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 |
|---|---|---|---|---|---|---|
1037_E. Trips | There are n persons who initially don't know each other. On each morning, two of them, who were not friends before, become friends.
We want to plan a trip for every evening of m days. On each trip, you have to select a group of people that will go on the trip. For every person, one of the following should hold:
* ... | {
"input": [
"4 4 2\n2 3\n1 2\n1 3\n1 4\n",
"5 8 2\n2 1\n4 2\n5 4\n5 2\n4 3\n5 1\n4 1\n3 2\n",
"5 7 2\n1 5\n3 2\n2 5\n3 4\n1 2\n5 3\n1 3\n"
],
"output": [
"0\n0\n3\n3\n",
"0\n0\n0\n3\n3\n4\n4\n5\n",
"0\n0\n0\n0\n3\n4\n4\n"
]
} | {
"input": [
"16 20 2\n10 3\n5 3\n10 5\n12 7\n7 6\n9 12\n9 6\n1 10\n11 16\n11 1\n16 2\n10 2\n14 4\n15 14\n4 13\n13 15\n1 8\n7 15\n1 7\n8 15\n",
"2 1 1\n2 1\n"
],
"output": [
"0\n0\n3\n3\n3\n3\n7\n7\n7\n7\n7\n11\n11\n11\n11\n15\n15\n15\n15\n16\n",
"2\n"
]
} | IN-CORRECT | java | //package contests.CF1037;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.*;
public class E {
static ArrayList<Integer>[] adj;
static int[] vis, going;
static int NO = 3, YES = 2, VIS = 1;
static int k;
pub... |
1037_E. Trips | There are n persons who initially don't know each other. On each morning, two of them, who were not friends before, become friends.
We want to plan a trip for every evening of m days. On each trip, you have to select a group of people that will go on the trip. For every person, one of the following should hold:
* ... | {
"input": [
"4 4 2\n2 3\n1 2\n1 3\n1 4\n",
"5 8 2\n2 1\n4 2\n5 4\n5 2\n4 3\n5 1\n4 1\n3 2\n",
"5 7 2\n1 5\n3 2\n2 5\n3 4\n1 2\n5 3\n1 3\n"
],
"output": [
"0\n0\n3\n3\n",
"0\n0\n0\n3\n3\n4\n4\n5\n",
"0\n0\n0\n0\n3\n4\n4\n"
]
} | {
"input": [
"16 20 2\n10 3\n5 3\n10 5\n12 7\n7 6\n9 12\n9 6\n1 10\n11 16\n11 1\n16 2\n10 2\n14 4\n15 14\n4 13\n13 15\n1 8\n7 15\n1 7\n8 15\n",
"2 1 1\n2 1\n"
],
"output": [
"0\n0\n3\n3\n3\n3\n7\n7\n7\n7\n7\n11\n11\n11\n11\n15\n15\n15\n15\n16\n",
"2\n"
]
} | IN-CORRECT | java | import java.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.OutputStream;
import java.io.PrintWriter;
import java.io.BufferedWriter;
import java.util.Set;
import java.util.InputMismatchException;
import java.io.IOException;
import java.util.ArrayList;
import java.util.HashSet;
im... |
1037_E. Trips | There are n persons who initially don't know each other. On each morning, two of them, who were not friends before, become friends.
We want to plan a trip for every evening of m days. On each trip, you have to select a group of people that will go on the trip. For every person, one of the following should hold:
* ... | {
"input": [
"4 4 2\n2 3\n1 2\n1 3\n1 4\n",
"5 8 2\n2 1\n4 2\n5 4\n5 2\n4 3\n5 1\n4 1\n3 2\n",
"5 7 2\n1 5\n3 2\n2 5\n3 4\n1 2\n5 3\n1 3\n"
],
"output": [
"0\n0\n3\n3\n",
"0\n0\n0\n3\n3\n4\n4\n5\n",
"0\n0\n0\n0\n3\n4\n4\n"
]
} | {
"input": [
"16 20 2\n10 3\n5 3\n10 5\n12 7\n7 6\n9 12\n9 6\n1 10\n11 16\n11 1\n16 2\n10 2\n14 4\n15 14\n4 13\n13 15\n1 8\n7 15\n1 7\n8 15\n",
"2 1 1\n2 1\n"
],
"output": [
"0\n0\n3\n3\n3\n3\n7\n7\n7\n7\n7\n11\n11\n11\n11\n15\n15\n15\n15\n16\n",
"2\n"
]
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
long long read() {
long long x = 0;
char f = 1, 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 write(long lo... |
1037_E. Trips | There are n persons who initially don't know each other. On each morning, two of them, who were not friends before, become friends.
We want to plan a trip for every evening of m days. On each trip, you have to select a group of people that will go on the trip. For every person, one of the following should hold:
* ... | {
"input": [
"4 4 2\n2 3\n1 2\n1 3\n1 4\n",
"5 8 2\n2 1\n4 2\n5 4\n5 2\n4 3\n5 1\n4 1\n3 2\n",
"5 7 2\n1 5\n3 2\n2 5\n3 4\n1 2\n5 3\n1 3\n"
],
"output": [
"0\n0\n3\n3\n",
"0\n0\n0\n3\n3\n4\n4\n5\n",
"0\n0\n0\n0\n3\n4\n4\n"
]
} | {
"input": [
"16 20 2\n10 3\n5 3\n10 5\n12 7\n7 6\n9 12\n9 6\n1 10\n11 16\n11 1\n16 2\n10 2\n14 4\n15 14\n4 13\n13 15\n1 8\n7 15\n1 7\n8 15\n",
"2 1 1\n2 1\n"
],
"output": [
"0\n0\n3\n3\n3\n3\n7\n7\n7\n7\n7\n11\n11\n11\n11\n15\n15\n15\n15\n16\n",
"2\n"
]
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
const int INF = 0x3f3f3f3f;
const int MAXN = 200005, MOD = 1e9 + 7;
struct ii {
long long a, b;
bool operator<(ii o) const { return tie(a, b) < tie(o.a, o.b); }
};
int N, M, K, ans, valid[MAXN];
ii arr[MAXN];
vector<int> edge[MAXN];
void gagalin(int now) {
ans--;
fo... |
1037_E. Trips | There are n persons who initially don't know each other. On each morning, two of them, who were not friends before, become friends.
We want to plan a trip for every evening of m days. On each trip, you have to select a group of people that will go on the trip. For every person, one of the following should hold:
* ... | {
"input": [
"4 4 2\n2 3\n1 2\n1 3\n1 4\n",
"5 8 2\n2 1\n4 2\n5 4\n5 2\n4 3\n5 1\n4 1\n3 2\n",
"5 7 2\n1 5\n3 2\n2 5\n3 4\n1 2\n5 3\n1 3\n"
],
"output": [
"0\n0\n3\n3\n",
"0\n0\n0\n3\n3\n4\n4\n5\n",
"0\n0\n0\n0\n3\n4\n4\n"
]
} | {
"input": [
"16 20 2\n10 3\n5 3\n10 5\n12 7\n7 6\n9 12\n9 6\n1 10\n11 16\n11 1\n16 2\n10 2\n14 4\n15 14\n4 13\n13 15\n1 8\n7 15\n1 7\n8 15\n",
"2 1 1\n2 1\n"
],
"output": [
"0\n0\n3\n3\n3\n3\n7\n7\n7\n7\n7\n11\n11\n11\n11\n15\n15\n15\n15\n16\n",
"2\n"
]
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 3;
int n, m, k, deg[N], x[N], y[N], ans[N], cnt;
set<int> G[N];
set<int>::iterator it;
void Del() {
queue<int> q;
for (int i = 1; i <= n; i++)
if (deg[i] && deg[i] < k) q.push(i);
while (!q.empty()) {
int u = q.front();
q.pop();
cnt... |
1037_E. Trips | There are n persons who initially don't know each other. On each morning, two of them, who were not friends before, become friends.
We want to plan a trip for every evening of m days. On each trip, you have to select a group of people that will go on the trip. For every person, one of the following should hold:
* ... | {
"input": [
"4 4 2\n2 3\n1 2\n1 3\n1 4\n",
"5 8 2\n2 1\n4 2\n5 4\n5 2\n4 3\n5 1\n4 1\n3 2\n",
"5 7 2\n1 5\n3 2\n2 5\n3 4\n1 2\n5 3\n1 3\n"
],
"output": [
"0\n0\n3\n3\n",
"0\n0\n0\n3\n3\n4\n4\n5\n",
"0\n0\n0\n0\n3\n4\n4\n"
]
} | {
"input": [
"16 20 2\n10 3\n5 3\n10 5\n12 7\n7 6\n9 12\n9 6\n1 10\n11 16\n11 1\n16 2\n10 2\n14 4\n15 14\n4 13\n13 15\n1 8\n7 15\n1 7\n8 15\n",
"2 1 1\n2 1\n"
],
"output": [
"0\n0\n3\n3\n3\n3\n7\n7\n7\n7\n7\n11\n11\n11\n11\n15\n15\n15\n15\n16\n",
"2\n"
]
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
const int maxn = 1e5 + 10;
set<pair<int, int> > sq;
set<int> to[maxn];
int n, m, k;
int a[maxn];
struct node {
int l, r;
} q[maxn];
void update() {
while (sq.size() > 0 && (*(sq.begin())).first < k) {
auto x = *(sq.begin());
sq.erase(sq.begin());
for (auto j... |
1037_E. Trips | There are n persons who initially don't know each other. On each morning, two of them, who were not friends before, become friends.
We want to plan a trip for every evening of m days. On each trip, you have to select a group of people that will go on the trip. For every person, one of the following should hold:
* ... | {
"input": [
"4 4 2\n2 3\n1 2\n1 3\n1 4\n",
"5 8 2\n2 1\n4 2\n5 4\n5 2\n4 3\n5 1\n4 1\n3 2\n",
"5 7 2\n1 5\n3 2\n2 5\n3 4\n1 2\n5 3\n1 3\n"
],
"output": [
"0\n0\n3\n3\n",
"0\n0\n0\n3\n3\n4\n4\n5\n",
"0\n0\n0\n0\n3\n4\n4\n"
]
} | {
"input": [
"16 20 2\n10 3\n5 3\n10 5\n12 7\n7 6\n9 12\n9 6\n1 10\n11 16\n11 1\n16 2\n10 2\n14 4\n15 14\n4 13\n13 15\n1 8\n7 15\n1 7\n8 15\n",
"2 1 1\n2 1\n"
],
"output": [
"0\n0\n3\n3\n3\n3\n7\n7\n7\n7\n7\n11\n11\n11\n11\n15\n15\n15\n15\n16\n",
"2\n"
]
} | IN-CORRECT | java | //package com.company;
import java.io.*;
import java.util.*;
public class Main {
static long TIME_START, TIME_END;
public static void main(String[] args) throws IOException {
Scanner sc = new Scanner(System.in);
// Scanner sc = new Scanner(new FileInputStream("Test.in"));
PrintWriter p... |
1037_E. Trips | There are n persons who initially don't know each other. On each morning, two of them, who were not friends before, become friends.
We want to plan a trip for every evening of m days. On each trip, you have to select a group of people that will go on the trip. For every person, one of the following should hold:
* ... | {
"input": [
"4 4 2\n2 3\n1 2\n1 3\n1 4\n",
"5 8 2\n2 1\n4 2\n5 4\n5 2\n4 3\n5 1\n4 1\n3 2\n",
"5 7 2\n1 5\n3 2\n2 5\n3 4\n1 2\n5 3\n1 3\n"
],
"output": [
"0\n0\n3\n3\n",
"0\n0\n0\n3\n3\n4\n4\n5\n",
"0\n0\n0\n0\n3\n4\n4\n"
]
} | {
"input": [
"16 20 2\n10 3\n5 3\n10 5\n12 7\n7 6\n9 12\n9 6\n1 10\n11 16\n11 1\n16 2\n10 2\n14 4\n15 14\n4 13\n13 15\n1 8\n7 15\n1 7\n8 15\n",
"2 1 1\n2 1\n"
],
"output": [
"0\n0\n3\n3\n3\n3\n7\n7\n7\n7\n7\n11\n11\n11\n11\n15\n15\n15\n15\n16\n",
"2\n"
]
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
template <typename T>
inline T const &MAX(T const &a, T const &b) {
return a > b ? a : b;
}
template <typename T>
inline T const &MIN(T const &a, T const &b) {
return a < b ? a : b;
}
inline void add(long long &a, long long b) {
a += b;
if (a >= 1000000007) a -= 1000000007;
}
inline voi... |
1037_E. Trips | There are n persons who initially don't know each other. On each morning, two of them, who were not friends before, become friends.
We want to plan a trip for every evening of m days. On each trip, you have to select a group of people that will go on the trip. For every person, one of the following should hold:
* ... | {
"input": [
"4 4 2\n2 3\n1 2\n1 3\n1 4\n",
"5 8 2\n2 1\n4 2\n5 4\n5 2\n4 3\n5 1\n4 1\n3 2\n",
"5 7 2\n1 5\n3 2\n2 5\n3 4\n1 2\n5 3\n1 3\n"
],
"output": [
"0\n0\n3\n3\n",
"0\n0\n0\n3\n3\n4\n4\n5\n",
"0\n0\n0\n0\n3\n4\n4\n"
]
} | {
"input": [
"16 20 2\n10 3\n5 3\n10 5\n12 7\n7 6\n9 12\n9 6\n1 10\n11 16\n11 1\n16 2\n10 2\n14 4\n15 14\n4 13\n13 15\n1 8\n7 15\n1 7\n8 15\n",
"2 1 1\n2 1\n"
],
"output": [
"0\n0\n3\n3\n3\n3\n7\n7\n7\n7\n7\n11\n11\n11\n11\n15\n15\n15\n15\n16\n",
"2\n"
]
} | IN-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.Arrays;
import java.io.FilterInputStream;
import java.io.BufferedInputStream;
import java.util.TreeSet;
import java.io.InputStream;
/**
* @author khokharnikunj8
... |
1037_E. Trips | There are n persons who initially don't know each other. On each morning, two of them, who were not friends before, become friends.
We want to plan a trip for every evening of m days. On each trip, you have to select a group of people that will go on the trip. For every person, one of the following should hold:
* ... | {
"input": [
"4 4 2\n2 3\n1 2\n1 3\n1 4\n",
"5 8 2\n2 1\n4 2\n5 4\n5 2\n4 3\n5 1\n4 1\n3 2\n",
"5 7 2\n1 5\n3 2\n2 5\n3 4\n1 2\n5 3\n1 3\n"
],
"output": [
"0\n0\n3\n3\n",
"0\n0\n0\n3\n3\n4\n4\n5\n",
"0\n0\n0\n0\n3\n4\n4\n"
]
} | {
"input": [
"16 20 2\n10 3\n5 3\n10 5\n12 7\n7 6\n9 12\n9 6\n1 10\n11 16\n11 1\n16 2\n10 2\n14 4\n15 14\n4 13\n13 15\n1 8\n7 15\n1 7\n8 15\n",
"2 1 1\n2 1\n"
],
"output": [
"0\n0\n3\n3\n3\n3\n7\n7\n7\n7\n7\n11\n11\n11\n11\n15\n15\n15\n15\n16\n",
"2\n"
]
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
const int maxn = 2e5 + 10;
vector<int> g[maxn];
int edge[maxn][3];
int res[maxn];
map<pair<int, int>, int> ma;
set<pair<int, int> > s;
int deg[maxn];
int n, m, k;
void add(int x, int y) {
g[x].push_back(y);
deg[y]++;
}
void del(int x) {
int si = g[x].size();
for (in... |
1037_E. Trips | There are n persons who initially don't know each other. On each morning, two of them, who were not friends before, become friends.
We want to plan a trip for every evening of m days. On each trip, you have to select a group of people that will go on the trip. For every person, one of the following should hold:
* ... | {
"input": [
"4 4 2\n2 3\n1 2\n1 3\n1 4\n",
"5 8 2\n2 1\n4 2\n5 4\n5 2\n4 3\n5 1\n4 1\n3 2\n",
"5 7 2\n1 5\n3 2\n2 5\n3 4\n1 2\n5 3\n1 3\n"
],
"output": [
"0\n0\n3\n3\n",
"0\n0\n0\n3\n3\n4\n4\n5\n",
"0\n0\n0\n0\n3\n4\n4\n"
]
} | {
"input": [
"16 20 2\n10 3\n5 3\n10 5\n12 7\n7 6\n9 12\n9 6\n1 10\n11 16\n11 1\n16 2\n10 2\n14 4\n15 14\n4 13\n13 15\n1 8\n7 15\n1 7\n8 15\n",
"2 1 1\n2 1\n"
],
"output": [
"0\n0\n3\n3\n3\n3\n7\n7\n7\n7\n7\n11\n11\n11\n11\n15\n15\n15\n15\n16\n",
"2\n"
]
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
struct edge {
int v, id;
};
map<int, int> D;
vector<edge> E[200005];
int ans[200005], n, m, u, v, k, x[200005], y[200005];
bool vis[200005];
queue<int> Q;
int main() {
scanf("%d%d%d", &n, &m, &k);
for (int i = 1; i <= m; i++) {
scanf("%d%d", &x[i], &y[i]);
E[x... |
1037_E. Trips | There are n persons who initially don't know each other. On each morning, two of them, who were not friends before, become friends.
We want to plan a trip for every evening of m days. On each trip, you have to select a group of people that will go on the trip. For every person, one of the following should hold:
* ... | {
"input": [
"4 4 2\n2 3\n1 2\n1 3\n1 4\n",
"5 8 2\n2 1\n4 2\n5 4\n5 2\n4 3\n5 1\n4 1\n3 2\n",
"5 7 2\n1 5\n3 2\n2 5\n3 4\n1 2\n5 3\n1 3\n"
],
"output": [
"0\n0\n3\n3\n",
"0\n0\n0\n3\n3\n4\n4\n5\n",
"0\n0\n0\n0\n3\n4\n4\n"
]
} | {
"input": [
"16 20 2\n10 3\n5 3\n10 5\n12 7\n7 6\n9 12\n9 6\n1 10\n11 16\n11 1\n16 2\n10 2\n14 4\n15 14\n4 13\n13 15\n1 8\n7 15\n1 7\n8 15\n",
"2 1 1\n2 1\n"
],
"output": [
"0\n0\n3\n3\n3\n3\n7\n7\n7\n7\n7\n11\n11\n11\n11\n15\n15\n15\n15\n16\n",
"2\n"
]
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
int n, m, k;
vector<int> used;
vector<int> selected;
vector<vector<int>> g;
void dfs(int cur) {
used[cur] = 1;
selected[cur] = 1;
int cnt = 0;
for (auto t : g[cur]) {
if (!used[t]) {
dfs(t);
}
if (selected[t]) {
cnt++;
}
}
if (cnt < k... |
1037_E. Trips | There are n persons who initially don't know each other. On each morning, two of them, who were not friends before, become friends.
We want to plan a trip for every evening of m days. On each trip, you have to select a group of people that will go on the trip. For every person, one of the following should hold:
* ... | {
"input": [
"4 4 2\n2 3\n1 2\n1 3\n1 4\n",
"5 8 2\n2 1\n4 2\n5 4\n5 2\n4 3\n5 1\n4 1\n3 2\n",
"5 7 2\n1 5\n3 2\n2 5\n3 4\n1 2\n5 3\n1 3\n"
],
"output": [
"0\n0\n3\n3\n",
"0\n0\n0\n3\n3\n4\n4\n5\n",
"0\n0\n0\n0\n3\n4\n4\n"
]
} | {
"input": [
"16 20 2\n10 3\n5 3\n10 5\n12 7\n7 6\n9 12\n9 6\n1 10\n11 16\n11 1\n16 2\n10 2\n14 4\n15 14\n4 13\n13 15\n1 8\n7 15\n1 7\n8 15\n",
"2 1 1\n2 1\n"
],
"output": [
"0\n0\n3\n3\n3\n3\n7\n7\n7\n7\n7\n11\n11\n11\n11\n15\n15\n15\n15\n16\n",
"2\n"
]
} | IN-CORRECT | java | import java.io.*;
import java.math.BigInteger;
import java.util.*;
public class Test {
static int readInt() {
int ans = 0;
boolean neg = false;
try {
boolean start = false;
for (int c = 0; (c = System.in.read()) != -1; ) {
if (c == '-') {
... |
1037_E. Trips | There are n persons who initially don't know each other. On each morning, two of them, who were not friends before, become friends.
