n-1 lines contains two integers u and v (1 β€ u, v β€ n, u β 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"
]
}
|
CORRECT
|
java
|
import java.io.*;
import java.util.*;
public class Main {
public static final boolean DO_LOG = false;
static class Task {
Logger log = new Logger();
PrintWriter out;
MyScanner in;
public void solve(MyScanner in, PrintWriter out) {
int n = in.nextInt();
int[] d = new int[n + 1];
for (int i = 0; i < n -1 ; i++) {
int fr = in.nextInt();
int to = in.nextInt();
d[fr]++;
d[to]++;
}
for (int i = 1; i <= n; i++) {
log.info("d[" + i + "] = " + d[i]);
}
for (int i = 1; i <= n; i++) {
if (d[i] == 2) {
out.println("NO");
return;
}
}
out.println("YES");
}
}
public static void main(String[] args) {
MyScanner in = new MyScanner();
PrintWriter out = new PrintWriter(System.out);
Task solver = new Task();
solver.solve(in, out);
out.close();
}
public static class Logger {
public void info(String s) {
if (DO_LOG) {
System.out.println(s);
}
}
}
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 β€ n β€ 10^5) β the number of nodes.
Each of the next n-1 lines contains two integers u and v (1 β€ u, v β€ n, u β 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"
]
}
|
CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
int deg[100005];
int main() {
int n;
scanf("%d", &n);
for (int i = 1; i < n; i++) {
int u, v;
scanf("%d%d", &u, &v);
deg[u]++;
deg[v]++;
}
for (int i = 1; i <= n; i++) {
if (deg[i] == 2) {
printf("NO\n");
return 0;
}
}
printf("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 β€ n β€ 10^5) β the number of nodes.
Each of the next n-1 lines contains two integers u and v (1 β€ u, v β€ n, u β 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"
]
}
|
CORRECT
|
java
|
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.List;
import java.util.StringTokenizer;
public class D {
List<Integer>[] g;
void solve(){
int n = readInt();
g = new List[n];
for(int i = 0;i<n;i++){
g[i] = new ArrayList<>();
}
for(int i = 0;i<n - 1;i++){
int from = readInt() - 1;
int to = readInt() - 1;
g[from].add(to);
g[to].add(from);
}
for(int i = 0;i<n;i++){
for(int j : g[i]){
if(g[j].size() == 2){
out.print("NO");
return;
}
}
}
out.print("YES");
}
public static void main(String[] args) {
new D().run();
}
void run(){
init();
solve();
out.close();
}
BufferedReader in;
PrintWriter out;
StringTokenizer tok = new StringTokenizer("");
void init(){
in = new BufferedReader(new InputStreamReader(System.in));
out = new PrintWriter(System.out);
}
String readLine(){
try{
return in.readLine();
}catch(Exception ex){
throw new RuntimeException(ex);
}
}
String readString(){
while(!tok.hasMoreTokens()){
String nextLine = readLine();
if(nextLine == null) return null;
tok = new StringTokenizer(nextLine);
}
return tok.nextToken();
}
int readInt(){
return Integer.parseInt(readString());
}
long readLong(){
return Long.parseLong(readString());
}
double readDouble(){
return Double.parseDouble(readString());
}
}
|
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 β€ n β€ 10^5) β the number of nodes.
Each of the next n-1 lines contains two integers u and v (1 β€ u, v β€ n, u β 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"
]
}
|
CORRECT
|
java
|
//make sure to make new file!
import java.io.*;
import java.util.*;
public class D572A{
public static void main(String[] args)throws IOException{
BufferedReader f = new BufferedReader(new InputStreamReader(System.in));
PrintWriter out = new PrintWriter(System.out);
int n = Integer.parseInt(f.readLine());
ArrayList<ArrayList<Integer>> adj = new ArrayList<ArrayList<Integer>>(n+1);
for(int k = 0; k <= n; k++) adj.add(new ArrayList<Integer>());
for(int k = 0; k < n-1; k++){
StringTokenizer st = new StringTokenizer(f.readLine());
int a = Integer.parseInt(st.nextToken());
int b = Integer.parseInt(st.nextToken());
adj.get(a).add(b);
adj.get(b).add(a);
}
for(int k = 1; k <= n; k++){
if(adj.get(k).size()==2){
out.println("NO");
out.close();
System.exit(0);
}
}
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 β€ n β€ 10^5) β the number of nodes.
