prompt string | response string |
|---|---|
Problem: In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to wor... | #include <bits/stdc++.h>
using namespace std;
const int MAXN = 2E5 + 10;
int n;
vector<pair<int, int> > E[MAXN];
vector<vector<int> > ans;
void DFS(int u, int pre, int lastC) {
int c = -1;
for (int i = 0; i < E[u].size(); ++i)
if (E[u][i].first != pre) {
if (++c == lastC) ++c;
if (ans.size() <= c) a... |
Problem: In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to wor... |
import java.io.BufferedReader;
import java.io.File;
import java.io.FileInputStream;
import java.io.FileNotFoundException;
import java.io.FileOutputStream;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.math.BigInteger;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Bit... |
Problem: In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to wor... | #include <bits/stdc++.h>
using namespace std;
const long double eps = 1e-9, pi = acos(-1.0);
const long long inf = 2000000009;
int n, a, b, p = 0, c;
vector<pair<int, int> > v;
vector<vector<int> > g, ans;
vector<int> used;
map<pair<int, int>, int> M;
void dfs(int v, int p) {
used[v] = 1;
int c = 1, l;
if (p == -... |
Problem: In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to wor... | #include <bits/stdc++.h>
using namespace std;
struct r {
int to, n;
};
int max1 = 0;
bool vis[300000] = {0};
vector<r> graph[200001];
vector<int> ans[200001];
void dfs(int v, int c) {
if (vis[v]) return;
vis[v] = 1;
int p = 0;
for (int i = 0; i < graph[v].size(); i++) {
int to = graph[v][i].to;
if (vi... |
Problem: In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to wor... | #include <bits/stdc++.h>
using namespace std;
#pragma warning(disable : 4996)
vector<pair<int, int>> X[200009];
int n, a, b, maxn;
bool used[200009];
vector<int> ret[200009];
void dfs(int pos, int t) {
if (used[pos] == true) return;
used[pos] = true;
int G = 1;
for (pair<int, int> i : X[pos]) {
if (used[i.f... |
Problem: In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to wor... | n=input()
e=[map(int,raw_input().split()) for i in range(n-1)]
head=[-1] * 2 * n
next=[0] * 2 * n
for i in range(n-1):
a,b=e[i]
next[i]=head[a];head[a]=i;
next[i+n]=head[b];head[b]=i+n;
ans=[[] for i in range(n)]
maxtime=0
q=[[0,0,0] for i in range(n)]
qh,qe=0,1
q[0]=[1,0,0]
while qh<qe:
now,fa,t=q[qh]
... |
Problem: In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to wor... | #include <bits/stdc++.h>
using namespace ::std;
const int MOD = 1000000007, INF = 1e9, MAX = 2e6;
vector<long long int> v;
vector<long long int> g[200005];
vector<long long int> ans[200005];
long long int marked[200005];
long long int global_size = 0;
map<pair<long long int, long long int>, long long int> m;
void dfs(i... |
Problem: In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to wor... | #include <bits/stdc++.h>
using namespace std;
long long int max(long long int a, long long int b) {
if (a > b)
return a;
else
return b;
}
long long int min(long long int a, long long int b) {
if (a < b)
return a;
else
return b;
}
const int dx[4] = {-1, 1, 0, 0};
const int dy[4] = {0, 0, -1, 1};
... |
Problem: In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to wor... | #include <bits/stdc++.h>
using namespace std;
vector<pair<int, int>> e[200005];
vector<int> ans[200005];
int n, all;
void dfs(int fa, int u, int k, bool selected) {
int day = selected;
for (auto i : e[u]) {
int v = i.first, id = i.second;
if (fa != v) {
if (++day == k) {
day++;
}
a... |
Problem: In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to wor... | #include <bits/stdc++.h>
using namespace std;
int gn;
array<int, 2> ge[200010];
vector<vector<int> > aj(200010);
int vis[200010];
inline int ng(int vn, int ve) {
if (ge[ve][0] == vn)
return ge[ve][1];
else
return ge[ve][0];
}
vector<vector<int> > gm(200010);
void dfs(int vn, int vp) {
vis[vn] = 1;
int v... |
Problem: In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to wor... | #include <bits/stdc++.h>