We want to plan a trip for every evening of m days. On each trip, you have to select a group of people that will go on the trip. For every person, one of the following should hold:
* ... | {
"input": [
"4 4 2\n2 3\n1 2\n1 3\n1 4\n",
"5 8 2\n2 1\n4 2\n5 4\n5 2\n4 3\n5 1\n4 1\n3 2\n",
"5 7 2\n1 5\n3 2\n2 5\n3 4\n1 2\n5 3\n1 3\n"
],
"output": [
"0\n0\n3\n3\n",
"0\n0\n0\n3\n3\n4\n4\n5\n",
"0\n0\n0\n0\n3\n4\n4\n"
]
} | {
"input": [
"16 20 2\n10 3\n5 3\n10 5\n12 7\n7 6\n9 12\n9 6\n1 10\n11 16\n11 1\n16 2\n10 2\n14 4\n15 14\n4 13\n13 15\n1 8\n7 15\n1 7\n8 15\n",
"2 1 1\n2 1\n"
],
"output": [
"0\n0\n3\n3\n3\n3\n7\n7\n7\n7\n7\n11\n11\n11\n11\n15\n15\n15\n15\n16\n",
"2\n"
]
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
int p[300005], c[300005], a[300005];
int main() {
ios_base::sync_with_stdio(false);
int n, i, u, v, j, f;
cin >> n;
for (i = 0; i < n - 1; i++) {
cin >> u >> v;
p[v] = u;
c[u]++;
}
for (i = 0; i < n; i++) {
cin >> a[i];
}
i = 0;
j = 1;
f ... |
1037_E. Trips | There are n persons who initially don't know each other. On each morning, two of them, who were not friends before, become friends.
We want to plan a trip for every evening of m days. On each trip, you have to select a group of people that will go on the trip. For every person, one of the following should hold:
* ... | {
"input": [
"4 4 2\n2 3\n1 2\n1 3\n1 4\n",
"5 8 2\n2 1\n4 2\n5 4\n5 2\n4 3\n5 1\n4 1\n3 2\n",
"5 7 2\n1 5\n3 2\n2 5\n3 4\n1 2\n5 3\n1 3\n"
],
"output": [
"0\n0\n3\n3\n",
"0\n0\n0\n3\n3\n4\n4\n5\n",
"0\n0\n0\n0\n3\n4\n4\n"
]
} | {
"input": [
"16 20 2\n10 3\n5 3\n10 5\n12 7\n7 6\n9 12\n9 6\n1 10\n11 16\n11 1\n16 2\n10 2\n14 4\n15 14\n4 13\n13 15\n1 8\n7 15\n1 7\n8 15\n",
"2 1 1\n2 1\n"
],
"output": [
"0\n0\n3\n3\n3\n3\n7\n7\n7\n7\n7\n11\n11\n11\n11\n15\n15\n15\n15\n16\n",
"2\n"
]
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
const int maxn = (int)2e5 + 5;
int n, m, k;
vector<pair<int, int> > out[maxn];
int la[maxn];
int pos[maxn];
int deg[maxn];
int add[maxn];
bool dead[maxn];
set<pair<int, int>, greater<pair<int, int> > > s;
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
cin >> n... |
1037_E. Trips | There are n persons who initially don't know each other. On each morning, two of them, who were not friends before, become friends.
We want to plan a trip for every evening of m days. On each trip, you have to select a group of people that will go on the trip. For every person, one of the following should hold:
* ... | {
"input": [
"4 4 2\n2 3\n1 2\n1 3\n1 4\n",
"5 8 2\n2 1\n4 2\n5 4\n5 2\n4 3\n5 1\n4 1\n3 2\n",
"5 7 2\n1 5\n3 2\n2 5\n3 4\n1 2\n5 3\n1 3\n"
],
"output": [
"0\n0\n3\n3\n",
"0\n0\n0\n3\n3\n4\n4\n5\n",
"0\n0\n0\n0\n3\n4\n4\n"
]
} | {
"input": [
"16 20 2\n10 3\n5 3\n10 5\n12 7\n7 6\n9 12\n9 6\n1 10\n11 16\n11 1\n16 2\n10 2\n14 4\n15 14\n4 13\n13 15\n1 8\n7 15\n1 7\n8 15\n",
"2 1 1\n2 1\n"
],
"output": [
"0\n0\n3\n3\n3\n3\n7\n7\n7\n7\n7\n11\n11\n11\n11\n15\n15\n15\n15\n16\n",
"2\n"
]
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
const long long LINF = 1e18;
const int INF = 1e9;
const int M = 1e9 + 7;
const double EPS = 1e-9;
using namespace std;
vector<vector<pair<int, int> > > g;
int main(int argc, const char* argv[]) {
std::ios_base::sync_with_stdio(0);
int n, m, k;
cin >> n >> m >> k;
g.resize(n);
int u[20... |
1037_E. Trips | There are n persons who initially don't know each other. On each morning, two of them, who were not friends before, become friends.
We want to plan a trip for every evening of m days. On each trip, you have to select a group of people that will go on the trip. For every person, one of the following should hold:
* ... | {
"input": [
"4 4 2\n2 3\n1 2\n1 3\n1 4\n",
"5 8 2\n2 1\n4 2\n5 4\n5 2\n4 3\n5 1\n4 1\n3 2\n",
"5 7 2\n1 5\n3 2\n2 5\n3 4\n1 2\n5 3\n1 3\n"
],
"output": [
"0\n0\n3\n3\n",
"0\n0\n0\n3\n3\n4\n4\n5\n",
"0\n0\n0\n0\n3\n4\n4\n"
]
} | {
"input": [
"16 20 2\n10 3\n5 3\n10 5\n12 7\n7 6\n9 12\n9 6\n1 10\n11 16\n11 1\n16 2\n10 2\n14 4\n15 14\n4 13\n13 15\n1 8\n7 15\n1 7\n8 15\n",
"2 1 1\n2 1\n"
],
"output": [
"0\n0\n3\n3\n3\n3\n7\n7\n7\n7\n7\n11\n11\n11\n11\n15\n15\n15\n15\n16\n",
"2\n"
]
} | IN-CORRECT | python3 | def findSolution(graph,edges,m,k,n):
toBeRemoved = []
numOfConnections = [len(lst) for lst in graph]
for i in range(0,n):
if numOfConnections[i]<k:
toBeRemoved.append(i)
while len(toBeRemoved) >0:
node = toBeRemoved.pop(0)
for i in graph[node]:
numOfConnec... |
1037_E. Trips | There are n persons who initially don't know each other. On each morning, two of them, who were not friends before, become friends.
We want to plan a trip for every evening of m days. On each trip, you have to select a group of people that will go on the trip. For every person, one of the following should hold:
* ... | {
"input": [
"4 4 2\n2 3\n1 2\n1 3\n1 4\n",
"5 8 2\n2 1\n4 2\n5 4\n5 2\n4 3\n5 1\n4 1\n3 2\n",
"5 7 2\n1 5\n3 2\n2 5\n3 4\n1 2\n5 3\n1 3\n"
],
"output": [
"0\n0\n3\n3\n",
"0\n0\n0\n3\n3\n4\n4\n5\n",
"0\n0\n0\n0\n3\n4\n4\n"
]
} | {
"input": [
"16 20 2\n10 3\n5 3\n10 5\n12 7\n7 6\n9 12\n9 6\n1 10\n11 16\n11 1\n16 2\n10 2\n14 4\n15 14\n4 13\n13 15\n1 8\n7 15\n1 7\n8 15\n",
"2 1 1\n2 1\n"
],
"output": [
"0\n0\n3\n3\n3\n3\n7\n7\n7\n7\n7\n11\n11\n11\n11\n15\n15\n15\n15\n16\n",
"2\n"
]
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
int n, m, k, ans[200005], cnt[200005];
vector<int> a[200005];
pair<int, int> ed[200005];
set<pair<int, int> > s;
set<pair<int, int> >::iterator it;
bool dd[200005];
void Delete() {
while (s.size()) {
it = s.begin();
pair<int, int> temp = *it;
if (temp.first >=... |
1037_E. Trips | There are n persons who initially don't know each other. On each morning, two of them, who were not friends before, become friends.
We want to plan a trip for every evening of m days. On each trip, you have to select a group of people that will go on the trip. For every person, one of the following should hold:
* ... | {
"input": [
"4 4 2\n2 3\n1 2\n1 3\n1 4\n",
"5 8 2\n2 1\n4 2\n5 4\n5 2\n4 3\n5 1\n4 1\n3 2\n",
"5 7 2\n1 5\n3 2\n2 5\n3 4\n1 2\n5 3\n1 3\n"
],
"output": [
"0\n0\n3\n3\n",
"0\n0\n0\n3\n3\n4\n4\n5\n",
"0\n0\n0\n0\n3\n4\n4\n"
]
} | {
"input": [
"16 20 2\n10 3\n5 3\n10 5\n12 7\n7 6\n9 12\n9 6\n1 10\n11 16\n11 1\n16 2\n10 2\n14 4\n15 14\n4 13\n13 15\n1 8\n7 15\n1 7\n8 15\n",
"2 1 1\n2 1\n"
],
"output": [
"0\n0\n3\n3\n3\n3\n7\n7\n7\n7\n7\n11\n11\n11\n11\n15\n15\n15\n15\n16\n",
"2\n"
]
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
const int N = 1000007, inf = 0x3f3f3f3f;
set<int> e[N];
int n, m, k;
pair<int, int> edges[N];
set<int> st;
int main() {
ios::sync_with_stdio(false);
srand(time(NULL));
cin >> n >> m >> k;
for (int i = 0; i < m; ++i) {
int u, v;
cin >> u >> v;
e[u].insert... |
1037_E. Trips | There are n persons who initially don't know each other. On each morning, two of them, who were not friends before, become friends.
We want to plan a trip for every evening of m days. On each trip, you have to select a group of people that will go on the trip. For every person, one of the following should hold:
* ... | {
"input": [
"4 4 2\n2 3\n1 2\n1 3\n1 4\n",
"5 8 2\n2 1\n4 2\n5 4\n5 2\n4 3\n5 1\n4 1\n3 2\n",
"5 7 2\n1 5\n3 2\n2 5\n3 4\n1 2\n5 3\n1 3\n"
],
"output": [
"0\n0\n3\n3\n",
"0\n0\n0\n3\n3\n4\n4\n5\n",
"0\n0\n0\n0\n3\n4\n4\n"
]
} | {
"input": [
"16 20 2\n10 3\n5 3\n10 5\n12 7\n7 6\n9 12\n9 6\n1 10\n11 16\n11 1\n16 2\n10 2\n14 4\n15 14\n4 13\n13 15\n1 8\n7 15\n1 7\n8 15\n",
"2 1 1\n2 1\n"
],
"output": [
"0\n0\n3\n3\n3\n3\n7\n7\n7\n7\n7\n11\n11\n11\n11\n15\n15\n15\n15\n16\n",
"2\n"
]
} | IN-CORRECT | cpp | #include<bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
using namespace std;
#define ordered_set tree<int, null_type,less<int>, rb_tree_tag,tree_order_statistics_node_update>
#define ll long long int
#define ld long double
#define ff first
#defin... |
1037_E. Trips | There are n persons who initially don't know each other. On each morning, two of them, who were not friends before, become friends.
We want to plan a trip for every evening of m days. On each trip, you have to select a group of people that will go on the trip. For every person, one of the following should hold:
* ... | {
"input": [
"4 4 2\n2 3\n1 2\n1 3\n1 4\n",
"5 8 2\n2 1\n4 2\n5 4\n5 2\n4 3\n5 1\n4 1\n3 2\n",
"5 7 2\n1 5\n3 2\n2 5\n3 4\n1 2\n5 3\n1 3\n"
],
"output": [
"0\n0\n3\n3\n",
"0\n0\n0\n3\n3\n4\n4\n5\n",
"0\n0\n0\n0\n3\n4\n4\n"
]
} | {
"input": [
"16 20 2\n10 3\n5 3\n10 5\n12 7\n7 6\n9 12\n9 6\n1 10\n11 16\n11 1\n16 2\n10 2\n14 4\n15 14\n4 13\n13 15\n1 8\n7 15\n1 7\n8 15\n",
"2 1 1\n2 1\n"
],
"output": [
"0\n0\n3\n3\n3\n3\n7\n7\n7\n7\n7\n11\n11\n11\n11\n15\n15\n15\n15\n16\n",
"2\n"
]
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
const int N = 3e5 + 7;
const long long INF = 1e18 + 7;
pair<int, int> a[N];
int d[N], ans[N];
int n, m, k;
set<pair<int, int> > s;
vector<pair<int, int> > g[N];
int main() {
cin >> n >> m >> k;
for (int i = 1; i <= m; i++) {
scanf("%d%d", &a[i].first, &a[i].second);... |
1037_E. Trips | There are n persons who initially don't know each other. On each morning, two of them, who were not friends before, become friends.
We want to plan a trip for every evening of m days. On each trip, you have to select a group of people that will go on the trip. For every person, one of the following should hold:
* ... | {
"input": [
"4 4 2\n2 3\n1 2\n1 3\n1 4\n",
"5 8 2\n2 1\n4 2\n5 4\n5 2\n4 3\n5 1\n4 1\n3 2\n",
"5 7 2\n1 5\n3 2\n2 5\n3 4\n1 2\n5 3\n1 3\n"
],
"output": [
"0\n0\n3\n3\n",
"0\n0\n0\n3\n3\n4\n4\n5\n",
"0\n0\n0\n0\n3\n4\n4\n"
]
} | {
"input": [
"16 20 2\n10 3\n5 3\n10 5\n12 7\n7 6\n9 12\n9 6\n1 10\n11 16\n11 1\n16 2\n10 2\n14 4\n15 14\n4 13\n13 15\n1 8\n7 15\n1 7\n8 15\n",
"2 1 1\n2 1\n"
],
"output": [
"0\n0\n3\n3\n3\n3\n7\n7\n7\n7\n7\n11\n11\n11\n11\n15\n15\n15\n15\n16\n",
"2\n"
]
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 3;
int n, m, k, deg[N], x[N], y[N], ans[N], cnt;
set<int> G[N];
set<int>::iterator it;
void Del() {
queue<int> q;
for (int i = 1; i <= n; i++)
if (deg[i] > 0 && deg[i] < k) q.push(i);
while (!q.empty()) {
int u = q.front();
q.pop();
... |
1037_E. Trips | There are n persons who initially don't know each other. On each morning, two of them, who were not friends before, become friends.
We want to plan a trip for every evening of m days. On each trip, you have to select a group of people that will go on the trip. For every person, one of the following should hold:
* ... | {
"input": [
"4 4 2\n2 3\n1 2\n1 3\n1 4\n",
"5 8 2\n2 1\n4 2\n5 4\n5 2\n4 3\n5 1\n4 1\n3 2\n",
"5 7 2\n1 5\n3 2\n2 5\n3 4\n1 2\n5 3\n1 3\n"
],
"output": [
"0\n0\n3\n3\n",
"0\n0\n0\n3\n3\n4\n4\n5\n",
"0\n0\n0\n0\n3\n4\n4\n"
]
} | {
"input": [
"16 20 2\n10 3\n5 3\n10 5\n12 7\n7 6\n9 12\n9 6\n1 10\n11 16\n11 1\n16 2\n10 2\n14 4\n15 14\n4 13\n13 15\n1 8\n7 15\n1 7\n8 15\n",
"2 1 1\n2 1\n"
],
"output": [
"0\n0\n3\n3\n3\n3\n7\n7\n7\n7\n7\n11\n11\n11\n11\n15\n15\n15\n15\n16\n",
"2\n"
]
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
const int N = 2E5 + 10;
set<int> adj[N];
set<pair<int, int> > S;
int n, m, k;
int x[N], y[N];
int ans[N];
void Read_Input() {
scanf("%d%d%d", &n, &m, &k);
for (int i = 1; i <= m; i++) {
scanf("%d%d", &x[i], &y[i]);
adj[x[i]].insert(y[i]);
adj[y[i]].insert(x[... |
1037_E. Trips | There are n persons who initially don't know each other. On each morning, two of them, who were not friends before, become friends.
We want to plan a trip for every evening of m days. On each trip, you have to select a group of people that will go on the trip. For every person, one of the following should hold:
* ... | {
"input": [
"4 4 2\n2 3\n1 2\n1 3\n1 4\n",
"5 8 2\n2 1\n4 2\n5 4\n5 2\n4 3\n5 1\n4 1\n3 2\n",
"5 7 2\n1 5\n3 2\n2 5\n3 4\n1 2\n5 3\n1 3\n"
],
"output": [
"0\n0\n3\n3\n",
"0\n0\n0\n3\n3\n4\n4\n5\n",
"0\n0\n0\n0\n3\n4\n4\n"
]
} | {
"input": [
"16 20 2\n10 3\n5 3\n10 5\n12 7\n7 6\n9 12\n9 6\n1 10\n11 16\n11 1\n16 2\n10 2\n14 4\n15 14\n4 13\n13 15\n1 8\n7 15\n1 7\n8 15\n",
"2 1 1\n2 1\n"
],
"output": [
"0\n0\n3\n3\n3\n3\n7\n7\n7\n7\n7\n11\n11\n11\n11\n15\n15\n15\n15\n16\n",
"2\n"
]
} | IN-CORRECT | java | import java.io.*;
import java.util.*;
public class Main {
static int deg[];
static Set<Integer> s;
static Set[] graph;
static int n,k;
public static void main(String[] args) {
FastReader sc = new FastReader();
OutputWriter out = new OutputWriter(System.out);
n = sc.nextInt();
int m = sc.nextInt();
k = s... |
1037_E. Trips | There are n persons who initially don't know each other. On each morning, two of them, who were not friends before, become friends.
We want to plan a trip for every evening of m days. On each trip, you have to select a group of people that will go on the trip. For every person, one of the following should hold:
* ... | {
"input": [
"4 4 2\n2 3\n1 2\n1 3\n1 4\n",
"5 8 2\n2 1\n4 2\n5 4\n5 2\n4 3\n5 1\n4 1\n3 2\n",
"5 7 2\n1 5\n3 2\n2 5\n3 4\n1 2\n5 3\n1 3\n"
],
"output": [
"0\n0\n3\n3\n",
"0\n0\n0\n3\n3\n4\n4\n5\n",
"0\n0\n0\n0\n3\n4\n4\n"
]
} | {
"input": [
"16 20 2\n10 3\n5 3\n10 5\n12 7\n7 6\n9 12\n9 6\n1 10\n11 16\n11 1\n16 2\n10 2\n14 4\n15 14\n4 13\n13 15\n1 8\n7 15\n1 7\n8 15\n",
"2 1 1\n2 1\n"
],
"output": [
"0\n0\n3\n3\n3\n3\n7\n7\n7\n7\n7\n11\n11\n11\n11\n15\n15\n15\n15\n16\n",
"2\n"
]
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
const int MAXN = 2e5 + 5;
int n, m, k, x[MAXN], y[MAXN], deg[MAXN], cnt;
vector<pair<int, int> > E[MAXN];
bool removed[MAXN];
void ukloni(int x, int tijme) {
if (!removed[x] && deg[x] < k) {
removed[x] = true;
cnt--;
for (auto e : E[x])
if (e.second < ti... |
1037_E. Trips | There are n persons who initially don't know each other. On each morning, two of them, who were not friends before, become friends.