Each of the next n-1 lines contains two integers u and v (1 β€ u, v β€ n, u β 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"
]
}
|
CORRECT
|
python3
|
n = int(input())
nodes = [[] for _ in range(n)]
edges = []
for _ in range(n-1):
u, v = sorted(map(int, input().split(" ")))
u -= 1
v -= 1
nodes[u].append(v)
nodes[v].append(u)
edge = [u, v]
edges.append(edge)
leaf_nodes = [node for node in nodes if len(node) == 1]
num_ads = [len(node) for node in nodes]
if 2 in num_ads:
out = "NO"
else:
out = "YES"
#print(num_ads)
#for u, v in edges:
# num_ad_u = num_ads[u]
# num_ad_v = num_ads[v]
# #if num_ad_u == 1 or num_ad_v == 1:
# # continue
# if num_ad_u == 2 or num_ad_v == 2:
# out = "NO"
# break
print(out)
|
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 β€ n β€ 10^5) β the number of nodes.
Each of the next n-1 lines contains two integers u and v (1 β€ u, v β€ n, u β 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"
]
}
|
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)
flag = 1
for i in hash.keys():
if len(hash[i]) == 2:
flag = 0
break
if flag:
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 β€ n β€ 10^5) β the number of nodes.
Each of the next n-1 lines contains two integers u and v (1 β€ u, v β€ n, u β 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"
]
}
|
CORRECT
|
cpp
|
#include <bits/stdc++.h>
using ll = int64_t;
using namespace std;
const int N = 1e6 + 5;
int n;
int a[N];
basic_string<int> g[N];
int32_t main() {
cin.tie(0)->sync_with_stdio(0);
cin >> n;
for (int i = 1, u, v; i < n; i++) {
cin >> u >> v;
g[u] += v;
g[v] += u;
}
for (int i = 1; i <= n; i++) {
if (g[i].size() == 2) {
cout << "NO";
return 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 β€ n β€ 10^5) β the number of nodes.
Each of the next n-1 lines contains two integers u and v (1 β€ u, v β€ n, u β 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"
]
}
|
CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
int n;
int n1, n2;
int grau[100010];
vector<pair<int, int> > edges;
int main() {
scanf("%d", &n);
for (int g = 0; g < n - 1; g++) {
scanf("%d %d", &n1, &n2);
grau[n1]++;
grau[n2]++;
}
for (int g = 1; g <= n; g++) {
if (grau[g] == 2) {
printf("NO\n");
return 0;
}
}
printf("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 β€ n β€ 10^5) β the number of nodes.
Each of the next n-1 lines contains two integers u and v (1 β€ u, v β€ n, u β 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"
]
}
|
CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
int dx8[] = {1, 1, 0, -1, -1, -1, 0, 1};
int dy8[] = {0, 1, 1, 1, 0, -1, -1, -1};
int dx4[] = {1, 0, -1, 0};
int dy4[] = {0, 1, 0, -1};
template <class A, class B>
ostream& operator<<(ostream& out, const pair<A, B>& a) {
return out << "(" << a.first << "," << a.second << ")";
}
template <class A>
ostream& operator<<(ostream& out, const vector<A>& a) {
for (const A& it : a) out << it << " ";
return out;
}
template <class A, class B>
istream& operator>>(istream& in, pair<A, B>& a) {
return in >> a.first >> a.second;
}
template <class A>
istream& operator>>(istream& in, vector<A>& a) {
for (A& i : a) in >> i;
return in;
}
vector<int>* inputG(int n, int m) {
vector<int>* edges = new vector<int>[n + 1];
for (int i = 0; i < m; i++) {
int sv, ev;
cin >> sv >> ev;
edges[sv].push_back(ev);
edges[ev].push_back(sv);
}
return edges;
}
void solve() {
int n;
cin >> n;
vector<int>* edges = inputG(n, n - 1);
for (int i = 1; i <= n; i++) {
if (edges[i].size() == 2) {
cout << "NO"
<< "\n";
;
return;
}
}
cout << "YES"
<< "\n";
;
}
int main() {
ios_base::sync_with_stdio(false), cin.tie(NULL), cout.tie(NULL);
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 β€ n β€ 10^5) β the number of nodes.