We want to plan a trip for every evening of m days. On each trip, you have to select a group of people that will go on the trip. For every person, one of the following should hold:
* ... | {
"input": [
"4 4 2\n2 3\n1 2\n1 3\n1 4\n",
"5 8 2\n2 1\n4 2\n5 4\n5 2\n4 3\n5 1\n4 1\n3 2\n",
"5 7 2\n1 5\n3 2\n2 5\n3 4\n1 2\n5 3\n1 3\n"
],
"output": [
"0\n0\n3\n3\n",
"0\n0\n0\n3\n3\n4\n4\n5\n",
"0\n0\n0\n0\n3\n4\n4\n"
]
} | {
"input": [
"16 20 2\n10 3\n5 3\n10 5\n12 7\n7 6\n9 12\n9 6\n1 10\n11 16\n11 1\n16 2\n10 2\n14 4\n15 14\n4 13\n13 15\n1 8\n7 15\n1 7\n8 15\n",
"2 1 1\n2 1\n"
],
"output": [
"0\n0\n3\n3\n3\n3\n7\n7\n7\n7\n7\n11\n11\n11\n11\n15\n15\n15\n15\n16\n",
"2\n"
]
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
struct con {
int x, y;
} a[200002];
struct node {
int id, d;
} p[200002];
int head[200002], ver[200002 * 2], nxt[200002 * 2], l;
int n, m, k, i, j, d[200002], ans[200002], sum;
bool cut[200002];
void insert(int x, int y) {
l++;
ver[l] = y;
nxt[l] = head[x];
head... |
1037_E. Trips | There are n persons who initially don't know each other. On each morning, two of them, who were not friends before, become friends.
We want to plan a trip for every evening of m days. On each trip, you have to select a group of people that will go on the trip. For every person, one of the following should hold:
* ... | {
"input": [
"4 4 2\n2 3\n1 2\n1 3\n1 4\n",
"5 8 2\n2 1\n4 2\n5 4\n5 2\n4 3\n5 1\n4 1\n3 2\n",
"5 7 2\n1 5\n3 2\n2 5\n3 4\n1 2\n5 3\n1 3\n"
],
"output": [
"0\n0\n3\n3\n",
"0\n0\n0\n3\n3\n4\n4\n5\n",
"0\n0\n0\n0\n3\n4\n4\n"
]
} | {
"input": [
"16 20 2\n10 3\n5 3\n10 5\n12 7\n7 6\n9 12\n9 6\n1 10\n11 16\n11 1\n16 2\n10 2\n14 4\n15 14\n4 13\n13 15\n1 8\n7 15\n1 7\n8 15\n",
"2 1 1\n2 1\n"
],
"output": [
"0\n0\n3\n3\n3\n3\n7\n7\n7\n7\n7\n11\n11\n11\n11\n15\n15\n15\n15\n16\n",
"2\n"
]
} | IN-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.Arrays;
import java.io.FilterInputStream;
import java.io.BufferedInputStream;
import java.util.TreeSet;
import java.io.InputStream;
/**
* @author khokharnikunj8
... |
1037_E. Trips | There are n persons who initially don't know each other. On each morning, two of them, who were not friends before, become friends.
We want to plan a trip for every evening of m days. On each trip, you have to select a group of people that will go on the trip. For every person, one of the following should hold:
* ... | {
"input": [
"4 4 2\n2 3\n1 2\n1 3\n1 4\n",
"5 8 2\n2 1\n4 2\n5 4\n5 2\n4 3\n5 1\n4 1\n3 2\n",
"5 7 2\n1 5\n3 2\n2 5\n3 4\n1 2\n5 3\n1 3\n"
],
"output": [
"0\n0\n3\n3\n",
"0\n0\n0\n3\n3\n4\n4\n5\n",
"0\n0\n0\n0\n3\n4\n4\n"
]
} | {
"input": [
"16 20 2\n10 3\n5 3\n10 5\n12 7\n7 6\n9 12\n9 6\n1 10\n11 16\n11 1\n16 2\n10 2\n14 4\n15 14\n4 13\n13 15\n1 8\n7 15\n1 7\n8 15\n",
"2 1 1\n2 1\n"
],
"output": [
"0\n0\n3\n3\n3\n3\n7\n7\n7\n7\n7\n11\n11\n11\n11\n15\n15\n15\n15\n16\n",
"2\n"
]
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
struct edge {
int u, v;
};
struct adj {
int v, key;
};
int n, m, k;
vector<edge> edges;
set<pair<int, int>> s;
stack<int> res;
bool deleted[200000] = {false};
int deg[200000] = {0};
vector<adj> graph[200000];
int main() {
ios_base::sync_with_stdio(false);
cin.tie(nu... |
1037_E. Trips | There are n persons who initially don't know each other. On each morning, two of them, who were not friends before, become friends.
We want to plan a trip for every evening of m days. On each trip, you have to select a group of people that will go on the trip. For every person, one of the following should hold:
* ... | {
"input": [
"4 4 2\n2 3\n1 2\n1 3\n1 4\n",
"5 8 2\n2 1\n4 2\n5 4\n5 2\n4 3\n5 1\n4 1\n3 2\n",
"5 7 2\n1 5\n3 2\n2 5\n3 4\n1 2\n5 3\n1 3\n"
],
"output": [
"0\n0\n3\n3\n",
"0\n0\n0\n3\n3\n4\n4\n5\n",
"0\n0\n0\n0\n3\n4\n4\n"
]
} | {
"input": [
"16 20 2\n10 3\n5 3\n10 5\n12 7\n7 6\n9 12\n9 6\n1 10\n11 16\n11 1\n16 2\n10 2\n14 4\n15 14\n4 13\n13 15\n1 8\n7 15\n1 7\n8 15\n",
"2 1 1\n2 1\n"
],
"output": [
"0\n0\n3\n3\n3\n3\n7\n7\n7\n7\n7\n11\n11\n11\n11\n15\n15\n15\n15\n16\n",
"2\n"
]
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
const int MAXN = 200011;
struct edge {
int x, y, pre, nxt;
} E[MAXN << 1];
vector<int> G[MAXN];
int e[MAXN];
int head[MAXN], vis[MAXN], deg[MAXN], ans[MAXN];
int n, m, k, tot = -1, cnt;
inline void AddEdge(int x, int y) {
E[++tot] = (edge){x, y, -1, head[x]};
if (head... |
1037_E. Trips | There are n persons who initially don't know each other. On each morning, two of them, who were not friends before, become friends.
We want to plan a trip for every evening of m days. On each trip, you have to select a group of people that will go on the trip. For every person, one of the following should hold:
* ... | {
"input": [
"4 4 2\n2 3\n1 2\n1 3\n1 4\n",
"5 8 2\n2 1\n4 2\n5 4\n5 2\n4 3\n5 1\n4 1\n3 2\n",
"5 7 2\n1 5\n3 2\n2 5\n3 4\n1 2\n5 3\n1 3\n"
],
"output": [
"0\n0\n3\n3\n",
"0\n0\n0\n3\n3\n4\n4\n5\n",
"0\n0\n0\n0\n3\n4\n4\n"
]
} | {
"input": [
"16 20 2\n10 3\n5 3\n10 5\n12 7\n7 6\n9 12\n9 6\n1 10\n11 16\n11 1\n16 2\n10 2\n14 4\n15 14\n4 13\n13 15\n1 8\n7 15\n1 7\n8 15\n",
"2 1 1\n2 1\n"
],
"output": [
"0\n0\n3\n3\n3\n3\n7\n7\n7\n7\n7\n11\n11\n11\n11\n15\n15\n15\n15\n16\n",
"2\n"
]
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
int n, Head[400010], Next[400010], Go[400010], A[400010], Fa[400010], D[400010],
Cnt = 0;
void addedge(int x, int y) {
Go[++Cnt] = y;
Next[Cnt] = Head[x];
Head[x] = Cnt;
}
void DFS(int x = 1) {
for (int T = Head[x]; T; T = Next[T])
if (Go[T] != Fa[x]) Fa[Go[... |
1037_E. Trips | There are n persons who initially don't know each other. On each morning, two of them, who were not friends before, become friends.
We want to plan a trip for every evening of m days. On each trip, you have to select a group of people that will go on the trip. For every person, one of the following should hold:
* ... | {
"input": [
"4 4 2\n2 3\n1 2\n1 3\n1 4\n",
"5 8 2\n2 1\n4 2\n5 4\n5 2\n4 3\n5 1\n4 1\n3 2\n",
"5 7 2\n1 5\n3 2\n2 5\n3 4\n1 2\n5 3\n1 3\n"
],
"output": [
"0\n0\n3\n3\n",
"0\n0\n0\n3\n3\n4\n4\n5\n",
"0\n0\n0\n0\n3\n4\n4\n"
]
} | {
"input": [
"16 20 2\n10 3\n5 3\n10 5\n12 7\n7 6\n9 12\n9 6\n1 10\n11 16\n11 1\n16 2\n10 2\n14 4\n15 14\n4 13\n13 15\n1 8\n7 15\n1 7\n8 15\n",
"2 1 1\n2 1\n"
],
"output": [
"0\n0\n3\n3\n3\n3\n7\n7\n7\n7\n7\n11\n11\n11\n11\n15\n15\n15\n15\n16\n",
"2\n"
]
} | IN-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.Arrays;
import java.io.FilterInputStream;
import java.io.BufferedInputStream;
import java.util.TreeSet;
import java.io.InputStream;
/**
* @author khokharnikunj8
... |
1037_E. Trips | There are n persons who initially don't know each other. On each morning, two of them, who were not friends before, become friends.
We want to plan a trip for every evening of m days. On each trip, you have to select a group of people that will go on the trip. For every person, one of the following should hold:
* ... | {
"input": [
"4 4 2\n2 3\n1 2\n1 3\n1 4\n",
"5 8 2\n2 1\n4 2\n5 4\n5 2\n4 3\n5 1\n4 1\n3 2\n",
"5 7 2\n1 5\n3 2\n2 5\n3 4\n1 2\n5 3\n1 3\n"
],
"output": [
"0\n0\n3\n3\n",
"0\n0\n0\n3\n3\n4\n4\n5\n",
"0\n0\n0\n0\n3\n4\n4\n"
]
} | {
"input": [
"16 20 2\n10 3\n5 3\n10 5\n12 7\n7 6\n9 12\n9 6\n1 10\n11 16\n11 1\n16 2\n10 2\n14 4\n15 14\n4 13\n13 15\n1 8\n7 15\n1 7\n8 15\n",
"2 1 1\n2 1\n"
],
"output": [
"0\n0\n3\n3\n3\n3\n7\n7\n7\n7\n7\n11\n11\n11\n11\n15\n15\n15\n15\n16\n",
"2\n"
]
} | IN-CORRECT | cpp | // amigos y no más
// dile a la jardinera que traigo flores
// += O(logn) ; + = O(n)
//#pragma GCC optimize("Ofast")
//#pragma GCC optimize("unroll-loops")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#include<bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp> // Common file
... |
1037_E. Trips | There are n persons who initially don't know each other. On each morning, two of them, who were not friends before, become friends.
We want to plan a trip for every evening of m days. On each trip, you have to select a group of people that will go on the trip. For every person, one of the following should hold:
* ... | {
"input": [
"4 4 2\n2 3\n1 2\n1 3\n1 4\n",
"5 8 2\n2 1\n4 2\n5 4\n5 2\n4 3\n5 1\n4 1\n3 2\n",
"5 7 2\n1 5\n3 2\n2 5\n3 4\n1 2\n5 3\n1 3\n"
],
"output": [
"0\n0\n3\n3\n",
"0\n0\n0\n3\n3\n4\n4\n5\n",
"0\n0\n0\n0\n3\n4\n4\n"
]
} | {
"input": [
"16 20 2\n10 3\n5 3\n10 5\n12 7\n7 6\n9 12\n9 6\n1 10\n11 16\n11 1\n16 2\n10 2\n14 4\n15 14\n4 13\n13 15\n1 8\n7 15\n1 7\n8 15\n",
"2 1 1\n2 1\n"
],
"output": [
"0\n0\n3\n3\n3\n3\n7\n7\n7\n7\n7\n11\n11\n11\n11\n15\n15\n15\n15\n16\n",
"2\n"
]
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
int n, m, k, ans, d[200010], u[200010], v[200010], t[200010];
std::set<int> next[200010];
inline void remove(int x) {
if (!d[x]) return;
ans--;
d[x] = 0;
for (auto y : next[x]) {
next[y].erase(x);
if (d[y] && --d[y] < k) remove(y);
}
next[x].clear();
}
int main() {
scanf("... |
1037_E. Trips | There are n persons who initially don't know each other. On each morning, two of them, who were not friends before, become friends.
We want to plan a trip for every evening of m days. On each trip, you have to select a group of people that will go on the trip. For every person, one of the following should hold:
* ... | {
"input": [
"4 4 2\n2 3\n1 2\n1 3\n1 4\n",
"5 8 2\n2 1\n4 2\n5 4\n5 2\n4 3\n5 1\n4 1\n3 2\n",
"5 7 2\n1 5\n3 2\n2 5\n3 4\n1 2\n5 3\n1 3\n"
],
"output": [
"0\n0\n3\n3\n",
"0\n0\n0\n3\n3\n4\n4\n5\n",
"0\n0\n0\n0\n3\n4\n4\n"
]
} | {
"input": [
"16 20 2\n10 3\n5 3\n10 5\n12 7\n7 6\n9 12\n9 6\n1 10\n11 16\n11 1\n16 2\n10 2\n14 4\n15 14\n4 13\n13 15\n1 8\n7 15\n1 7\n8 15\n",
"2 1 1\n2 1\n"
],
"output": [
"0\n0\n3\n3\n3\n3\n7\n7\n7\n7\n7\n11\n11\n11\n11\n15\n15\n15\n15\n16\n",
"2\n"
]
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
vector<int> ad[200005];
set<int> s;
vector<int> r;
set<int> w;
int k;
int ff(int x) {
int q, e, count1 = 0, cc = 1;
for (int i = 0; i < ad[x].size(); i++) {
q = ad[x][i];
if (w.find(q) != w.end()) {
continue;
cc++;
}
if (s.find(q) == s.end())... |
1037_E. Trips | There are n persons who initially don't know each other. On each morning, two of them, who were not friends before, become friends.
We want to plan a trip for every evening of m days. On each trip, you have to select a group of people that will go on the trip. For every person, one of the following should hold:
* ... | {
"input": [
"4 4 2\n2 3\n1 2\n1 3\n1 4\n",
"5 8 2\n2 1\n4 2\n5 4\n5 2\n4 3\n5 1\n4 1\n3 2\n",
"5 7 2\n1 5\n3 2\n2 5\n3 4\n1 2\n5 3\n1 3\n"
],
"output": [
"0\n0\n3\n3\n",
"0\n0\n0\n3\n3\n4\n4\n5\n",
"0\n0\n0\n0\n3\n4\n4\n"
]
} | {
"input": [
"16 20 2\n10 3\n5 3\n10 5\n12 7\n7 6\n9 12\n9 6\n1 10\n11 16\n11 1\n16 2\n10 2\n14 4\n15 14\n4 13\n13 15\n1 8\n7 15\n1 7\n8 15\n",
"2 1 1\n2 1\n"
],
"output": [
"0\n0\n3\n3\n3\n3\n7\n7\n7\n7\n7\n11\n11\n11\n11\n15\n15\n15\n15\n16\n",
"2\n"
]
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
template <class T>
using v2d = vector<vector<T>>;
template <class T>
bool uin(T &a, T b) {
return a > b ? (a = b, true) : false;
}
template <class T>
bool uax(T &a, T b) {
return a < b ? (a = b, true) : false;
}
mt19937 rng(chrono::system_clock::now().time_since_epoch()... |
1037_E. Trips | There are n persons who initially don't know each other. On each morning, two of them, who were not friends before, become friends.
We want to plan a trip for every evening of m days. On each trip, you have to select a group of people that will go on the trip. For every person, one of the following should hold:
* ... | {
"input": [
"4 4 2\n2 3\n1 2\n1 3\n1 4\n",
"5 8 2\n2 1\n4 2\n5 4\n5 2\n4 3\n5 1\n4 1\n3 2\n",
"5 7 2\n1 5\n3 2\n2 5\n3 4\n1 2\n5 3\n1 3\n"
],
"output": [
"0\n0\n3\n3\n",
"0\n0\n0\n3\n3\n4\n4\n5\n",
"0\n0\n0\n0\n3\n4\n4\n"
]
} | {
"input": [
"16 20 2\n10 3\n5 3\n10 5\n12 7\n7 6\n9 12\n9 6\n1 10\n11 16\n11 1\n16 2\n10 2\n14 4\n15 14\n4 13\n13 15\n1 8\n7 15\n1 7\n8 15\n",
"2 1 1\n2 1\n"
],
"output": [
"0\n0\n3\n3\n3\n3\n7\n7\n7\n7\n7\n11\n11\n11\n11\n15\n15\n15\n15\n16\n",
"2\n"
]
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
int ANS = 0;
int n, m, k;
vector<vector<int> > adj;
vector<int> gr, friends, trip, vis;
int main() {
ios::sync_with_stdio(0);
cin.tie(0);
int n, m, k;
cin >> n >> m >> k;
adj = vector<vector<int> >(n);
gr = friends = trip = vector<int>(n);
for (int i = 0; i < ... |
1037_E. Trips | There are n persons who initially don't know each other. On each morning, two of them, who were not friends before, become friends.
We want to plan a trip for every evening of m days. On each trip, you have to select a group of people that will go on the trip. For every person, one of the following should hold:
* ... | {
"input": [
"4 4 2\n2 3\n1 2\n1 3\n1 4\n",
"5 8 2\n2 1\n4 2\n5 4\n5 2\n4 3\n5 1\n4 1\n3 2\n",
"5 7 2\n1 5\n3 2\n2 5\n3 4\n1 2\n5 3\n1 3\n"
],
"output": [
"0\n0\n3\n3\n",
"0\n0\n0\n3\n3\n4\n4\n5\n",
"0\n0\n0\n0\n3\n4\n4\n"
]
} | {
"input": [
"16 20 2\n10 3\n5 3\n10 5\n12 7\n7 6\n9 12\n9 6\n1 10\n11 16\n11 1\n16 2\n10 2\n14 4\n15 14\n4 13\n13 15\n1 8\n7 15\n1 7\n8 15\n",
"2 1 1\n2 1\n"
],
"output": [
"0\n0\n3\n3\n3\n3\n7\n7\n7\n7\n7\n11\n11\n11\n11\n15\n15\n15\n15\n16\n",
"2\n"
]
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
const int N = 2e5 + 5;
using namespace std;
int n, m, k;
int res[N];
int x[N], y[N];
int bac[N];
set<int> a[N];
set<pair<int, int> > S;
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
cin >> n >> m >> k;
for (int i = 1; i <= m; i++) {
cin >> x[i] >> y[i];
a... |
1037_E. Trips | There are n persons who initially don't know each other. On each morning, two of them, who were not friends before, become friends.
We want to plan a trip for every evening of m days. On each trip, you have to select a group of people that will go on the trip. For every person, one of the following should hold:
* ... | {
"input": [
"4 4 2\n2 3\n1 2\n1 3\n1 4\n",
"5 8 2\n2 1\n4 2\n5 4\n5 2\n4 3\n5 1\n4 1\n3 2\n",
"5 7 2\n1 5\n3 2\n2 5\n3 4\n1 2\n5 3\n1 3\n"
],
"output": [
"0\n0\n3\n3\n",
"0\n0\n0\n3\n3\n4\n4\n5\n",
"0\n0\n0\n0\n3\n4\n4\n"
]
} | {
"input": [
"16 20 2\n10 3\n5 3\n10 5\n12 7\n7 6\n9 12\n9 6\n1 10\n11 16\n11 1\n16 2\n10 2\n14 4\n15 14\n4 13\n13 15\n1 8\n7 15\n1 7\n8 15\n",
"2 1 1\n2 1\n"
],
"output": [
"0\n0\n3\n3\n3\n3\n7\n7\n7\n7\n7\n11\n11\n11\n11\n15\n15\n15\n15\n16\n",
"2\n"
]
} | IN-CORRECT | python3 | from collections import defaultdict
n, m, k = map(int, input().split())
xys = [tuple(map(int, input().split())) for _ in range(m)][::-1]
gv = defaultdict(set)
for x, y in xys:
gv[x].add(y)
gv[y].add(x)
q = [u for u, vs in tuple(gv.items()) if len(vs) < k]
i = 0
while i < len(q):
u = q[i]
i += 1
... |
1037_E. Trips | There are n persons who initially don't know each other. On each morning, two of them, who were not friends before, become friends.
We want to plan a trip for every evening of m days. On each trip, you have to select a group of people that will go on the trip. For every person, one of the following should hold:
* ... | {
"input": [
"4 4 2\n2 3\n1 2\n1 3\n1 4\n",
"5 8 2\n2 1\n4 2\n5 4\n5 2\n4 3\n5 1\n4 1\n3 2\n",
"5 7 2\n1 5\n3 2\n2 5\n3 4\n1 2\n5 3\n1 3\n"
],
"output": [
"0\n0\n3\n3\n",
"0\n0\n0\n3\n3\n4\n4\n5\n",
"0\n0\n0\n0\n3\n4\n4\n"
]
} | {
"input": [
"16 20 2\n10 3\n5 3\n10 5\n12 7\n7 6\n9 12\n9 6\n1 10\n11 16\n11 1\n16 2\n10 2\n14 4\n15 14\n4 13\n13 15\n1 8\n7 15\n1 7\n8 15\n",
"2 1 1\n2 1\n"
],
"output": [
"0\n0\n3\n3\n3\n3\n7\n7\n7\n7\n7\n11\n11\n11\n11\n15\n15\n15\n15\n16\n",
"2\n"
]
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
const int ms = 200200;
int n, m, k;
int ans = 0;
int size[ms], par[ms];
std::pair<int, int> edges[ms];
std::vector<int> use[ms], dels[ms];
int deg[ms];
bool del[ms];
int getPar(int x) { return x == par[x] ? x : par[x] = getPar(par[x]); }
void makeUnion(int a, int b) {
a = getPar(a);
b = get... |
1037_E. Trips | There are n persons who initially don't know each other. On each morning, two of them, who were not friends before, become friends.
We want to plan a trip for every evening of m days. On each trip, you have to select a group of people that will go on the trip. For every person, one of the following should hold:
* ... | {
"input": [
"4 4 2\n2 3\n1 2\n1 3\n1 4\n",
"5 8 2\n2 1\n4 2\n5 4\n5 2\n4 3\n5 1\n4 1\n3 2\n",
"5 7 2\n1 5\n3 2\n2 5\n3 4\n1 2\n5 3\n1 3\n"
],
"output": [
"0\n0\n3\n3\n",
"0\n0\n0\n3\n3\n4\n4\n5\n",
"0\n0\n0\n0\n3\n4\n4\n"
]
} | {
"input": [
"16 20 2\n10 3\n5 3\n10 5\n12 7\n7 6\n9 12\n9 6\n1 10\n11 16\n11 1\n16 2\n10 2\n14 4\n15 14\n4 13\n13 15\n1 8\n7 15\n1 7\n8 15\n",
"2 1 1\n2 1\n"
],
"output": [
"0\n0\n3\n3\n3\n3\n7\n7\n7\n7\n7\n11\n11\n11\n11\n15\n15\n15\n15\n16\n",
"2\n"
]
} | IN-CORRECT | java | /* Author: Ronak Agarwal, reader part not written by me*/
import java.io.* ; import java.util.* ;
import static java.lang.Math.min ; import static java.lang.Math.max ;
import static java.lang.Math.abs ; import static java.lang.Math.log ;
import static java.lang.Math.pow ; import static java.lang.Math.sqrt ;
/* Thread i... |
1037_E. Trips | There are n persons who initially don't know each other. On each morning, two of them, who were not friends before, become friends.
We want to plan a trip for every evening of m days. On each trip, you have to select a group of people that will go on the trip. For every person, one of the following should hold:
* ... | {
"input": [
"4 4 2\n2 3\n1 2\n1 3\n1 4\n",
"5 8 2\n2 1\n4 2\n5 4\n5 2\n4 3\n5 1\n4 1\n3 2\n",
"5 7 2\n1 5\n3 2\n2 5\n3 4\n1 2\n5 3\n1 3\n"
],
"output": [
"0\n0\n3\n3\n",
"0\n0\n0\n3\n3\n4\n4\n5\n",
"0\n0\n0\n0\n3\n4\n4\n"
]
} | {
"input": [
"16 20 2\n10 3\n5 3\n10 5\n12 7\n7 6\n9 12\n9 6\n1 10\n11 16\n11 1\n16 2\n10 2\n14 4\n15 14\n4 13\n13 15\n1 8\n7 15\n1 7\n8 15\n",
"2 1 1\n2 1\n"
],
"output": [
"0\n0\n3\n3\n3\n3\n7\n7\n7\n7\n7\n11\n11\n11\n11\n15\n15\n15\n15\n16\n",
"2\n"
]
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
signed main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
long long n, m, k;
cin >> n >> m >> k;
vector<vector<pair<long long, long long> > > friends(n);
vector<long long> pointers(n);
for (long long i = 0; i < m; i++) {
long long a, b;
cin >> a... |
1037_E. Trips | There are n persons who initially don't know each other. On each morning, two of them, who were not friends before, become friends.
We want to plan a trip for every evening of m days. On each trip, you have to select a group of people that will go on the trip. For every person, one of the following should hold:
* ... | {
"input": [
"4 4 2\n2 3\n1 2\n1 3\n1 4\n",
"5 8 2\n2 1\n4 2\n5 4\n5 2\n4 3\n5 1\n4 1\n3 2\n",
"5 7 2\n1 5\n3 2\n2 5\n3 4\n1 2\n5 3\n1 3\n"
],
"output": [
"0\n0\n3\n3\n",
"0\n0\n0\n3\n3\n4\n4\n5\n",
"0\n0\n0\n0\n3\n4\n4\n"
]
} | {
"input": [
"16 20 2\n10 3\n5 3\n10 5\n12 7\n7 6\n9 12\n9 6\n1 10\n11 16\n11 1\n16 2\n10 2\n14 4\n15 14\n4 13\n13 15\n1 8\n7 15\n1 7\n8 15\n",
"2 1 1\n2 1\n"
],
"output": [
"0\n0\n3\n3\n3\n3\n7\n7\n7\n7\n7\n11\n11\n11\n11\n15\n15\n15\n15\n16\n",
"2\n"
]
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
template <class T1, class T2>
ostream& operator<<(ostream& out, pair<T1, T2>& p) {
out << p.first << ' ' << p.second;
}
template <class T>
istream& operator>>(istream& in, vector<T>& v) {
for (auto& x : v) in >> x;
return in;
}
template <class T>
ostream& operator<<(o... |
1037_E. Trips | There are n persons who initially don't know each other. On each morning, two of them, who were not friends before, become friends.
We want to plan a trip for every evening of m days. On each trip, you have to select a group of people that will go on the trip. For every person, one of the following should hold:
* ... | {
"input": [
"4 4 2\n2 3\n1 2\n1 3\n1 4\n",
"5 8 2\n2 1\n4 2\n5 4\n5 2\n4 3\n5 1\n4 1\n3 2\n",
"5 7 2\n1 5\n3 2\n2 5\n3 4\n1 2\n5 3\n1 3\n"
],
"output": [
"0\n0\n3\n3\n",
"0\n0\n0\n3\n3\n4\n4\n5\n",
"0\n0\n0\n0\n3\n4\n4\n"
]
} | {
"input": [
"16 20 2\n10 3\n5 3\n10 5\n12 7\n7 6\n9 12\n9 6\n1 10\n11 16\n11 1\n16 2\n10 2\n14 4\n15 14\n4 13\n13 15\n1 8\n7 15\n1 7\n8 15\n",
"2 1 1\n2 1\n"
],
"output": [
"0\n0\n3\n3\n3\n3\n7\n7\n7\n7\n7\n11\n11\n11\n11\n15\n15\n15\n15\n16\n",
"2\n"
]
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
int n, m, k;
struct edge {
int u, v, nxt;
} e[200005 * 2];
int cnt, head[200005];
void adde(int u, int v) {
e[++cnt].u = u;
e[cnt].v = v;
e[cnt].nxt = head[u];
head[u] = cnt;
}
int vis[200005 * 2], d[200005], ans[200005], res;
struct ege2 {
int u, v;
} edge[2000... |
1037_E. Trips | There are n persons who initially don't know each other. On each morning, two of them, who were not friends before, become friends.
We want to plan a trip for every evening of m days. On each trip, you have to select a group of people that will go on the trip. For every person, one of the following should hold:
* ... | {
"input": [
"4 4 2\n2 3\n1 2\n1 3\n1 4\n",
"5 8 2\n2 1\n4 2\n5 4\n5 2\n4 3\n5 1\n4 1\n3 2\n",
"5 7 2\n1 5\n3 2\n2 5\n3 4\n1 2\n5 3\n1 3\n"
],
"output": [
"0\n0\n3\n3\n",
"0\n0\n0\n3\n3\n4\n4\n5\n",
"0\n0\n0\n0\n3\n4\n4\n"
]
} | {
"input": [
"16 20 2\n10 3\n5 3\n10 5\n12 7\n7 6\n9 12\n9 6\n1 10\n11 16\n11 1\n16 2\n10 2\n14 4\n15 14\n4 13\n13 15\n1 8\n7 15\n1 7\n8 15\n",
"2 1 1\n2 1\n"
],
"output": [
"0\n0\n3\n3\n3\n3\n7\n7\n7\n7\n7\n11\n11\n11\n11\n15\n15\n15\n15\n16\n",
"2\n"
]
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
const int maxn = 2e5 + 5;
int deg[maxn];
vector<int> G[maxn];
pair<int, int> o[maxn];
int ans[maxn], vis[maxn];
int main() {
int n, m, k, u, v;
cin >> n >> m >> k;
for (int i = 1; i <= m; i++) {
scanf("%d%d", &u, &v);
;
G[u].push_back(v);
G[v].push_bac... |
1037_E. Trips | There are n persons who initially don't know each other. On each morning, two of them, who were not friends before, become friends.
We want to plan a trip for every evening of m days. On each trip, you have to select a group of people that will go on the trip. For every person, one of the following should hold:
* ... | {
"input": [
"4 4 2\n2 3\n1 2\n1 3\n1 4\n",
"5 8 2\n2 1\n4 2\n5 4\n5 2\n4 3\n5 1\n4 1\n3 2\n",
"5 7 2\n1 5\n3 2\n2 5\n3 4\n1 2\n5 3\n1 3\n"
],
"output": [
"0\n0\n3\n3\n",
"0\n0\n0\n3\n3\n4\n4\n5\n",
"0\n0\n0\n0\n3\n4\n4\n"
]
} | {
"input": [
"16 20 2\n10 3\n5 3\n10 5\n12 7\n7 6\n9 12\n9 6\n1 10\n11 16\n11 1\n16 2\n10 2\n14 4\n15 14\n4 13\n13 15\n1 8\n7 15\n1 7\n8 15\n",
"2 1 1\n2 1\n"
],
"output": [
"0\n0\n3\n3\n3\n3\n7\n7\n7\n7\n7\n11\n11\n11\n11\n15\n15\n15\n15\n16\n",
"2\n"
]
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
int N, M, K;
vector<int> Adj[200005];
set<int> f[200005];
bool vis[200005];
int main() {
ios::sync_with_stdio(0);
cin >> N >> M >> K;
int u, v;
int ans = 0;
for (int i = 1; i <= M; i++) {
cin >> u >> v;
Adj[u].push_back(v);
Adj[v].push_back(u);
f[u... |
1037_E. Trips | There are n persons who initially don't know each other. On each morning, two of them, who were not friends before, become friends.
We want to plan a trip for every evening of m days. On each trip, you have to select a group of people that will go on the trip. For every person, one of the following should hold:
* ... | {
"input": [
"4 4 2\n2 3\n1 2\n1 3\n1 4\n",
"5 8 2\n2 1\n4 2\n5 4\n5 2\n4 3\n5 1\n4 1\n3 2\n",
"5 7 2\n1 5\n3 2\n2 5\n3 4\n1 2\n5 3\n1 3\n"
],
"output": [
"0\n0\n3\n3\n",
"0\n0\n0\n3\n3\n4\n4\n5\n",
"0\n0\n0\n0\n3\n4\n4\n"
]
} | {
"input": [
"16 20 2\n10 3\n5 3\n10 5\n12 7\n7 6\n9 12\n9 6\n1 10\n11 16\n11 1\n16 2\n10 2\n14 4\n15 14\n4 13\n13 15\n1 8\n7 15\n1 7\n8 15\n",
"2 1 1\n2 1\n"
],
"output": [
"0\n0\n3\n3\n3\n3\n7\n7\n7\n7\n7\n11\n11\n11\n11\n15\n15\n15\n15\n16\n",
"2\n"
]
} | IN-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.Arrays;
import java.io.FilterInputStream;
import java.io.BufferedInputStream;
import java.util.TreeSet;
import java.io.InputStream;
/**
* @author khokharnikunj8
... |
1037_E. Trips | There are n persons who initially don't know each other. On each morning, two of them, who were not friends before, become friends.
We want to plan a trip for every evening of m days. On each trip, you have to select a group of people that will go on the trip. For every person, one of the following should hold:
* ... | {
"input": [
"4 4 2\n2 3\n1 2\n1 3\n1 4\n",
"5 8 2\n2 1\n4 2\n5 4\n5 2\n4 3\n5 1\n4 1\n3 2\n",
"5 7 2\n1 5\n3 2\n2 5\n3 4\n1 2\n5 3\n1 3\n"
],
"output": [
"0\n0\n3\n3\n",
"0\n0\n0\n3\n3\n4\n4\n5\n",
"0\n0\n0\n0\n3\n4\n4\n"
]
} | {
"input": [
"16 20 2\n10 3\n5 3\n10 5\n12 7\n7 6\n9 12\n9 6\n1 10\n11 16\n11 1\n16 2\n10 2\n14 4\n15 14\n4 13\n13 15\n1 8\n7 15\n1 7\n8 15\n",
"2 1 1\n2 1\n"
],
"output": [
"0\n0\n3\n3\n3\n3\n7\n7\n7\n7\n7\n11\n11\n11\n11\n15\n15\n15\n15\n16\n",
"2\n"
]
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#define fio ios_base::sync_with_stdio(false)
#define pdl cout << "*" << endl
#define MOD 1000000007
#define INF 1000000000
#define INFLL 1000000000000000000ll
#define mp make_pair
#define pb push_back
#define ff first... |
1037_E. Trips | There are n persons who initially don't know each other. On each morning, two of them, who were not friends before, become friends.
We want to plan a trip for every evening of m days. On each trip, you have to select a group of people that will go on the trip. For every person, one of the following should hold:
* ... | {
"input": [
"4 4 2\n2 3\n1 2\n1 3\n1 4\n",
"5 8 2\n2 1\n4 2\n5 4\n5 2\n4 3\n5 1\n4 1\n3 2\n",
"5 7 2\n1 5\n3 2\n2 5\n3 4\n1 2\n5 3\n1 3\n"
],
"output": [
"0\n0\n3\n3\n",
"0\n0\n0\n3\n3\n4\n4\n5\n",
"0\n0\n0\n0\n3\n4\n4\n"
]
} | {
"input": [
"16 20 2\n10 3\n5 3\n10 5\n12 7\n7 6\n9 12\n9 6\n1 10\n11 16\n11 1\n16 2\n10 2\n14 4\n15 14\n4 13\n13 15\n1 8\n7 15\n1 7\n8 15\n",
"2 1 1\n2 1\n"
],
"output": [
"0\n0\n3\n3\n3\n3\n7\n7\n7\n7\n7\n11\n11\n11\n11\n15\n15\n15\n15\n16\n",
"2\n"
]
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
const int nmax = 2e5 + 5;
const int mod = 1e9 + 7;
vector<int> g[nmax];
int destroyed;
pair<int, int> ej[nmax];
int deg[nmax], ans[nmax];
bool vis[nmax];
void dhongsho(int u, int k, int p) {
if (vis[u]) return;
vis[u] = true;
destroyed++;
for (auto v : g[u]) {
i... |
1037_E. Trips | There are n persons who initially don't know each other. On each morning, two of them, who were not friends before, become friends.
We want to plan a trip for every evening of m days. On each trip, you have to select a group of people that will go on the trip. For every person, one of the following should hold:
* ... | {
"input": [
"4 4 2\n2 3\n1 2\n1 3\n1 4\n",
"5 8 2\n2 1\n4 2\n5 4\n5 2\n4 3\n5 1\n4 1\n3 2\n",
"5 7 2\n1 5\n3 2\n2 5\n3 4\n1 2\n5 3\n1 3\n"
],
"output": [
"0\n0\n3\n3\n",
"0\n0\n0\n3\n3\n4\n4\n5\n",
"0\n0\n0\n0\n3\n4\n4\n"
]
} | {
"input": [
"16 20 2\n10 3\n5 3\n10 5\n12 7\n7 6\n9 12\n9 6\n1 10\n11 16\n11 1\n16 2\n10 2\n14 4\n15 14\n4 13\n13 15\n1 8\n7 15\n1 7\n8 15\n",
"2 1 1\n2 1\n"
],
"output": [
"0\n0\n3\n3\n3\n3\n7\n7\n7\n7\n7\n11\n11\n11\n11\n15\n15\n15\n15\n16\n",
"2\n"
]
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
int main() {
ios_base::sync_with_stdio(false);
cin.tie(0);
int n, m, k;
cin >> n >> m >> k;
pair<int, int> vv[m];
set<int> adj[n];
int deg[n];
for (int i = 0; i < n; i++) deg[i] = 0;
for (int i = 0; i < m; i++) {
int x, y;
cin >> x >> y;
x--;
... |
1037_E. Trips | There are n persons who initially don't know each other. On each morning, two of them, who were not friends before, become friends.
We want to plan a trip for every evening of m days. On each trip, you have to select a group of people that will go on the trip. For every person, one of the following should hold:
* ... | {
"input": [
"4 4 2\n2 3\n1 2\n1 3\n1 4\n",
"5 8 2\n2 1\n4 2\n5 4\n5 2\n4 3\n5 1\n4 1\n3 2\n",
"5 7 2\n1 5\n3 2\n2 5\n3 4\n1 2\n5 3\n1 3\n"
],
"output": [
"0\n0\n3\n3\n",
"0\n0\n0\n3\n3\n4\n4\n5\n",
"0\n0\n0\n0\n3\n4\n4\n"
]
} | {
"input": [
"16 20 2\n10 3\n5 3\n10 5\n12 7\n7 6\n9 12\n9 6\n1 10\n11 16\n11 1\n16 2\n10 2\n14 4\n15 14\n4 13\n13 15\n1 8\n7 15\n1 7\n8 15\n",
"2 1 1\n2 1\n"
],
"output": [
"0\n0\n3\n3\n3\n3\n7\n7\n7\n7\n7\n11\n11\n11\n11\n15\n15\n15\n15\n16\n",
"2\n"
]
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
inline int read() {
int sum = 0, ff = 1;
char ch = getchar();
while (!isdigit(ch)) {
if (ch == '-') ff = -1;
ch = getchar();
}
while (isdigit(ch)) sum = sum * 10 + (ch ^ 48), ch = getchar();
return sum * ff;
}
const int mod = 1e9 + 7;
const int mo = 9982... |
1037_E. Trips | There are n persons who initially don't know each other. On each morning, two of them, who were not friends before, become friends.