Each of the next n-1 lines contains two integers u and v (1 β€ u, v β€ n, u β 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"
]
}
|
CORRECT
|
java
|
import java.io.*;
import java.lang.reflect.Array;
import java.util.ArrayList;
import java.util.Collections;
import java.util.StringTokenizer;
public class Controller {
public static void main(String[] args) throws IOException {
BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
PrintWriter writer = new PrintWriter(System.out);
int n = Integer.parseInt(reader.readLine());
int[] step = new int[n];
for(int i = 0; i < n-1; ++i){
StringTokenizer st = new StringTokenizer(reader.readLine());
int x = Integer.parseInt(st.nextToken())-1;
int y = Integer.parseInt(st.nextToken())-1;
step[x] ++;
step[y]++;
}
for(int i = 0; i < n; ++i){
if(step[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 β€ n β€ 10^5) β the number of nodes.
Each of the next n-1 lines contains two integers u and v (1 β€ u, v β€ n, u β 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"
]
}
|
CORRECT
|
python3
|
import sys
input = sys.stdin.readline
n = int(input())
a = [list(map(int, input().split())) for i in range(n-1)]
tree = [[] for i in range(n)]
for i in range(n - 1):
tree[a[i][0] - 1].append(a[i][1] - 1)
tree[a[i][1] - 1].append(a[i][0] - 1)
for i in range(n):
if len(tree[i]) == 2:
print("NO")
exit()
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 β€ n β€ 10^5) β the number of nodes.
Each of the next n-1 lines contains two integers u and v (1 β€ u, v β€ n, u β 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"
]
}
|
CORRECT
|
java
|
import java.util.*;
import java.lang.*;
import java.io.*;
public class Main
{
static final int MAX = 100100;
static int n;
static int[] deg = new int[MAX];
public static void main (String[] args) throws java.lang.Exception
{
InputStream inputStream = System.in;
OutputStream outputStream = System.out;
InputReader in = new InputReader(inputStream);
PrintWriter out = new PrintWriter(outputStream);
n = in.nextInt();
for(int i = 1; i < n; i++)
{
int v = in.nextInt();
int u = in.nextInt();
v--;
u--;
deg[v]++;
deg[u]++;
}
for(int v = 0; v < n; v++)
{
if(deg[v] == 2)
{
out.println("NO");
out.close();
return;
}
}
out.println("YES");
out.close();
}
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());
}
}
}
|
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 β€ n β€ 10^5) β the number of nodes.
Each of the next n-1 lines contains two integers u and v (1 β€ u, v β€ n, u β 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"
]
}
|
CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
int deg[200000];
int main() {
int n;
cin >> n;
for (int i = 0; i < n - 1; ++i) {
int u, v;
cin >> u >> v;
deg[u]++;
deg[v]++;
}
for (int i = 1; i <= n; ++i) {
if (deg[i] == 2) {
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 β€ n β€ 10^5) β the number of nodes.
Each of the next n-1 lines contains two integers u and v (1 β€ u, v β€ n, u β 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"
]
}
|
CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
int main() {
int x, y, n;
cin >> n;
vector<int> v[100010], barg;
for (int i = 0; i < n - 1; i++) {
cin >> x >> y;
v[x].push_back(y);
v[y].push_back(x);
}
for (int i = 1; i <= n; i++) {
if (v[i].size() == 2) {
cout << "NO";
return 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 β€ n β€ 10^5) β the number of nodes.
Each of the next n-1 lines contains two integers u and v (1 β€ u, v β€ n, u β 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"
]
}
|
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() == 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 β€ n β€ 10^5) β the number of nodes.
Each of the next n-1 lines contains two integers u and v (1 β€ u, v β€ n, u β 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"
]
}
|
CORRECT
|
python2
|
"""
Beautiful is better than ugly.
Explicit is better than implicit.
Simple is better than complex.
Complex is better than complicated.
Flat is better than nested.
Sparse is better than dense.