We want to plan a trip for every evening of m days. On each trip, you have to select a group of people that will go on the trip. For every person, one of the following should hold:
* ... | {
"input": [
"4 4 2\n2 3\n1 2\n1 3\n1 4\n",
"5 8 2\n2 1\n4 2\n5 4\n5 2\n4 3\n5 1\n4 1\n3 2\n",
"5 7 2\n1 5\n3 2\n2 5\n3 4\n1 2\n5 3\n1 3\n"
],
"output": [
"0\n0\n3\n3\n",
"0\n0\n0\n3\n3\n4\n4\n5\n",
"0\n0\n0\n0\n3\n4\n4\n"
]
} | {
"input": [
"16 20 2\n10 3\n5 3\n10 5\n12 7\n7 6\n9 12\n9 6\n1 10\n11 16\n11 1\n16 2\n10 2\n14 4\n15 14\n4 13\n13 15\n1 8\n7 15\n1 7\n8 15\n",
"2 1 1\n2 1\n"
],
"output": [
"0\n0\n3\n3\n3\n3\n7\n7\n7\n7\n7\n11\n11\n11\n11\n15\n15\n15\n15\n16\n",
"2\n"
]
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
const long long MOD = 1e9 + 7;
const long long N = 3e5 + 10;
long long x[N], y[N], deg[N];
vector<long long> vp[N];
int32_t main() {
ios_base::sync_with_stdio(false);
cin.tie(0);
long long n, m, k;
cin >> n >> m >> k;
for (long long i = 1; i <= m; i++) {
cin >... |
1037_E. Trips | There are n persons who initially don't know each other. On each morning, two of them, who were not friends before, become friends.
We want to plan a trip for every evening of m days. On each trip, you have to select a group of people that will go on the trip. For every person, one of the following should hold:
* ... | {
"input": [
"4 4 2\n2 3\n1 2\n1 3\n1 4\n",
"5 8 2\n2 1\n4 2\n5 4\n5 2\n4 3\n5 1\n4 1\n3 2\n",
"5 7 2\n1 5\n3 2\n2 5\n3 4\n1 2\n5 3\n1 3\n"
],
"output": [
"0\n0\n3\n3\n",
"0\n0\n0\n3\n3\n4\n4\n5\n",
"0\n0\n0\n0\n3\n4\n4\n"
]
} | {
"input": [
"16 20 2\n10 3\n5 3\n10 5\n12 7\n7 6\n9 12\n9 6\n1 10\n11 16\n11 1\n16 2\n10 2\n14 4\n15 14\n4 13\n13 15\n1 8\n7 15\n1 7\n8 15\n",
"2 1 1\n2 1\n"
],
"output": [
"0\n0\n3\n3\n3\n3\n7\n7\n7\n7\n7\n11\n11\n11\n11\n15\n15\n15\n15\n16\n",
"2\n"
]
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
const int N = 2e5 + 5;
using namespace std;
int n, m, k;
int res[N];
int x[N], y[N];
int bac[N];
set<int> a[N];
set<pair<int, int> > S;
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
cin >> n >> m >> k;
for (int i = 1; i <= m; i++) {
cin >> x[i] >> y[i];
a... |
1037_E. Trips | There are n persons who initially don't know each other. On each morning, two of them, who were not friends before, become friends.
We want to plan a trip for every evening of m days. On each trip, you have to select a group of people that will go on the trip. For every person, one of the following should hold:
* ... | {
"input": [
"4 4 2\n2 3\n1 2\n1 3\n1 4\n",
"5 8 2\n2 1\n4 2\n5 4\n5 2\n4 3\n5 1\n4 1\n3 2\n",
"5 7 2\n1 5\n3 2\n2 5\n3 4\n1 2\n5 3\n1 3\n"
],
"output": [
"0\n0\n3\n3\n",
"0\n0\n0\n3\n3\n4\n4\n5\n",
"0\n0\n0\n0\n3\n4\n4\n"
]
} | {
"input": [
"16 20 2\n10 3\n5 3\n10 5\n12 7\n7 6\n9 12\n9 6\n1 10\n11 16\n11 1\n16 2\n10 2\n14 4\n15 14\n4 13\n13 15\n1 8\n7 15\n1 7\n8 15\n",
"2 1 1\n2 1\n"
],
"output": [
"0\n0\n3\n3\n3\n3\n7\n7\n7\n7\n7\n11\n11\n11\n11\n15\n15\n15\n15\n16\n",
"2\n"
]
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
long long getBit(long long num, int idx) { return ((num >> idx) & 1ll) == 1ll; }
long long setBit1(long long num, int idx) { return num or (1ll << idx); }
long long setBit0(long long num, int idx) { return num & ~(1ll << idx); }
long long flipBit(long long num, int idx) { r... |
1037_E. Trips | There are n persons who initially don't know each other. On each morning, two of them, who were not friends before, become friends.
We want to plan a trip for every evening of m days. On each trip, you have to select a group of people that will go on the trip. For every person, one of the following should hold:
* ... | {
"input": [
"4 4 2\n2 3\n1 2\n1 3\n1 4\n",
"5 8 2\n2 1\n4 2\n5 4\n5 2\n4 3\n5 1\n4 1\n3 2\n",
"5 7 2\n1 5\n3 2\n2 5\n3 4\n1 2\n5 3\n1 3\n"
],
"output": [
"0\n0\n3\n3\n",
"0\n0\n0\n3\n3\n4\n4\n5\n",
"0\n0\n0\n0\n3\n4\n4\n"
]
} | {
"input": [
"16 20 2\n10 3\n5 3\n10 5\n12 7\n7 6\n9 12\n9 6\n1 10\n11 16\n11 1\n16 2\n10 2\n14 4\n15 14\n4 13\n13 15\n1 8\n7 15\n1 7\n8 15\n",
"2 1 1\n2 1\n"
],
"output": [
"0\n0\n3\n3\n3\n3\n7\n7\n7\n7\n7\n11\n11\n11\n11\n15\n15\n15\n15\n16\n",
"2\n"
]
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
const int ms = 200200;
int n, m, k;
int ans = 0;
int size[ms], par[ms];
int getPar(int x) { return x == par[x] ? x : getPar(par[x]); }
void makeUnion(int a, int b) {
a = getPar(a);
b = getPar(b);
if (a == b) return;
size[a] += size[b];
ans = std::max(size[a], ans);
par[b] = a;
}
int... |
1037_E. Trips | There are n persons who initially don't know each other. On each morning, two of them, who were not friends before, become friends.
We want to plan a trip for every evening of m days. On each trip, you have to select a group of people that will go on the trip. For every person, one of the following should hold:
* ... | {
"input": [
"4 4 2\n2 3\n1 2\n1 3\n1 4\n",
"5 8 2\n2 1\n4 2\n5 4\n5 2\n4 3\n5 1\n4 1\n3 2\n",
"5 7 2\n1 5\n3 2\n2 5\n3 4\n1 2\n5 3\n1 3\n"
],
"output": [
"0\n0\n3\n3\n",
"0\n0\n0\n3\n3\n4\n4\n5\n",
"0\n0\n0\n0\n3\n4\n4\n"
]
} | {
"input": [
"16 20 2\n10 3\n5 3\n10 5\n12 7\n7 6\n9 12\n9 6\n1 10\n11 16\n11 1\n16 2\n10 2\n14 4\n15 14\n4 13\n13 15\n1 8\n7 15\n1 7\n8 15\n",
"2 1 1\n2 1\n"
],
"output": [
"0\n0\n3\n3\n3\n3\n7\n7\n7\n7\n7\n11\n11\n11\n11\n15\n15\n15\n15\n16\n",
"2\n"
]
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
const int N = 2e5 + 5;
using namespace std;
int n, m, k;
int res[N];
int x[N], y[N];
int bac[N];
set<int> a[N];
set<pair<int, int> > S;
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
cin >> n >> m >> k;
for (int i = 1; i <= m; i++) {
cin >> x[i] >> y[i];
a... |
1037_E. Trips | There are n persons who initially don't know each other. On each morning, two of them, who were not friends before, become friends.
We want to plan a trip for every evening of m days. On each trip, you have to select a group of people that will go on the trip. For every person, one of the following should hold:
* ... | {
"input": [
"4 4 2\n2 3\n1 2\n1 3\n1 4\n",
"5 8 2\n2 1\n4 2\n5 4\n5 2\n4 3\n5 1\n4 1\n3 2\n",
"5 7 2\n1 5\n3 2\n2 5\n3 4\n1 2\n5 3\n1 3\n"
],
"output": [
"0\n0\n3\n3\n",
"0\n0\n0\n3\n3\n4\n4\n5\n",
"0\n0\n0\n0\n3\n4\n4\n"
]
} | {
"input": [
"16 20 2\n10 3\n5 3\n10 5\n12 7\n7 6\n9 12\n9 6\n1 10\n11 16\n11 1\n16 2\n10 2\n14 4\n15 14\n4 13\n13 15\n1 8\n7 15\n1 7\n8 15\n",
"2 1 1\n2 1\n"
],
"output": [
"0\n0\n3\n3\n3\n3\n7\n7\n7\n7\n7\n11\n11\n11\n11\n15\n15\n15\n15\n16\n",
"2\n"
]
} | IN-CORRECT | java | import java.io.*;
import java.util.*;
public class Trips {
public static void main(String[] args) {
FastScanner scan=new FastScanner();
PrintWriter out=new PrintWriter(System.out);
int n=scan.nextInt(), m=scan.nextInt();
k=scan.nextInt();
deg=new int[n];
a=new ArrayList[n];
for(int i=0;i<n;i++) a[i]=new... |
1037_E. Trips | There are n persons who initially don't know each other. On each morning, two of them, who were not friends before, become friends.
We want to plan a trip for every evening of m days. On each trip, you have to select a group of people that will go on the trip. For every person, one of the following should hold:
* ... | {
"input": [
"4 4 2\n2 3\n1 2\n1 3\n1 4\n",
"5 8 2\n2 1\n4 2\n5 4\n5 2\n4 3\n5 1\n4 1\n3 2\n",
"5 7 2\n1 5\n3 2\n2 5\n3 4\n1 2\n5 3\n1 3\n"
],
"output": [
"0\n0\n3\n3\n",
"0\n0\n0\n3\n3\n4\n4\n5\n",
"0\n0\n0\n0\n3\n4\n4\n"
]
} | {
"input": [
"16 20 2\n10 3\n5 3\n10 5\n12 7\n7 6\n9 12\n9 6\n1 10\n11 16\n11 1\n16 2\n10 2\n14 4\n15 14\n4 13\n13 15\n1 8\n7 15\n1 7\n8 15\n",
"2 1 1\n2 1\n"
],
"output": [
"0\n0\n3\n3\n3\n3\n7\n7\n7\n7\n7\n11\n11\n11\n11\n15\n15\n15\n15\n16\n",
"2\n"
]
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
vector<int> v, graph[200005];
vector<pair<long long, long long> > ed;
set<int> s;
multiset<int> s2;
int deg[200005], cnt[200005], ans[200005], vis[200005], out[200005];
map<pair<long long, long long>, int> mp;
void dfs(int u) {
vis[u] = 1;
v.push_back(u);
for (int j =... |
1037_E. Trips | There are n persons who initially don't know each other. On each morning, two of them, who were not friends before, become friends.
We want to plan a trip for every evening of m days. On each trip, you have to select a group of people that will go on the trip. For every person, one of the following should hold:
* ... | {
"input": [
"4 4 2\n2 3\n1 2\n1 3\n1 4\n",
"5 8 2\n2 1\n4 2\n5 4\n5 2\n4 3\n5 1\n4 1\n3 2\n",
"5 7 2\n1 5\n3 2\n2 5\n3 4\n1 2\n5 3\n1 3\n"
],
"output": [
"0\n0\n3\n3\n",
"0\n0\n0\n3\n3\n4\n4\n5\n",
"0\n0\n0\n0\n3\n4\n4\n"
]
} | {
"input": [
"16 20 2\n10 3\n5 3\n10 5\n12 7\n7 6\n9 12\n9 6\n1 10\n11 16\n11 1\n16 2\n10 2\n14 4\n15 14\n4 13\n13 15\n1 8\n7 15\n1 7\n8 15\n",
"2 1 1\n2 1\n"
],
"output": [
"0\n0\n3\n3\n3\n3\n7\n7\n7\n7\n7\n11\n11\n11\n11\n15\n15\n15\n15\n16\n",
"2\n"
]
} | IN-CORRECT | java | //package superior;
import java.util.*;
public class Tester {
static TreeSet<Integer> tr[];
static int k,n,m,cnt=0;
static Queue<Integer> q;
public static void main(String[] args) {
// TODO Auto-generated method stub
Scanner s=new Scanner(System.in);
n=s.nextInt();
m=s.nextInt();
k=s.nextInt();
... |
1037_E. Trips | There are n persons who initially don't know each other. On each morning, two of them, who were not friends before, become friends.
We want to plan a trip for every evening of m days. On each trip, you have to select a group of people that will go on the trip. For every person, one of the following should hold:
* ... | {
"input": [
"4 4 2\n2 3\n1 2\n1 3\n1 4\n",
"5 8 2\n2 1\n4 2\n5 4\n5 2\n4 3\n5 1\n4 1\n3 2\n",
"5 7 2\n1 5\n3 2\n2 5\n3 4\n1 2\n5 3\n1 3\n"
],
"output": [
"0\n0\n3\n3\n",
"0\n0\n0\n3\n3\n4\n4\n5\n",
"0\n0\n0\n0\n3\n4\n4\n"
]
} | {
"input": [
"16 20 2\n10 3\n5 3\n10 5\n12 7\n7 6\n9 12\n9 6\n1 10\n11 16\n11 1\n16 2\n10 2\n14 4\n15 14\n4 13\n13 15\n1 8\n7 15\n1 7\n8 15\n",
"2 1 1\n2 1\n"
],
"output": [
"0\n0\n3\n3\n3\n3\n7\n7\n7\n7\n7\n11\n11\n11\n11\n15\n15\n15\n15\n16\n",
"2\n"
]
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
#pragma comment(linker, "/stack:200000000")
#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#pragma GCC optimize("unroll-loops")
using namespace std;
const long long INF = 1000LL * 1000 * 1000 * 1000 * 1000 * 1000;
const int inf = 1000... |
1037_E. Trips | There are n persons who initially don't know each other. On each morning, two of them, who were not friends before, become friends.
We want to plan a trip for every evening of m days. On each trip, you have to select a group of people that will go on the trip. For every person, one of the following should hold:
* ... | {
"input": [
"4 4 2\n2 3\n1 2\n1 3\n1 4\n",
"5 8 2\n2 1\n4 2\n5 4\n5 2\n4 3\n5 1\n4 1\n3 2\n",
"5 7 2\n1 5\n3 2\n2 5\n3 4\n1 2\n5 3\n1 3\n"
],
"output": [
"0\n0\n3\n3\n",
"0\n0\n0\n3\n3\n4\n4\n5\n",
"0\n0\n0\n0\n3\n4\n4\n"
]
} | {
"input": [
"16 20 2\n10 3\n5 3\n10 5\n12 7\n7 6\n9 12\n9 6\n1 10\n11 16\n11 1\n16 2\n10 2\n14 4\n15 14\n4 13\n13 15\n1 8\n7 15\n1 7\n8 15\n",
"2 1 1\n2 1\n"
],
"output": [
"0\n0\n3\n3\n3\n3\n7\n7\n7\n7\n7\n11\n11\n11\n11\n15\n15\n15\n15\n16\n",
"2\n"
]
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 100;
int d[N], w[N];
vector<int> v[N];
int n, m, k, ans = 0;
void addW(int x) {
for (int to : v[x]) {
w[to]++;
if (w[to] == k) ans++;
}
}
void addD(int x) {
for (int to : v[x]) {
d[to]++;
if (d[to] == k) {
addW(to);
}
}
... |
1037_E. Trips | There are n persons who initially don't know each other. On each morning, two of them, who were not friends before, become friends.
We want to plan a trip for every evening of m days. On each trip, you have to select a group of people that will go on the trip. For every person, one of the following should hold:
* ... | {
"input": [
"4 4 2\n2 3\n1 2\n1 3\n1 4\n",
"5 8 2\n2 1\n4 2\n5 4\n5 2\n4 3\n5 1\n4 1\n3 2\n",
"5 7 2\n1 5\n3 2\n2 5\n3 4\n1 2\n5 3\n1 3\n"
],
"output": [
"0\n0\n3\n3\n",
"0\n0\n0\n3\n3\n4\n4\n5\n",
"0\n0\n0\n0\n3\n4\n4\n"
]
} | {
"input": [
"16 20 2\n10 3\n5 3\n10 5\n12 7\n7 6\n9 12\n9 6\n1 10\n11 16\n11 1\n16 2\n10 2\n14 4\n15 14\n4 13\n13 15\n1 8\n7 15\n1 7\n8 15\n",
"2 1 1\n2 1\n"
],
"output": [
"0\n0\n3\n3\n3\n3\n7\n7\n7\n7\n7\n11\n11\n11\n11\n15\n15\n15\n15\n16\n",
"2\n"
]
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
const int N = 2e5;
int vis[N + 2];
vector<int> com[N + 2];
map<int, int> mp[N + 2];
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
int n, m, k;
cin >> n >> m >> k;
for (int i = 1; i <= n; i++) com[i].push_back(i);
int ans = 0;
while (m--) {
int u, ... |
1037_E. Trips | There are n persons who initially don't know each other. On each morning, two of them, who were not friends before, become friends.
We want to plan a trip for every evening of m days. On each trip, you have to select a group of people that will go on the trip. For every person, one of the following should hold:
* ... | {
"input": [
"4 4 2\n2 3\n1 2\n1 3\n1 4\n",
"5 8 2\n2 1\n4 2\n5 4\n5 2\n4 3\n5 1\n4 1\n3 2\n",
"5 7 2\n1 5\n3 2\n2 5\n3 4\n1 2\n5 3\n1 3\n"
],
"output": [
"0\n0\n3\n3\n",
"0\n0\n0\n3\n3\n4\n4\n5\n",
"0\n0\n0\n0\n3\n4\n4\n"
]
} | {
"input": [
"16 20 2\n10 3\n5 3\n10 5\n12 7\n7 6\n9 12\n9 6\n1 10\n11 16\n11 1\n16 2\n10 2\n14 4\n15 14\n4 13\n13 15\n1 8\n7 15\n1 7\n8 15\n",
"2 1 1\n2 1\n"
],
"output": [
"0\n0\n3\n3\n3\n3\n7\n7\n7\n7\n7\n11\n11\n11\n11\n15\n15\n15\n15\n16\n",
"2\n"
]
} | IN-CORRECT | java | import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.util.HashSet;
import java.util.Set;
import java.util.StringTokenizer;
public class Main {
static int deg[];
static Set<Integer> s;
static Set<Integer>[] graph;
static int n,m,k;
pub... |
1037_E. Trips | There are n persons who initially don't know each other. On each morning, two of them, who were not friends before, become friends.
We want to plan a trip for every evening of m days. On each trip, you have to select a group of people that will go on the trip. For every person, one of the following should hold:
* ... | {
"input": [
"4 4 2\n2 3\n1 2\n1 3\n1 4\n",
"5 8 2\n2 1\n4 2\n5 4\n5 2\n4 3\n5 1\n4 1\n3 2\n",
"5 7 2\n1 5\n3 2\n2 5\n3 4\n1 2\n5 3\n1 3\n"
],
"output": [
"0\n0\n3\n3\n",
"0\n0\n0\n3\n3\n4\n4\n5\n",
"0\n0\n0\n0\n3\n4\n4\n"
]
} | {
"input": [
"16 20 2\n10 3\n5 3\n10 5\n12 7\n7 6\n9 12\n9 6\n1 10\n11 16\n11 1\n16 2\n10 2\n14 4\n15 14\n4 13\n13 15\n1 8\n7 15\n1 7\n8 15\n",
"2 1 1\n2 1\n"
],
"output": [
"0\n0\n3\n3\n3\n3\n7\n7\n7\n7\n7\n11\n11\n11\n11\n15\n15\n15\n15\n16\n",
"2\n"
]
} | IN-CORRECT | python2 | '''input
8 12 3
1 2
1 3
2 3
3 4
2 4
1 4
5 6
6 7
5 7
4 5
1 5
2 5
'''
n,m,k=[int(x) for x in raw_input().split()]
F=[0]*n
E=[0]*n
C=[set() for x in range(n)]
tot=0
M=[]
for i in range(m):
a,b=[int(x)-1 for x in raw_input().split()]
C[a].add(b)
C[b].add(a)
F[a]+=1
F[b]+=1
M.append((a,b))
M=M[::-1]
for i in ra... |
1037_E. Trips | There are n persons who initially don't know each other. On each morning, two of them, who were not friends before, become friends.
We want to plan a trip for every evening of m days. On each trip, you have to select a group of people that will go on the trip. For every person, one of the following should hold:
* ... | {
"input": [
"4 4 2\n2 3\n1 2\n1 3\n1 4\n",
"5 8 2\n2 1\n4 2\n5 4\n5 2\n4 3\n5 1\n4 1\n3 2\n",
"5 7 2\n1 5\n3 2\n2 5\n3 4\n1 2\n5 3\n1 3\n"
],
"output": [
"0\n0\n3\n3\n",
"0\n0\n0\n3\n3\n4\n4\n5\n",
"0\n0\n0\n0\n3\n4\n4\n"
]
} | {
"input": [
"16 20 2\n10 3\n5 3\n10 5\n12 7\n7 6\n9 12\n9 6\n1 10\n11 16\n11 1\n16 2\n10 2\n14 4\n15 14\n4 13\n13 15\n1 8\n7 15\n1 7\n8 15\n",
"2 1 1\n2 1\n"
],
"output": [
"0\n0\n3\n3\n3\n3\n7\n7\n7\n7\n7\n11\n11\n11\n11\n15\n15\n15\n15\n16\n",
"2\n"
]
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
int main() {
ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
int n, m, k;
cin >> n >> m >> k;
vector<int> from(m), to(m);
vector<vector<pair<int, int> > > adj(n);
for (int i = (int)(0); i < (int)(m); ++i) {
cin >> from[i] >> to[i];
from[i]--, to[i... |
1037_E. Trips | There are n persons who initially don't know each other. On each morning, two of them, who were not friends before, become friends.
We want to plan a trip for every evening of m days. On each trip, you have to select a group of people that will go on the trip. For every person, one of the following should hold:
* ... | {
"input": [
"4 4 2\n2 3\n1 2\n1 3\n1 4\n",
"5 8 2\n2 1\n4 2\n5 4\n5 2\n4 3\n5 1\n4 1\n3 2\n",
"5 7 2\n1 5\n3 2\n2 5\n3 4\n1 2\n5 3\n1 3\n"
],
"output": [
"0\n0\n3\n3\n",
"0\n0\n0\n3\n3\n4\n4\n5\n",
"0\n0\n0\n0\n3\n4\n4\n"
]
} | {
"input": [
"16 20 2\n10 3\n5 3\n10 5\n12 7\n7 6\n9 12\n9 6\n1 10\n11 16\n11 1\n16 2\n10 2\n14 4\n15 14\n4 13\n13 15\n1 8\n7 15\n1 7\n8 15\n",
"2 1 1\n2 1\n"
],
"output": [
"0\n0\n3\n3\n3\n3\n7\n7\n7\n7\n7\n11\n11\n11\n11\n15\n15\n15\n15\n16\n",
"2\n"
]
} | IN-CORRECT | python2 | '''input
4 4 2
2 3
1 2
1 3
1 4
'''
n,m,k=[int(x) for x in raw_input().split()]
F=[0]*n
E=[0]*n
C=[[] for x in range(n)]
tot=0
for i in range(m):
a,b=[int(x)-1 for x in raw_input().split()]
C[a].append(b)
C[b].append(a)
F[a]+=1
F[b]+=1
if F[a]==k:
for j in C[a]:
E[j]+=1
if E[j]==k:
tot+=1
elif F... |
1037_E. Trips | There are n persons who initially don't know each other. On each morning, two of them, who were not friends before, become friends.
We want to plan a trip for every evening of m days. On each trip, you have to select a group of people that will go on the trip. For every person, one of the following should hold:
* ... | {
"input": [
"4 4 2\n2 3\n1 2\n1 3\n1 4\n",
"5 8 2\n2 1\n4 2\n5 4\n5 2\n4 3\n5 1\n4 1\n3 2\n",
"5 7 2\n1 5\n3 2\n2 5\n3 4\n1 2\n5 3\n1 3\n"
],
"output": [
"0\n0\n3\n3\n",
"0\n0\n0\n3\n3\n4\n4\n5\n",
"0\n0\n0\n0\n3\n4\n4\n"
]
} | {
"input": [
"16 20 2\n10 3\n5 3\n10 5\n12 7\n7 6\n9 12\n9 6\n1 10\n11 16\n11 1\n16 2\n10 2\n14 4\n15 14\n4 13\n13 15\n1 8\n7 15\n1 7\n8 15\n",
"2 1 1\n2 1\n"
],
"output": [
"0\n0\n3\n3\n3\n3\n7\n7\n7\n7\n7\n11\n11\n11\n11\n15\n15\n15\n15\n16\n",
"2\n"
]
} | IN-CORRECT | java | //package superior;
import java.util.*;
public class Tester {
static TreeSet<Integer> tr[];
static int k,n,m,cnt=0;
static Queue<Integer> q;
public static void main(String[] args) {
// TODO Auto-generated method stub
Scanner s=new Scanner(System.in);
n=s.nextInt();
m=s.nextInt();
k=s.nextInt();
... |
1037_E. Trips | There are n persons who initially don't know each other. On each morning, two of them, who were not friends before, become friends.
We want to plan a trip for every evening of m days. On each trip, you have to select a group of people that will go on the trip. For every person, one of the following should hold:
* ... | {
"input": [
"4 4 2\n2 3\n1 2\n1 3\n1 4\n",
"5 8 2\n2 1\n4 2\n5 4\n5 2\n4 3\n5 1\n4 1\n3 2\n",
"5 7 2\n1 5\n3 2\n2 5\n3 4\n1 2\n5 3\n1 3\n"
],
"output": [
"0\n0\n3\n3\n",
"0\n0\n0\n3\n3\n4\n4\n5\n",
"0\n0\n0\n0\n3\n4\n4\n"
]
} | {
"input": [
"16 20 2\n10 3\n5 3\n10 5\n12 7\n7 6\n9 12\n9 6\n1 10\n11 16\n11 1\n16 2\n10 2\n14 4\n15 14\n4 13\n13 15\n1 8\n7 15\n1 7\n8 15\n",
"2 1 1\n2 1\n"
],
"output": [
"0\n0\n3\n3\n3\n3\n7\n7\n7\n7\n7\n11\n11\n11\n11\n15\n15\n15\n15\n16\n",
"2\n"
]
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
const long long mod = 1e9 + 7, inf = 1e9 + 3;
void update(int a, vector<set<int> >& g, set<pair<int, int> >& s, int k) {
for (int v : g[a]) {
s.erase({g[v].size(), v});
g[v].erase(a);
if (g[v].size() >= k) {
s.insert({g[v].size(), v});
}
}
}
int ma... |
1037_E. Trips | There are n persons who initially don't know each other. On each morning, two of them, who were not friends before, become friends.
We want to plan a trip for every evening of m days. On each trip, you have to select a group of people that will go on the trip. For every person, one of the following should hold:
* ... | {
"input": [
"4 4 2\n2 3\n1 2\n1 3\n1 4\n",
"5 8 2\n2 1\n4 2\n5 4\n5 2\n4 3\n5 1\n4 1\n3 2\n",
"5 7 2\n1 5\n3 2\n2 5\n3 4\n1 2\n5 3\n1 3\n"
],
"output": [
"0\n0\n3\n3\n",
"0\n0\n0\n3\n3\n4\n4\n5\n",
"0\n0\n0\n0\n3\n4\n4\n"
]
} | {
"input": [
"16 20 2\n10 3\n5 3\n10 5\n12 7\n7 6\n9 12\n9 6\n1 10\n11 16\n11 1\n16 2\n10 2\n14 4\n15 14\n4 13\n13 15\n1 8\n7 15\n1 7\n8 15\n",
"2 1 1\n2 1\n"
],
"output": [
"0\n0\n3\n3\n3\n3\n7\n7\n7\n7\n7\n11\n11\n11\n11\n15\n15\n15\n15\n16\n",
"2\n"
]
} | IN-CORRECT | cpp | // clang-format off
// very important, much wow, such don't touch this
#define protected public
using iii = int; using yeee = iii;
#ifdef LOCAL
const iii DEBUG = 10;
#else
const iii DEBUG = -1;
#endif
#define DBG(x) if(DEBUG >= x)
#include <bits/stdc++.h>
using namespace std;
#define FOR(i, lo, hi) for(ll i = ll(lo)... |
1037_E. Trips | There are n persons who initially don't know each other. On each morning, two of them, who were not friends before, become friends.
We want to plan a trip for every evening of m days. On each trip, you have to select a group of people that will go on the trip. For every person, one of the following should hold:
* ... | {
"input": [
"4 4 2\n2 3\n1 2\n1 3\n1 4\n",
"5 8 2\n2 1\n4 2\n5 4\n5 2\n4 3\n5 1\n4 1\n3 2\n",
"5 7 2\n1 5\n3 2\n2 5\n3 4\n1 2\n5 3\n1 3\n"
],
"output": [
"0\n0\n3\n3\n",
"0\n0\n0\n3\n3\n4\n4\n5\n",
"0\n0\n0\n0\n3\n4\n4\n"
]
} | {
"input": [
"16 20 2\n10 3\n5 3\n10 5\n12 7\n7 6\n9 12\n9 6\n1 10\n11 16\n11 1\n16 2\n10 2\n14 4\n15 14\n4 13\n13 15\n1 8\n7 15\n1 7\n8 15\n",
"2 1 1\n2 1\n"
],
"output": [
"0\n0\n3\n3\n3\n3\n7\n7\n7\n7\n7\n11\n11\n11\n11\n15\n15\n15\n15\n16\n",
"2\n"
]
} | IN-CORRECT | cpp | // amigos y no más
// dile a la jardinera que traigo flores
// += O(logn) ; + = O(n)
//#pragma GCC optimize("Ofast")
//#pragma GCC optimize("unroll-loops")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#include<bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp> // Common file
... |
1037_E. Trips | There are n persons who initially don't know each other. On each morning, two of them, who were not friends before, become friends.
We want to plan a trip for every evening of m days. On each trip, you have to select a group of people that will go on the trip. For every person, one of the following should hold:
* ... | {
"input": [
"4 4 2\n2 3\n1 2\n1 3\n1 4\n",
"5 8 2\n2 1\n4 2\n5 4\n5 2\n4 3\n5 1\n4 1\n3 2\n",
"5 7 2\n1 5\n3 2\n2 5\n3 4\n1 2\n5 3\n1 3\n"
],
"output": [
"0\n0\n3\n3\n",
"0\n0\n0\n3\n3\n4\n4\n5\n",
"0\n0\n0\n0\n3\n4\n4\n"
]
} | {
"input": [
"16 20 2\n10 3\n5 3\n10 5\n12 7\n7 6\n9 12\n9 6\n1 10\n11 16\n11 1\n16 2\n10 2\n14 4\n15 14\n4 13\n13 15\n1 8\n7 15\n1 7\n8 15\n",
"2 1 1\n2 1\n"
],
"output": [
"0\n0\n3\n3\n3\n3\n7\n7\n7\n7\n7\n11\n11\n11\n11\n15\n15\n15\n15\n16\n",
"2\n"
]
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
const int N = 3e5 + 7;
const long long INF = 1e18 + 7;
pair<int, int> a[N];
int d[N], ans[N];
int n, m, k;
set<pair<int, int> > s;
vector<pair<int, int> > g[N];
int main() {
cin >> n >> m >> k;
for (int i = 1; i <= m; i++) {
scanf("%d%d", &a[i].first, &a[i].second);... |
1037_E. Trips | There are n persons who initially don't know each other. On each morning, two of them, who were not friends before, become friends.
We want to plan a trip for every evening of m days. On each trip, you have to select a group of people that will go on the trip. For every person, one of the following should hold:
* ... | {
"input": [
"4 4 2\n2 3\n1 2\n1 3\n1 4\n",
"5 8 2\n2 1\n4 2\n5 4\n5 2\n4 3\n5 1\n4 1\n3 2\n",
"5 7 2\n1 5\n3 2\n2 5\n3 4\n1 2\n5 3\n1 3\n"
],
"output": [
"0\n0\n3\n3\n",
"0\n0\n0\n3\n3\n4\n4\n5\n",
"0\n0\n0\n0\n3\n4\n4\n"
]
} | {
"input": [
"16 20 2\n10 3\n5 3\n10 5\n12 7\n7 6\n9 12\n9 6\n1 10\n11 16\n11 1\n16 2\n10 2\n14 4\n15 14\n4 13\n13 15\n1 8\n7 15\n1 7\n8 15\n",
"2 1 1\n2 1\n"
],
"output": [
"0\n0\n3\n3\n3\n3\n7\n7\n7\n7\n7\n11\n11\n11\n11\n15\n15\n15\n15\n16\n",
"2\n"
]
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
map<int, int> m[200010];
int s[200010];
int main() {
int n, a, b, i, j;
cin >> n;
for (i = 0; i < n - 1; i++) {
scanf("%d%d", &a, &b);
m[a][b] = 1;
m[b][a] = 1;
}
for (i = 0; i < n; i++) scanf("%d", &s[i]);
for (i = 0, j = 1; i < n; i++)
while (m... |
1037_E. Trips | There are n persons who initially don't know each other. On each morning, two of them, who were not friends before, become friends.
We want to plan a trip for every evening of m days. On each trip, you have to select a group of people that will go on the trip. For every person, one of the following should hold:
* ... | {
"input": [
"4 4 2\n2 3\n1 2\n1 3\n1 4\n",
"5 8 2\n2 1\n4 2\n5 4\n5 2\n4 3\n5 1\n4 1\n3 2\n",
"5 7 2\n1 5\n3 2\n2 5\n3 4\n1 2\n5 3\n1 3\n"
],
"output": [
"0\n0\n3\n3\n",
"0\n0\n0\n3\n3\n4\n4\n5\n",
"0\n0\n0\n0\n3\n4\n4\n"
]
} | {
"input": [
"16 20 2\n10 3\n5 3\n10 5\n12 7\n7 6\n9 12\n9 6\n1 10\n11 16\n11 1\n16 2\n10 2\n14 4\n15 14\n4 13\n13 15\n1 8\n7 15\n1 7\n8 15\n",
"2 1 1\n2 1\n"
],
"output": [
"0\n0\n3\n3\n3\n3\n7\n7\n7\n7\n7\n11\n11\n11\n11\n15\n15\n15\n15\n16\n",
"2\n"
]
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
using vi = vector<int>;
using vvi = vector<vi>;
using vvvi = vector<vvi>;
using ii = pair<int, int>;
using lu = unsigned long long;
using l = long long;
using vs = vector<string>;
using vii = vector<ii>;
using vl = vector<l>;
using vvl = vector<vl>;
using vvvl = vector<vvl>... |
1037_E. Trips | There are n persons who initially don't know each other. On each morning, two of them, who were not friends before, become friends.
We want to plan a trip for every evening of m days. On each trip, you have to select a group of people that will go on the trip. For every person, one of the following should hold:
* ... | {
"input": [
"4 4 2\n2 3\n1 2\n1 3\n1 4\n",
"5 8 2\n2 1\n4 2\n5 4\n5 2\n4 3\n5 1\n4 1\n3 2\n",
"5 7 2\n1 5\n3 2\n2 5\n3 4\n1 2\n5 3\n1 3\n"
],
"output": [
"0\n0\n3\n3\n",
"0\n0\n0\n3\n3\n4\n4\n5\n",
"0\n0\n0\n0\n3\n4\n4\n"
]
} | {
"input": [
"16 20 2\n10 3\n5 3\n10 5\n12 7\n7 6\n9 12\n9 6\n1 10\n11 16\n11 1\n16 2\n10 2\n14 4\n15 14\n4 13\n13 15\n1 8\n7 15\n1 7\n8 15\n",
"2 1 1\n2 1\n"
],
"output": [
"0\n0\n3\n3\n3\n3\n7\n7\n7\n7\n7\n11\n11\n11\n11\n15\n15\n15\n15\n16\n",
"2\n"
]
} | IN-CORRECT | java | //package superior;
import java.util.*;
public class Tester {
static TreeSet<Integer> tr[];
static int k,n,m,cnt=0;
static Queue<Integer> q;
public static void main(String[] args) {
// TODO Auto-generated method stub
Scanner s=new Scanner(System.in);
n=s.nextInt();
m=s.nextInt();
k=s.nextInt();
... |
1037_E. Trips | There are n persons who initially don't know each other. On each morning, two of them, who were not friends before, become friends.
We want to plan a trip for every evening of m days. On each trip, you have to select a group of people that will go on the trip. For every person, one of the following should hold:
* ... | {
"input": [
"4 4 2\n2 3\n1 2\n1 3\n1 4\n",
"5 8 2\n2 1\n4 2\n5 4\n5 2\n4 3\n5 1\n4 1\n3 2\n",
"5 7 2\n1 5\n3 2\n2 5\n3 4\n1 2\n5 3\n1 3\n"
],
"output": [
"0\n0\n3\n3\n",
"0\n0\n0\n3\n3\n4\n4\n5\n",
"0\n0\n0\n0\n3\n4\n4\n"
]
} | {
"input": [
"16 20 2\n10 3\n5 3\n10 5\n12 7\n7 6\n9 12\n9 6\n1 10\n11 16\n11 1\n16 2\n10 2\n14 4\n15 14\n4 13\n13 15\n1 8\n7 15\n1 7\n8 15\n",
"2 1 1\n2 1\n"
],
"output": [
"0\n0\n3\n3\n3\n3\n7\n7\n7\n7\n7\n11\n11\n11\n11\n15\n15\n15\n15\n16\n",
"2\n"
]
} | IN-CORRECT | java | import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.util.HashSet;
import java.util.Set;
import java.util.StringTokenizer;
public class Main {
static int deg[];
static Set<Integer> s;
static Set<Integer>[] graph;
static int n,m,k;
pub... |
1037_E. Trips | There are n persons who initially don't know each other. On each morning, two of them, who were not friends before, become friends.
We want to plan a trip for every evening of m days. On each trip, you have to select a group of people that will go on the trip. For every person, one of the following should hold:
* ... | {
"input": [
"4 4 2\n2 3\n1 2\n1 3\n1 4\n",
"5 8 2\n2 1\n4 2\n5 4\n5 2\n4 3\n5 1\n4 1\n3 2\n",
"5 7 2\n1 5\n3 2\n2 5\n3 4\n1 2\n5 3\n1 3\n"
],
"output": [
"0\n0\n3\n3\n",
"0\n0\n0\n3\n3\n4\n4\n5\n",
"0\n0\n0\n0\n3\n4\n4\n"
]
} | {
"input": [
"16 20 2\n10 3\n5 3\n10 5\n12 7\n7 6\n9 12\n9 6\n1 10\n11 16\n11 1\n16 2\n10 2\n14 4\n15 14\n4 13\n13 15\n1 8\n7 15\n1 7\n8 15\n",
"2 1 1\n2 1\n"
],
"output": [
"0\n0\n3\n3\n3\n3\n7\n7\n7\n7\n7\n11\n11\n11\n11\n15\n15\n15\n15\n16\n",
"2\n"
]
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
int n, m, k, ANS[200005];
set<int> v[200005];
pair<int, int> q[200005];
set<int> s;
void remove(int u) {
if (v[u].size() < k && s.erase(u))
for (auto &i : v[u]) {
v[i].erase(u);
remove(i);
}
}
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
... |
1037_E. Trips | There are n persons who initially don't know each other. On each morning, two of them, who were not friends before, become friends.