* Readability counts *
// Author : raj1307 - Raj Singh
// Date : 5.07.19
"""
from __future__ import division, print_function
import os,sys
from io import BytesIO, IOBase
if sys.version_info[0] < 3:
from __builtin__ import xrange as range
from future_builtins import ascii, filter, hex, map, oct, zip
else:
_str = str
str = lambda x=b"": x if type(x) is bytes else _str(x).encode()
def ii(): return int(input())
def si(): return input()
def mi(): return map(int,input().strip().split(" "))
def li(): return list(mi())
def dmain():
sys.setrecursionlimit(100000000)
threading.stack_size(40960000)
thread = threading.Thread(target=main)
thread.start()
from collections import deque, Counter, OrderedDict,defaultdict
#from heapq import nsmallest, nlargest, heapify,heappop ,heappush, heapreplace
#from math import ceil,floor,log,sqrt,factorial,pow,pi
#from bisect import bisect,bisect_left,bisect_right,insort,insort_left,insort_right
#from decimal import *,threading
#from itertools import permutations
abc='abcdefghijklmnopqrstuvwxyz'
abd={'a': 0, 'b': 1, 'c': 2, 'd': 3, 'e': 4, 'f': 5, 'g': 6, 'h': 7, 'i': 8, 'j': 9, 'k': 10, 'l': 11, 'm': 12, 'n': 13, 'o': 14, 'p': 15, 'q': 16, 'r': 17, 's': 18, 't': 19, 'u': 20, 'v': 21, 'w': 22, 'x': 23, 'y': 24, 'z': 25}
mod,MOD=1000000007,998244353
vow=['a','e','i','o','u']
dx,dy=[-1,1,0,0],[0,0,1,-1]
def getKey(item): return item[0]
def sort2(l):return sorted(l, key=getKey)
def d2(n,m,num):return [[num for x in range(m)] for y in range(n)]
def isPowerOfTwo (x): return (x and (not(x & (x - 1))) )
def decimalToBinary(n): return bin(n).replace("0b","")
def ntl(n):return [i for i in str(n)]
def powerMod(x,y,p):
res = 1
x %= p
while y > 0:
if y&1:
res = (res*x)%p
y = y>>1
x = (x*x)%p1
return res
def gcd(x, y):
while y:
x, y = y, x % y
return x
def isPrime(n) : # Check Prime Number or not
if (n <= 1) : return False
if (n <= 3) : return True
if (n % 2 == 0 or n % 3 == 0) : return False
i = 5
while(i * i <= n) :
if (n % i == 0 or n % (i + 2) == 0) :
return False
i = i + 6
return True
# For getting input from input.txt file
#sys.stdin = open('input.txt', 'r')
# Printing the Output to output.txt file
#sys.stdout = open('output.txt', 'w')
def main():
#for _ in range(ii()):
n=ii()
deg=[0]*(n+1)
for i in range(n-1):
x,y=mi()
deg[x]+=1
deg[y]+=1
for i in range(1,n+1):
if deg[i]==2:
print('NO')
exit()
print('YES')
# region fastio
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
if self.buffer.tell():
return self.buffer.read()
return os.read(self._fd, os.fstat(self._fd).st_size)
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
def print(*args, **kwargs):
"""Prints the values to a stream, or to sys.stdout by default."""
sep, file = kwargs.pop("sep", b" "), kwargs.pop("file", sys.stdout)
at_start = True
for x in args:
if not at_start:
file.write(sep)
file.write(str(x))
at_start = False
file.write(kwargs.pop("end", b"\n"))
if kwargs.pop("flush", False):
file.flush()
sys.stdin, sys.stdout = FastIO(sys.stdin), FastIO(sys.stdout)
input = lambda: sys.stdin.readline().rstrip(b"\r\n")
# endregion
if __name__ == "__main__":
main()
#dmain()
|
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 β€ n β€ 10^5) β the number of nodes.
Each of the next n-1 lines contains two integers u and v (1 β€ u, v β€ n, u β 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"
]
}
|
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++;
continue;
}
dfs(i, v);
}
if (v == 0) {
if (g[v].size() == 2) endProg();
} else {
if (g[v].size() == 2) 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 β€ n β€ 10^5) β the number of nodes.
Each of the next n-1 lines contains two integers u and v (1 β€ u, v β€ n, u β 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"
]
}
|
CORRECT
|
python2
|
from __future__ import division, print_function
def main():
n = int(input())
d = array_of(int, n)
for _ in range(n-1):
u, v = input_as_list()
d[u-1] += 1
d[v-1] += 1
print("NO" if 2 in d else "YES")
INF = float('inf')
MOD = 10**9 + 7
import os, sys
from atexit import register
from io import BytesIO
import itertools
if sys.version_info[0] < 3:
input = raw_input
range = xrange
filter = itertools.ifilter
map = itertools.imap
zip = itertools.izip
if "LOCAL_" in os.environ:
debug_print = print
else:
sys.stdin = BytesIO(os.read(0, os.fstat(0).st_size))
sys.stdout = BytesIO()
register(lambda: os.write(1, sys.stdout.getvalue()))
input = lambda: sys.stdin.readline().rstrip('\r\n')
debug_print = lambda *x, **y: None
def input_as_list():
return list(map(int, input().split()))
def array_of(f, *dim):
return [array_of(f, *dim[1:]) for _ in range(dim[0])] if dim else f()
main()
|
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 β€ n β€ 10^5) β the number of nodes.
Each of the next n-1 lines contains two integers u and v (1 β€ u, v β€ n, u β 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"
]
}
|
CORRECT
|
python3
|
import sys
num = int(input())
deg = [0]*(num+1)
for _ in range(1,num):
a, b = [int(i) for i in input().split()]
deg[a ] += 1
deg[b] += 1
for i in range(1, num+1):
if deg[i] == 2:
print("NO")
sys.exit()
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 β€ n β€ 10^5) β the number of nodes.
Each of the next n-1 lines contains two integers u and v (1 β€ u, v β€ n, u β 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"
]
}
|
CORRECT
|
python3
|
# -*- coding: utf-8 -*-
import sys
# from operator import itemgetter
# from fractions import gcd
# from math import ceil, floor
# from copy import deepcopy
# from itertools import accumulate
from collections import deque
# import math
# from functools import reduce
input = sys.stdin.readline
def ii(): return int(input())
def mi(): return map(int, input().rstrip().split())
def lmi(): return list(map(int, input().rstrip().split()))
def li(): return list(input().rstrip())
# template
# BEGIN CUT HERE
class Graph():
def __init__(self, n, Weighted=False, Directed=True, Matrix=False):
self.sz = n
self.is_Weighted = Weighted
self.is_Directed = Directed
self.is_Matrix = Matrix
if Matrix:
if Weighted:
self.graph = [[0 for _i in range(n)] for _j in range(n)]
else:
self.graph = [[0 for _i in range(n)] for _j in range(n)]
else:
self.graph = [[] for _i in range(n)]
def _weighted_add_edge(self, x, y, w):
if self.is_Matrix:
self.graph[x][y] = w
else:
self.graph[x].append((y, w))
def _unweighted_add_edge(self, x, y):
if self.is_Matrix:
self.graph[x][y] = 1
else:
self.graph[x].append(y)
def add_edge(self, x, y, *w):
if self.is_Directed:
if self.is_Weighted:
self._weighted_add_edge(x, y, w[0])
else:
self._unweighted_add_edge(x, y)
else:
if self.is_Weighted:
self._weighted_add_edge(x, y, w[0])
self._weighted_add_edge(y, x, w[0])
else:
self._unweighted_add_edge(x, y)
self._unweighted_add_edge(y, x)
def _convert_to_maxrix(self):
if self.is_Matrix:
return self
mat_g = self.__class__(
self.sz, Weighted=self.is_Weighted, Directed=self.is_Directed, Matrix=True)
if self.is_Weighted:
for i in range(self.sz):
for j in self.graph[i]:
mat_g.add_edge(i, j[0], j[1])
else:
for i in range(self.sz):
for j in self.graph[i]:
mat_g.add_edge(i, j)
return mat_g
def __getitem__(self, n):
return self.graph[n]
# def __setitem__(self, n, v):
# if not self.is_Weighted and not self.is_Matrix:
# self.graph[n] = v
def __str__(self):
return str([self.graph[i] for i in range(self.sz)])
def main():
n = ii()
g = Graph(n, Directed=False)
for i in range(n - 1):
s, t = mi()
g.add_edge(s-1, t-1)
for i in range(n):
if len(g[i]) == 2:
break
else:
print('YES')
sys.exit()
print('NO')
if __name__ == '__main__':
main()
|
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 β€ n β€ 10^5) β the number of nodes.