We want to plan a trip for every evening of m days. On each trip, you have to select a group of people that will go on the trip. For every person, one of the following should hold:
* ... | {
"input": [
"4 4 2\n2 3\n1 2\n1 3\n1 4\n",
"5 8 2\n2 1\n4 2\n5 4\n5 2\n4 3\n5 1\n4 1\n3 2\n",
"5 7 2\n1 5\n3 2\n2 5\n3 4\n1 2\n5 3\n1 3\n"
],
"output": [
"0\n0\n3\n3\n",
"0\n0\n0\n3\n3\n4\n4\n5\n",
"0\n0\n0\n0\n3\n4\n4\n"
]
} | {
"input": [
"16 20 2\n10 3\n5 3\n10 5\n12 7\n7 6\n9 12\n9 6\n1 10\n11 16\n11 1\n16 2\n10 2\n14 4\n15 14\n4 13\n13 15\n1 8\n7 15\n1 7\n8 15\n",
"2 1 1\n2 1\n"
],
"output": [
"0\n0\n3\n3\n3\n3\n7\n7\n7\n7\n7\n11\n11\n11\n11\n15\n15\n15\n15\n16\n",
"2\n"
]
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
int u[400100], v[400100], fir[200100], nxt[400100], cnt = 0, n, m, k;
int dis[200100], ans = 0;
void addedge(int ui, int vi) {
++cnt;
u[cnt] = ui;
v[cnt] = vi;
nxt[cnt] = fir[ui];
fir[ui] = cnt;
}
void dfs(int u) {
for (int i = fir[u]; i; i = nxt[i]) {
if (!... |
1037_E. Trips | There are n persons who initially don't know each other. On each morning, two of them, who were not friends before, become friends.
We want to plan a trip for every evening of m days. On each trip, you have to select a group of people that will go on the trip. For every person, one of the following should hold:
* ... | {
"input": [
"4 4 2\n2 3\n1 2\n1 3\n1 4\n",
"5 8 2\n2 1\n4 2\n5 4\n5 2\n4 3\n5 1\n4 1\n3 2\n",
"5 7 2\n1 5\n3 2\n2 5\n3 4\n1 2\n5 3\n1 3\n"
],
"output": [
"0\n0\n3\n3\n",
"0\n0\n0\n3\n3\n4\n4\n5\n",
"0\n0\n0\n0\n3\n4\n4\n"
]
} | {
"input": [
"16 20 2\n10 3\n5 3\n10 5\n12 7\n7 6\n9 12\n9 6\n1 10\n11 16\n11 1\n16 2\n10 2\n14 4\n15 14\n4 13\n13 15\n1 8\n7 15\n1 7\n8 15\n",
"2 1 1\n2 1\n"
],
"output": [
"0\n0\n3\n3\n3\n3\n7\n7\n7\n7\n7\n11\n11\n11\n11\n15\n15\n15\n15\n16\n",
"2\n"
]
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
int res[200010], x[200010], y[200010];
vector<int> v;
set<int> a[200010];
bool vis[200010];
int main() {
int n, m, k;
while (cin >> n >> m >> k) {
for (int i = 1; i <= m; ++i) {
cin >> x[i] >> y[i];
a[x[i]].insert(y[i]);
a[y[i]].insert(x[i]);
}... |
1037_E. Trips | There are n persons who initially don't know each other. On each morning, two of them, who were not friends before, become friends.
We want to plan a trip for every evening of m days. On each trip, you have to select a group of people that will go on the trip. For every person, one of the following should hold:
* ... | {
"input": [
"4 4 2\n2 3\n1 2\n1 3\n1 4\n",
"5 8 2\n2 1\n4 2\n5 4\n5 2\n4 3\n5 1\n4 1\n3 2\n",
"5 7 2\n1 5\n3 2\n2 5\n3 4\n1 2\n5 3\n1 3\n"
],
"output": [
"0\n0\n3\n3\n",
"0\n0\n0\n3\n3\n4\n4\n5\n",
"0\n0\n0\n0\n3\n4\n4\n"
]
} | {
"input": [
"16 20 2\n10 3\n5 3\n10 5\n12 7\n7 6\n9 12\n9 6\n1 10\n11 16\n11 1\n16 2\n10 2\n14 4\n15 14\n4 13\n13 15\n1 8\n7 15\n1 7\n8 15\n",
"2 1 1\n2 1\n"
],
"output": [
"0\n0\n3\n3\n3\n3\n7\n7\n7\n7\n7\n11\n11\n11\n11\n15\n15\n15\n15\n16\n",
"2\n"
]
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
template <typename T, typename Pr = less<T>>
using pq = priority_queue<T, vector<T>, Pr>;
using i64 = long long int;
using ii = pair<int, int>;
using ii64 = pair<i64, i64>;
vector<int> edge[200005];
int deg[200005];
int degk[200005];
int degkk[200005];
int ans = 0;
int k;
v... |
1037_E. Trips | There are n persons who initially don't know each other. On each morning, two of them, who were not friends before, become friends.
We want to plan a trip for every evening of m days. On each trip, you have to select a group of people that will go on the trip. For every person, one of the following should hold:
* ... | {
"input": [
"4 4 2\n2 3\n1 2\n1 3\n1 4\n",
"5 8 2\n2 1\n4 2\n5 4\n5 2\n4 3\n5 1\n4 1\n3 2\n",
"5 7 2\n1 5\n3 2\n2 5\n3 4\n1 2\n5 3\n1 3\n"
],
"output": [
"0\n0\n3\n3\n",
"0\n0\n0\n3\n3\n4\n4\n5\n",
"0\n0\n0\n0\n3\n4\n4\n"
]
} | {
"input": [
"16 20 2\n10 3\n5 3\n10 5\n12 7\n7 6\n9 12\n9 6\n1 10\n11 16\n11 1\n16 2\n10 2\n14 4\n15 14\n4 13\n13 15\n1 8\n7 15\n1 7\n8 15\n",
"2 1 1\n2 1\n"
],
"output": [
"0\n0\n3\n3\n3\n3\n7\n7\n7\n7\n7\n11\n11\n11\n11\n15\n15\n15\n15\n16\n",
"2\n"
]
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
const int N = 2e5 + 5;
using namespace std;
int n, m, k;
int res[N];
int x[N], y[N];
int bac[N];
set<int> a[N];
set<pair<int, int> > S;
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
cin >> n >> m >> k;
for (int i = 1; i <= m; i++) {
cin >> x[i] >> y[i];
a... |
1037_E. Trips | There are n persons who initially don't know each other. On each morning, two of them, who were not friends before, become friends.
We want to plan a trip for every evening of m days. On each trip, you have to select a group of people that will go on the trip. For every person, one of the following should hold:
* ... | {
"input": [
"4 4 2\n2 3\n1 2\n1 3\n1 4\n",
"5 8 2\n2 1\n4 2\n5 4\n5 2\n4 3\n5 1\n4 1\n3 2\n",
"5 7 2\n1 5\n3 2\n2 5\n3 4\n1 2\n5 3\n1 3\n"
],
"output": [
"0\n0\n3\n3\n",
"0\n0\n0\n3\n3\n4\n4\n5\n",
"0\n0\n0\n0\n3\n4\n4\n"
]
} | {
"input": [
"16 20 2\n10 3\n5 3\n10 5\n12 7\n7 6\n9 12\n9 6\n1 10\n11 16\n11 1\n16 2\n10 2\n14 4\n15 14\n4 13\n13 15\n1 8\n7 15\n1 7\n8 15\n",
"2 1 1\n2 1\n"
],
"output": [
"0\n0\n3\n3\n3\n3\n7\n7\n7\n7\n7\n11\n11\n11\n11\n15\n15\n15\n15\n16\n",
"2\n"
]
} | IN-CORRECT | python3 | maxN =200005
G = [None] * maxN
s = set()
k = [0] * 1
def delete(v):
if len(G[v]) < k[0] and (v in s):
s.remove(v)
for u in G[v]:
G[u].discard(v)
delete(u)
def main():
n,m,k[0] = map(int,input().split())
edges = [None] * (m + 1)
ans = [0] * m
for i in ra... |
1037_E. Trips | There are n persons who initially don't know each other. On each morning, two of them, who were not friends before, become friends.
We want to plan a trip for every evening of m days. On each trip, you have to select a group of people that will go on the trip. For every person, one of the following should hold:
* ... | {
"input": [
"4 4 2\n2 3\n1 2\n1 3\n1 4\n",
"5 8 2\n2 1\n4 2\n5 4\n5 2\n4 3\n5 1\n4 1\n3 2\n",
"5 7 2\n1 5\n3 2\n2 5\n3 4\n1 2\n5 3\n1 3\n"
],
"output": [
"0\n0\n3\n3\n",
"0\n0\n0\n3\n3\n4\n4\n5\n",
"0\n0\n0\n0\n3\n4\n4\n"
]
} | {
"input": [
"16 20 2\n10 3\n5 3\n10 5\n12 7\n7 6\n9 12\n9 6\n1 10\n11 16\n11 1\n16 2\n10 2\n14 4\n15 14\n4 13\n13 15\n1 8\n7 15\n1 7\n8 15\n",
"2 1 1\n2 1\n"
],
"output": [
"0\n0\n3\n3\n3\n3\n7\n7\n7\n7\n7\n11\n11\n11\n11\n15\n15\n15\n15\n16\n",
"2\n"
]
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
const int Max = 2e5 + 3;
vector<int> adj[Max];
set<int> adjNew[Max];
int cnt[Max] = {0}, degree[Max] = {0};
bool visited[Max] = {0}, going[Max] = {0};
int main() {
ios::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
int n, m, k;
cin >> n >> m >> k;
int u, v;
... |
1037_E. Trips | There are n persons who initially don't know each other. On each morning, two of them, who were not friends before, become friends.
We want to plan a trip for every evening of m days. On each trip, you have to select a group of people that will go on the trip. For every person, one of the following should hold:
* ... | {
"input": [
"4 4 2\n2 3\n1 2\n1 3\n1 4\n",
"5 8 2\n2 1\n4 2\n5 4\n5 2\n4 3\n5 1\n4 1\n3 2\n",
"5 7 2\n1 5\n3 2\n2 5\n3 4\n1 2\n5 3\n1 3\n"
],
"output": [
"0\n0\n3\n3\n",
"0\n0\n0\n3\n3\n4\n4\n5\n",
"0\n0\n0\n0\n3\n4\n4\n"
]
} | {
"input": [
"16 20 2\n10 3\n5 3\n10 5\n12 7\n7 6\n9 12\n9 6\n1 10\n11 16\n11 1\n16 2\n10 2\n14 4\n15 14\n4 13\n13 15\n1 8\n7 15\n1 7\n8 15\n",
"2 1 1\n2 1\n"
],
"output": [
"0\n0\n3\n3\n3\n3\n7\n7\n7\n7\n7\n11\n11\n11\n11\n15\n15\n15\n15\n16\n",
"2\n"
]
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
#pragma GCC optimize("unroll-loops")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
using namespace std;
typedef long long ll;
typedef double lf;
typedef pair<int,int> ii;
#define REP(i,n) for(int i=0;i<n;i++)
#define REP1(i,n) for(ll i=1;i<=n;i++)
#define RST(i,n... |
1037_E. Trips | There are n persons who initially don't know each other. On each morning, two of them, who were not friends before, become friends.
We want to plan a trip for every evening of m days. On each trip, you have to select a group of people that will go on the trip. For every person, one of the following should hold:
* ... | {
"input": [
"4 4 2\n2 3\n1 2\n1 3\n1 4\n",
"5 8 2\n2 1\n4 2\n5 4\n5 2\n4 3\n5 1\n4 1\n3 2\n",
"5 7 2\n1 5\n3 2\n2 5\n3 4\n1 2\n5 3\n1 3\n"
],
"output": [
"0\n0\n3\n3\n",
"0\n0\n0\n3\n3\n4\n4\n5\n",
"0\n0\n0\n0\n3\n4\n4\n"
]
} | {
"input": [
"16 20 2\n10 3\n5 3\n10 5\n12 7\n7 6\n9 12\n9 6\n1 10\n11 16\n11 1\n16 2\n10 2\n14 4\n15 14\n4 13\n13 15\n1 8\n7 15\n1 7\n8 15\n",
"2 1 1\n2 1\n"
],
"output": [
"0\n0\n3\n3\n3\n3\n7\n7\n7\n7\n7\n11\n11\n11\n11\n15\n15\n15\n15\n16\n",
"2\n"
]
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
int x[1000006], y[1000006], deg[1000006], ans[1000006];
set<pair<int, int> > my;
vector<int> v[1000006];
bool is[1000006];
map<pair<int, int>, int> my2;
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
int n, m, k;
cin >> n >> m >> k;
int i;
for (i ... |
1037_E. Trips | There are n persons who initially don't know each other. On each morning, two of them, who were not friends before, become friends.
We want to plan a trip for every evening of m days. On each trip, you have to select a group of people that will go on the trip. For every person, one of the following should hold:
* ... | {
"input": [
"4 4 2\n2 3\n1 2\n1 3\n1 4\n",
"5 8 2\n2 1\n4 2\n5 4\n5 2\n4 3\n5 1\n4 1\n3 2\n",
"5 7 2\n1 5\n3 2\n2 5\n3 4\n1 2\n5 3\n1 3\n"
],
"output": [
"0\n0\n3\n3\n",
"0\n0\n0\n3\n3\n4\n4\n5\n",
"0\n0\n0\n0\n3\n4\n4\n"
]
} | {
"input": [
"16 20 2\n10 3\n5 3\n10 5\n12 7\n7 6\n9 12\n9 6\n1 10\n11 16\n11 1\n16 2\n10 2\n14 4\n15 14\n4 13\n13 15\n1 8\n7 15\n1 7\n8 15\n",
"2 1 1\n2 1\n"
],
"output": [
"0\n0\n3\n3\n3\n3\n7\n7\n7\n7\n7\n11\n11\n11\n11\n15\n15\n15\n15\n16\n",
"2\n"
]
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
int N, M, K;
int X[200001], Y[200001];
vector<int> adj[200001];
set<int> adj2[200001];
int goadj[200001];
bool going[200001];
set<pair<int, int>> st;
int ans = 0;
vector<int> ants;
void dfs(int node) {
going[node] = false;
queue<int> todo;
ans--;
for (set<int>::iter... |
1037_E. Trips | There are n persons who initially don't know each other. On each morning, two of them, who were not friends before, become friends.
We want to plan a trip for every evening of m days. On each trip, you have to select a group of people that will go on the trip. For every person, one of the following should hold:
* ... | {
"input": [
"4 4 2\n2 3\n1 2\n1 3\n1 4\n",
"5 8 2\n2 1\n4 2\n5 4\n5 2\n4 3\n5 1\n4 1\n3 2\n",
"5 7 2\n1 5\n3 2\n2 5\n3 4\n1 2\n5 3\n1 3\n"
],
"output": [
"0\n0\n3\n3\n",
"0\n0\n0\n3\n3\n4\n4\n5\n",
"0\n0\n0\n0\n3\n4\n4\n"
]
} | {
"input": [
"16 20 2\n10 3\n5 3\n10 5\n12 7\n7 6\n9 12\n9 6\n1 10\n11 16\n11 1\n16 2\n10 2\n14 4\n15 14\n4 13\n13 15\n1 8\n7 15\n1 7\n8 15\n",
"2 1 1\n2 1\n"
],
"output": [
"0\n0\n3\n3\n3\n3\n7\n7\n7\n7\n7\n11\n11\n11\n11\n15\n15\n15\n15\n16\n",
"2\n"
]
} | IN-CORRECT | java | import java.io.FileInputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.OutputStream;
import java.lang.reflect.Array;
import java.nio.charset.Charset;
import java.util.*;
public class CFContest {
public static void main(String[] args) throws Exception {
boolean local = System... |
1037_E. Trips | There are n persons who initially don't know each other. On each morning, two of them, who were not friends before, become friends.
We want to plan a trip for every evening of m days. On each trip, you have to select a group of people that will go on the trip. For every person, one of the following should hold:
* ... | {
"input": [
"4 4 2\n2 3\n1 2\n1 3\n1 4\n",
"5 8 2\n2 1\n4 2\n5 4\n5 2\n4 3\n5 1\n4 1\n3 2\n",
"5 7 2\n1 5\n3 2\n2 5\n3 4\n1 2\n5 3\n1 3\n"
],
"output": [
"0\n0\n3\n3\n",
"0\n0\n0\n3\n3\n4\n4\n5\n",
"0\n0\n0\n0\n3\n4\n4\n"
]
} | {
"input": [
"16 20 2\n10 3\n5 3\n10 5\n12 7\n7 6\n9 12\n9 6\n1 10\n11 16\n11 1\n16 2\n10 2\n14 4\n15 14\n4 13\n13 15\n1 8\n7 15\n1 7\n8 15\n",
"2 1 1\n2 1\n"
],
"output": [
"0\n0\n3\n3\n3\n3\n7\n7\n7\n7\n7\n11\n11\n11\n11\n15\n15\n15\n15\n16\n",
"2\n"
]
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
long long const MAXN = 200005;
vector<long long> v[MAXN];
set<long long> g[MAXN];
long long deg[MAXN];
set<pair<long long, long long>> st;
pair<long long, long long> edge[MAXN];
long long ans[MAXN];
bool used[MAXN];
int main() {
ios::sync_with_stdio(0);
cin.tie(0);
co... |
1037_E. Trips | There are n persons who initially don't know each other. On each morning, two of them, who were not friends before, become friends.
We want to plan a trip for every evening of m days. On each trip, you have to select a group of people that will go on the trip. For every person, one of the following should hold:
* ... | {
"input": [
"4 4 2\n2 3\n1 2\n1 3\n1 4\n",
"5 8 2\n2 1\n4 2\n5 4\n5 2\n4 3\n5 1\n4 1\n3 2\n",
"5 7 2\n1 5\n3 2\n2 5\n3 4\n1 2\n5 3\n1 3\n"
],
"output": [
"0\n0\n3\n3\n",
"0\n0\n0\n3\n3\n4\n4\n5\n",
"0\n0\n0\n0\n3\n4\n4\n"
]
} | {
"input": [
"16 20 2\n10 3\n5 3\n10 5\n12 7\n7 6\n9 12\n9 6\n1 10\n11 16\n11 1\n16 2\n10 2\n14 4\n15 14\n4 13\n13 15\n1 8\n7 15\n1 7\n8 15\n",
"2 1 1\n2 1\n"
],
"output": [
"0\n0\n3\n3\n3\n3\n7\n7\n7\n7\n7\n11\n11\n11\n11\n15\n15\n15\n15\n16\n",
"2\n"
]
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
int toNumber(string s) {
int Number;
if (!(istringstream(s) >> Number)) Number = 0;
return Number;
}
string toString(int number) {
ostringstream ostr;
ostr << number;
return ostr.str();
}
long long ele(long long a, long long b) {
if (b == 0) return 1;
if (b ... |
1037_E. Trips | There are n persons who initially don't know each other. On each morning, two of them, who were not friends before, become friends.