Each of the next n-1 lines contains two integers u and v (1 β€ u, v β€ n, u β 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"
]
}
|
CORRECT
|
python3
|
from collections import defaultdict
n = int(input())
tr = defaultdict(int)
for i in range(n-1):
u, v = map(int, input().split())
tr[u]+=1
tr[v]+=1
for i in tr:
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 β€ n β€ 10^5) β the number of nodes.
Each of the next n-1 lines contains two integers u and v (1 β€ u, v β€ n, u β 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"
]
}
|
CORRECT
|
java
|
import java.util.*;
import java.io.*;
public class Solution1{
public static Integer INT(String s){
return Integer.parseInt(s);
}
public static Long LONG(String s){
return Long.parseLong(s);
}
//====================================================================================================================
static LinkedList<Integer> adj[];
public static void main(String args[])throws IOException{
BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
String line[]=br.readLine().split("\\s");
int n=INT(line[0]);
adj=new LinkedList[n+1];
for(int i=0; i<=n; i++)
adj[i]=new LinkedList<>();
for(int i=1; i<n; i++){
line=br.readLine().split("\\s");
int u=INT(line[0]),
v=INT(line[1]);
adj[u].add(v);
adj[v].add(u);
}
boolean flag=true;
for(int i=1; i<=n; i++)
if(adj[i].size()==2)
flag=false;
if(flag)
System.out.println("YES");
else
System.out.println("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 β€ n β€ 10^5) β the number of nodes.
Each of the next n-1 lines contains two integers u and v (1 β€ u, v β€ n, u β 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"
]
}
|
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 = 1; 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 β€ n β€ 10^5) β the number of nodes.
Each of the next n-1 lines contains two integers u and v (1 β€ u, v β€ n, u β 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"
]
}
|
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++) {
int a = sc.nextInt()-1, b = sc.nextInt()-1;
paths[a].add(b);
paths[b].add(a);
}
for (int i = 0 ; i < paths.length ; i++) {
if (paths[i].size() == 2) {
System.out.println("NO"); return;
}
}
System.out.println("YES");
}
/*
3
1 2
1 3
*/
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 β€ n β€ 10^5) β the number of nodes.
Each of the next n-1 lines contains two integers u and v (1 β€ u, v β€ n, u β 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"
]
}
|
CORRECT
|
java
|
import java.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.io.IOException;
import java.io.InputStream;
/**
* Built using CHelper plug-in
* Actual solution is at the top
*
* @author Aman Kumar Singh
*/
public class Main {
public static void main(String[] args) {
InputStream inputStream = System.in;
OutputStream outputStream = System.out;
InputReader in = new InputReader(inputStream);
PrintWriter out = new PrintWriter(outputStream);
D1AddOnATree solver = new D1AddOnATree();
solver.solve(1, in, out);
out.close();
}
static class D1AddOnATree {
PrintWriter out;
InputReader in;
public void solve(int testNumber, InputReader in, PrintWriter out) {
this.out = out;
this.in = in;
int n = ni();
int[] deg = new int[n];
int i = 0;
for (i = 0; i < n - 1; i++) {
deg[ni() - 1]++;
deg[ni() - 1]++;
}
int cnt = 0;
for (i = 0; i < n; i++) {
if (deg[i] == 1)
cnt++;
}
if (n == 2) {
pn("YES");
return;
}
boolean ok = true;
for (i = 0; i < n; i++) {
if (deg[i] == 2)
ok = false;
}
if (ok && cnt > 2)
pn("YES");
else
pn("NO");
}
int ni() {
return in.nextInt();
}
void pn(Object o) {
out.println(o);
}
}
static class InputReader {
private InputStream stream;
private byte[] buf = new byte[1024];
private int curChar;
private int numChars;
public InputReader(InputStream stream) {
this.stream = stream;
}
public int read() {
if (numChars == -1)
throw new UnknownError();
if (curChar >= numChars) {
curChar = 0;
try {
numChars = stream.read(buf);
} catch (IOException e) {
throw new UnknownError();
}
if (numChars <= 0)
return -1;
}
return buf[curChar++];
}
public int nextInt() {
return Integer.parseInt(next());
}
public String next() {
int c = read();
while (isSpaceChar(c))
c = read();
StringBuffer res = new StringBuffer();
do {
res.appendCodePoint(c);
c = read();
} while (!isSpaceChar(c));
return res.toString();
}
private boolean isSpaceChar(int c) {
return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
}
}
}
|
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.