We want to plan a trip for every evening of m days. On each trip, you have to select a group of people that will go on the trip. For every person, one of the following should hold:
* ... | {
"input": [
"4 4 2\n2 3\n1 2\n1 3\n1 4\n",
"5 8 2\n2 1\n4 2\n5 4\n5 2\n4 3\n5 1\n4 1\n3 2\n",
"5 7 2\n1 5\n3 2\n2 5\n3 4\n1 2\n5 3\n1 3\n"
],
"output": [
"0\n0\n3\n3\n",
"0\n0\n0\n3\n3\n4\n4\n5\n",
"0\n0\n0\n0\n3\n4\n4\n"
]
} | {
"input": [
"16 20 2\n10 3\n5 3\n10 5\n12 7\n7 6\n9 12\n9 6\n1 10\n11 16\n11 1\n16 2\n10 2\n14 4\n15 14\n4 13\n13 15\n1 8\n7 15\n1 7\n8 15\n",
"2 1 1\n2 1\n"
],
"output": [
"0\n0\n3\n3\n3\n3\n7\n7\n7\n7\n7\n11\n11\n11\n11\n15\n15\n15\n15\n16\n",
"2\n"
]
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
struct P {
int x, y, z;
bool operator<(const P &a) const { return x < a.x; }
};
vector<int> v[2];
int a, c, i, b, n, m, k, d;
int o[222111];
int l[111];
int j[11111];
int e;
int dx[10] = {0, 1, 0, -1, 1, 1, -1, -1}, dy[10] = {1, 0, -1, 0, 1, -1, 1, -1},
dz[10] = {0,... |
1037_E. Trips | There are n persons who initially don't know each other. On each morning, two of them, who were not friends before, become friends.
We want to plan a trip for every evening of m days. On each trip, you have to select a group of people that will go on the trip. For every person, one of the following should hold:
* ... | {
"input": [
"4 4 2\n2 3\n1 2\n1 3\n1 4\n",
"5 8 2\n2 1\n4 2\n5 4\n5 2\n4 3\n5 1\n4 1\n3 2\n",
"5 7 2\n1 5\n3 2\n2 5\n3 4\n1 2\n5 3\n1 3\n"
],
"output": [
"0\n0\n3\n3\n",
"0\n0\n0\n3\n3\n4\n4\n5\n",
"0\n0\n0\n0\n3\n4\n4\n"
]
} | {
"input": [
"16 20 2\n10 3\n5 3\n10 5\n12 7\n7 6\n9 12\n9 6\n1 10\n11 16\n11 1\n16 2\n10 2\n14 4\n15 14\n4 13\n13 15\n1 8\n7 15\n1 7\n8 15\n",
"2 1 1\n2 1\n"
],
"output": [
"0\n0\n3\n3\n3\n3\n7\n7\n7\n7\n7\n11\n11\n11\n11\n15\n15\n15\n15\n16\n",
"2\n"
]
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
const int MAXN = 2e5 + 5;
vector<pair<int, int> > e;
set<pair<int, int> > f;
vector<int> adj[MAXN];
int n, m, k, deg[MAXN], num;
bool kicked[MAXN];
vector<int> ans;
int main() {
scanf("%d %d %d", &n, &m, &k);
for (int i = 0; i < m; ++i) {
int u, v;
scanf("%d %d"... |
1037_E. Trips | There are n persons who initially don't know each other. On each morning, two of them, who were not friends before, become friends.
We want to plan a trip for every evening of m days. On each trip, you have to select a group of people that will go on the trip. For every person, one of the following should hold:
* ... | {
"input": [
"4 4 2\n2 3\n1 2\n1 3\n1 4\n",
"5 8 2\n2 1\n4 2\n5 4\n5 2\n4 3\n5 1\n4 1\n3 2\n",
"5 7 2\n1 5\n3 2\n2 5\n3 4\n1 2\n5 3\n1 3\n"
],
"output": [
"0\n0\n3\n3\n",
"0\n0\n0\n3\n3\n4\n4\n5\n",
"0\n0\n0\n0\n3\n4\n4\n"
]
} | {
"input": [
"16 20 2\n10 3\n5 3\n10 5\n12 7\n7 6\n9 12\n9 6\n1 10\n11 16\n11 1\n16 2\n10 2\n14 4\n15 14\n4 13\n13 15\n1 8\n7 15\n1 7\n8 15\n",
"2 1 1\n2 1\n"
],
"output": [
"0\n0\n3\n3\n3\n3\n7\n7\n7\n7\n7\n11\n11\n11\n11\n15\n15\n15\n15\n16\n",
"2\n"
]
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
inline int read() {
int k = 1, sum = 0;
char c = getchar();
for (; c < '0' || c > '9'; c = getchar())
if (c == '-') k = -1;
for (; c >= '0' && c <= '9'; c = getchar()) sum = sum * 10 + c - '0';
return sum * k;
}
int n, m, k;
int in[200000 + 10];
set<int> son[2... |
1037_E. Trips | There are n persons who initially don't know each other. On each morning, two of them, who were not friends before, become friends.
We want to plan a trip for every evening of m days. On each trip, you have to select a group of people that will go on the trip. For every person, one of the following should hold:
* ... | {
"input": [
"4 4 2\n2 3\n1 2\n1 3\n1 4\n",
"5 8 2\n2 1\n4 2\n5 4\n5 2\n4 3\n5 1\n4 1\n3 2\n",
"5 7 2\n1 5\n3 2\n2 5\n3 4\n1 2\n5 3\n1 3\n"
],
"output": [
"0\n0\n3\n3\n",
"0\n0\n0\n3\n3\n4\n4\n5\n",
"0\n0\n0\n0\n3\n4\n4\n"
]
} | {
"input": [
"16 20 2\n10 3\n5 3\n10 5\n12 7\n7 6\n9 12\n9 6\n1 10\n11 16\n11 1\n16 2\n10 2\n14 4\n15 14\n4 13\n13 15\n1 8\n7 15\n1 7\n8 15\n",
"2 1 1\n2 1\n"
],
"output": [
"0\n0\n3\n3\n3\n3\n7\n7\n7\n7\n7\n11\n11\n11\n11\n15\n15\n15\n15\n16\n",
"2\n"
]
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
signed main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
long long n, m, k;
cin >> n >> m >> k;
vector<vector<pair<long long, long long> > > friends(n);
vector<long long> pointers(n);
for (long long i = 0; i < m; i++) {
long long a, b;
cin >> a... |
1037_E. Trips | There are n persons who initially don't know each other. On each morning, two of them, who were not friends before, become friends.
We want to plan a trip for every evening of m days. On each trip, you have to select a group of people that will go on the trip. For every person, one of the following should hold:
* ... | {
"input": [
"4 4 2\n2 3\n1 2\n1 3\n1 4\n",
"5 8 2\n2 1\n4 2\n5 4\n5 2\n4 3\n5 1\n4 1\n3 2\n",
"5 7 2\n1 5\n3 2\n2 5\n3 4\n1 2\n5 3\n1 3\n"
],
"output": [
"0\n0\n3\n3\n",
"0\n0\n0\n3\n3\n4\n4\n5\n",
"0\n0\n0\n0\n3\n4\n4\n"
]
} | {
"input": [
"16 20 2\n10 3\n5 3\n10 5\n12 7\n7 6\n9 12\n9 6\n1 10\n11 16\n11 1\n16 2\n10 2\n14 4\n15 14\n4 13\n13 15\n1 8\n7 15\n1 7\n8 15\n",
"2 1 1\n2 1\n"
],
"output": [
"0\n0\n3\n3\n3\n3\n7\n7\n7\n7\n7\n11\n11\n11\n11\n15\n15\n15\n15\n16\n",
"2\n"
]
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
const int maxn = 1e5 + 10;
set<pair<int, int> > sq;
set<int> to[maxn];
int n, m, k;
int a[maxn];
struct node {
int l, r;
} q[maxn];
void update() {
while (sq.size() > 0 && (*(sq.begin())).first < k) {
auto x = *(sq.begin());
sq.erase(sq.begin());
for (auto j... |
1037_E. Trips | There are n persons who initially don't know each other. On each morning, two of them, who were not friends before, become friends.
We want to plan a trip for every evening of m days. On each trip, you have to select a group of people that will go on the trip. For every person, one of the following should hold:
* ... | {
"input": [
"4 4 2\n2 3\n1 2\n1 3\n1 4\n",
"5 8 2\n2 1\n4 2\n5 4\n5 2\n4 3\n5 1\n4 1\n3 2\n",
"5 7 2\n1 5\n3 2\n2 5\n3 4\n1 2\n5 3\n1 3\n"
],
"output": [
"0\n0\n3\n3\n",
"0\n0\n0\n3\n3\n4\n4\n5\n",
"0\n0\n0\n0\n3\n4\n4\n"
]
} | {
"input": [
"16 20 2\n10 3\n5 3\n10 5\n12 7\n7 6\n9 12\n9 6\n1 10\n11 16\n11 1\n16 2\n10 2\n14 4\n15 14\n4 13\n13 15\n1 8\n7 15\n1 7\n8 15\n",
"2 1 1\n2 1\n"
],
"output": [
"0\n0\n3\n3\n3\n3\n7\n7\n7\n7\n7\n11\n11\n11\n11\n15\n15\n15\n15\n16\n",
"2\n"
]
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
vector<vector<int> > G;
set<pair<int, int> > S;
stack<int> results;
vector<pair<int, int> > E;
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
int n, m, k;
cin >> n >> m >> k;
G.resize(n);
for (int i = 0; i < m; i++) {
int u, v;
cin >> u >>... |
1037_E. Trips | There are n persons who initially don't know each other. On each morning, two of them, who were not friends before, become friends.
We want to plan a trip for every evening of m days. On each trip, you have to select a group of people that will go on the trip. For every person, one of the following should hold:
* ... | {
"input": [
"4 4 2\n2 3\n1 2\n1 3\n1 4\n",
"5 8 2\n2 1\n4 2\n5 4\n5 2\n4 3\n5 1\n4 1\n3 2\n",
"5 7 2\n1 5\n3 2\n2 5\n3 4\n1 2\n5 3\n1 3\n"
],
"output": [
"0\n0\n3\n3\n",
"0\n0\n0\n3\n3\n4\n4\n5\n",
"0\n0\n0\n0\n3\n4\n4\n"
]
} | {
"input": [
"16 20 2\n10 3\n5 3\n10 5\n12 7\n7 6\n9 12\n9 6\n1 10\n11 16\n11 1\n16 2\n10 2\n14 4\n15 14\n4 13\n13 15\n1 8\n7 15\n1 7\n8 15\n",
"2 1 1\n2 1\n"
],
"output": [
"0\n0\n3\n3\n3\n3\n7\n7\n7\n7\n7\n11\n11\n11\n11\n15\n15\n15\n15\n16\n",
"2\n"
]
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
int n, m, k;
struct edge {
int u, v, nxt;
} e[200005 * 2];
int cnt, head[200005];
void adde(int u, int v) {
e[++cnt].u = u;
e[cnt].v = v;
e[cnt].nxt = head[u];
head[u] = cnt;
}
int vis[200005 * 2], d[200005], ans[200005], res;
struct ege2 {
int u, v;
} edge[2000... |
1037_E. Trips | There are n persons who initially don't know each other. On each morning, two of them, who were not friends before, become friends.
We want to plan a trip for every evening of m days. On each trip, you have to select a group of people that will go on the trip. For every person, one of the following should hold:
* ... | {
"input": [
"4 4 2\n2 3\n1 2\n1 3\n1 4\n",
"5 8 2\n2 1\n4 2\n5 4\n5 2\n4 3\n5 1\n4 1\n3 2\n",
"5 7 2\n1 5\n3 2\n2 5\n3 4\n1 2\n5 3\n1 3\n"
],
"output": [
"0\n0\n3\n3\n",
"0\n0\n0\n3\n3\n4\n4\n5\n",
"0\n0\n0\n0\n3\n4\n4\n"
]
} | {
"input": [
"16 20 2\n10 3\n5 3\n10 5\n12 7\n7 6\n9 12\n9 6\n1 10\n11 16\n11 1\n16 2\n10 2\n14 4\n15 14\n4 13\n13 15\n1 8\n7 15\n1 7\n8 15\n",
"2 1 1\n2 1\n"
],
"output": [
"0\n0\n3\n3\n3\n3\n7\n7\n7\n7\n7\n11\n11\n11\n11\n15\n15\n15\n15\n16\n",
"2\n"
]
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
int n, m, k;
vector<int> used;
vector<int> selected;
vector<vector<int>> g;
bool dfs(int cur) {
used[cur] = 1;
selected[cur] = 1;
int cnt = 0;
for (auto t : g[cur]) {
if (!used[t]) {
dfs(t);
}
if (selected[t]) {
cnt++;
}
}
if (cnt < k... |
1037_E. Trips | There are n persons who initially don't know each other. On each morning, two of them, who were not friends before, become friends.
We want to plan a trip for every evening of m days. On each trip, you have to select a group of people that will go on the trip. For every person, one of the following should hold:
* ... | {
"input": [
"4 4 2\n2 3\n1 2\n1 3\n1 4\n",
"5 8 2\n2 1\n4 2\n5 4\n5 2\n4 3\n5 1\n4 1\n3 2\n",
"5 7 2\n1 5\n3 2\n2 5\n3 4\n1 2\n5 3\n1 3\n"
],
"output": [
"0\n0\n3\n3\n",
"0\n0\n0\n3\n3\n4\n4\n5\n",
"0\n0\n0\n0\n3\n4\n4\n"
]
} | {
"input": [
"16 20 2\n10 3\n5 3\n10 5\n12 7\n7 6\n9 12\n9 6\n1 10\n11 16\n11 1\n16 2\n10 2\n14 4\n15 14\n4 13\n13 15\n1 8\n7 15\n1 7\n8 15\n",
"2 1 1\n2 1\n"
],
"output": [
"0\n0\n3\n3\n3\n3\n7\n7\n7\n7\n7\n11\n11\n11\n11\n15\n15\n15\n15\n16\n",
"2\n"
]
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
int n, m, d[200005], k, res, vs[200005], ans[200005], pa;
map<int, int> cnt[200005];
vector<int> ke[200005];
pair<int, int> a[200005];
void dfs(int u) {
d[u]--;
res--;
vs[u] = 1;
for (int v : ke[u])
if (vs[v] == 0) {
cnt[u][v] = cnt[v][u] = 1;
d[v]--... |
1037_E. Trips | There are n persons who initially don't know each other. On each morning, two of them, who were not friends before, become friends.
We want to plan a trip for every evening of m days. On each trip, you have to select a group of people that will go on the trip. For every person, one of the following should hold:
* ... | {
"input": [
"4 4 2\n2 3\n1 2\n1 3\n1 4\n",
"5 8 2\n2 1\n4 2\n5 4\n5 2\n4 3\n5 1\n4 1\n3 2\n",
"5 7 2\n1 5\n3 2\n2 5\n3 4\n1 2\n5 3\n1 3\n"
],
"output": [
"0\n0\n3\n3\n",
"0\n0\n0\n3\n3\n4\n4\n5\n",
"0\n0\n0\n0\n3\n4\n4\n"
]
} | {
"input": [
"16 20 2\n10 3\n5 3\n10 5\n12 7\n7 6\n9 12\n9 6\n1 10\n11 16\n11 1\n16 2\n10 2\n14 4\n15 14\n4 13\n13 15\n1 8\n7 15\n1 7\n8 15\n",
"2 1 1\n2 1\n"
],
"output": [
"0\n0\n3\n3\n3\n3\n7\n7\n7\n7\n7\n11\n11\n11\n11\n15\n15\n15\n15\n16\n",
"2\n"
]
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
const int N = 212345;
int n, m, k;
vector<int> g[N];
set<pair<int, int> > st;
int deg[N];
int main(void) {
ios_base::sync_with_stdio(false);
cin >> n >> m >> k;
vector<pair<int, int> > edg;
set<pair<int, int> > act;
for (int i = 0; i < m; i++) {
int a, b;
... |
1037_E. Trips | There are n persons who initially don't know each other. On each morning, two of them, who were not friends before, become friends.
We want to plan a trip for every evening of m days. On each trip, you have to select a group of people that will go on the trip. For every person, one of the following should hold:
* ... | {
"input": [
"4 4 2\n2 3\n1 2\n1 3\n1 4\n",
"5 8 2\n2 1\n4 2\n5 4\n5 2\n4 3\n5 1\n4 1\n3 2\n",
"5 7 2\n1 5\n3 2\n2 5\n3 4\n1 2\n5 3\n1 3\n"
],
"output": [
"0\n0\n3\n3\n",
"0\n0\n0\n3\n3\n4\n4\n5\n",
"0\n0\n0\n0\n3\n4\n4\n"
]
} | {
"input": [
"16 20 2\n10 3\n5 3\n10 5\n12 7\n7 6\n9 12\n9 6\n1 10\n11 16\n11 1\n16 2\n10 2\n14 4\n15 14\n4 13\n13 15\n1 8\n7 15\n1 7\n8 15\n",
"2 1 1\n2 1\n"
],
"output": [
"0\n0\n3\n3\n3\n3\n7\n7\n7\n7\n7\n11\n11\n11\n11\n15\n15\n15\n15\n16\n",
"2\n"
]
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
template <class C, class E>
inline bool contains(const C& container, const E& element) {
return container.find(element) != container.end();
}
template <class T>
inline void checkmin(T& a, T b) {
if (b < a) a = b;
}
template <class T>
inline void checkmax(T& a, T b) {
if (b > a) a = b;
}
u... |
1037_E. Trips | There are n persons who initially don't know each other. On each morning, two of them, who were not friends before, become friends.
We want to plan a trip for every evening of m days. On each trip, you have to select a group of people that will go on the trip. For every person, one of the following should hold:
* ... | {
"input": [
"4 4 2\n2 3\n1 2\n1 3\n1 4\n",
"5 8 2\n2 1\n4 2\n5 4\n5 2\n4 3\n5 1\n4 1\n3 2\n",
"5 7 2\n1 5\n3 2\n2 5\n3 4\n1 2\n5 3\n1 3\n"
],
"output": [
"0\n0\n3\n3\n",
"0\n0\n0\n3\n3\n4\n4\n5\n",
"0\n0\n0\n0\n3\n4\n4\n"
]
} | {
"input": [
"16 20 2\n10 3\n5 3\n10 5\n12 7\n7 6\n9 12\n9 6\n1 10\n11 16\n11 1\n16 2\n10 2\n14 4\n15 14\n4 13\n13 15\n1 8\n7 15\n1 7\n8 15\n",
"2 1 1\n2 1\n"
],
"output": [
"0\n0\n3\n3\n3\n3\n7\n7\n7\n7\n7\n11\n11\n11\n11\n15\n15\n15\n15\n16\n",
"2\n"
]
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
int x[200001], y[200001];
vector<int> edge[200001];
int vis[200001];
int count_col[200001];
int ans[200001];
int count_edge[200001];
void dfs(int u, int c) {
vis[u] = c;
count_col[c]++;
int i, j = edge[u].size(), v;
for (i = 0; i < j; i++) {
v = edge[u][i];
... |
1037_E. Trips | There are n persons who initially don't know each other. On each morning, two of them, who were not friends before, become friends.
We want to plan a trip for every evening of m days. On each trip, you have to select a group of people that will go on the trip. For every person, one of the following should hold:
* ... | {
"input": [
"4 4 2\n2 3\n1 2\n1 3\n1 4\n",
"5 8 2\n2 1\n4 2\n5 4\n5 2\n4 3\n5 1\n4 1\n3 2\n",
"5 7 2\n1 5\n3 2\n2 5\n3 4\n1 2\n5 3\n1 3\n"
],
"output": [
"0\n0\n3\n3\n",
"0\n0\n0\n3\n3\n4\n4\n5\n",
"0\n0\n0\n0\n3\n4\n4\n"
]
} | {
"input": [
"16 20 2\n10 3\n5 3\n10 5\n12 7\n7 6\n9 12\n9 6\n1 10\n11 16\n11 1\n16 2\n10 2\n14 4\n15 14\n4 13\n13 15\n1 8\n7 15\n1 7\n8 15\n",
"2 1 1\n2 1\n"
],
"output": [
"0\n0\n3\n3\n3\n3\n7\n7\n7\n7\n7\n11\n11\n11\n11\n15\n15\n15\n15\n16\n",
"2\n"
]
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
const int mx = 10000;
set<int> se[mx];
int del[mx] = {0};
int dep[mx] = {0};
int ans[mx];
struct node {
int fr, to;
} ed[mx];
int cnt;
int n, m, k;
void dfs(int st) {
if (del[st] || dep[st] >= k) return;
queue<int> q;
q.push(st);
del[st] = 1;
--cnt;
while (!q.... |
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