using namespace std;
signed main() {
ios_base::sync_with_stdio(false);
cin.tie(nullptr);
int n;
cin >> n;
vector<vector<pair<int, int> > > graph(n + 1);
for (int e = 1; e < n; e++) {
int u, v;
cin >> u >> v;
graph[u].emplace_back(v, e);
graph[v].emplace_back(u, e);
... |
Problem: In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to wor... | import java.util.*;
import java.io.*;
import java.math.*;
import java.text.*;
public class Main {
static PrintWriter out;
static Reader in;
public static void main(String[] args) throws IOException {
input_output();
//out = new PrintWriter(System.out);
//in = new Reader(new FileInpu... |
Problem: In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to wor... | #include <bits/stdc++.h>
const int MaxN = 2e5;
using namespace std;
vector<pair<int, int> > G[MaxN + 5];
vector<int> res[MaxN + 5];
int n, ans;
void Dfs(int x, int fa, int t) {
int day = 0;
for (int i = 0; i < G[x].size(); i++) {
int v = G[x][i].first, pos = G[x][i].second;
if (v == fa) continue;
day++;... |
Problem: In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to wor... | #include <bits/stdc++.h>
using namespace std;
const int inf = 0x3f3f3f3f;
const long double pi = acos(-1.0);
const string debug_line = "yolo";
const long double PI = acos((long double)-1.0);
const long double eps = 1e-6;
unsigned long long getRand() {
return ((unsigned long long)rand() << 40) +
((unsigned lo... |
Problem: In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to wor... | #include <bits/stdc++.h>
using namespace std;
template <typename T>
void read(T &x) {
x = 0;
char ch = getchar();
long long f = 1;
while (!isdigit(ch)) {
if (ch == '-') f *= -1;
ch = getchar();
}
while (isdigit(ch)) {
x = x * 10 + ch - 48;
ch = getchar();
}
x *= f;
}
template <typename T... |
Problem: In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to wor... | #include <bits/stdc++.h>
#pragma GCC optimize("Ofast")
#pragma GCC target("avx,avx2,fma")
using namespace std;
bool ispoweroftwo(long long n) { return n & (!(n & (n - 1))); }
long long mod = 1000000007;
long long dx[] = {1, 0, -1, 0};
long long dy[] = {0, -1, 0, 1};
bool test = 0;
const long long inf = 1e18;
const long... |
Problem: In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to wor... | #include <bits/stdc++.h>
using namespace std;
void dfs(long long p, vector<vector<long long> > &answ,
vector<vector<pair<long long, long long> > > &g, long long cur,
vector<long long> ¬_use, vector<long long> &step) {
for (long long i = 0; i < g[cur].size(); i++) {
if (g[cur][i].first != p) {... |
Problem: In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to wor... | #include <bits/stdc++.h>
using namespace std;
#pragma GCC optimize("-Ofast")
const int N = 1e5 + 5;
const int mod = 1e9 + 7;
vector<vector<pair<long long int, long long int> > > adj;
vector<vector<long long int> > repair;
long long int modexp(long long int a, long long int b, long long int mod) {
long long int x = 1;... |
Problem: In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to wor... | #include <bits/stdc++.h>
long long mpow(long long a, long long n, long long mod) {
long long ret = 1;
long long b = a;
while (n) {
if (n & 1) ret = (ret * b) % mod;
b = (b * b) % mod;
n >>= 1;
}
return (long long)ret;
}
using namespace std;
vector<int> ans[(int)(5e5 + 500)];
vector<pair<int, int> ... |
Problem: In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to wor... | #include <bits/stdc++.h>
using namespace std;
const int N = 2 * 100 * 1000 + 2;
int n, k;
vector<pair<int, int>> adj[N];
vector<int> ans[N];
void dfs(int u, int par, int d) {
for (int i = 0, cnt = 0; i < adj[u].size(); i++) {
int v = adj[u][i].first, num = adj[u][i].second;
if (v != par) {
cnt += 1 + (c... |
Problem: In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to wor... | #include <bits/stdc++.h>
using namespace std;
const int INF = 1000000000;
const int MOD = 1000000007;
const double EPS = 1e-9;
vector<vector<pair<int, int>>> g;
vector<int> donotwant;
vector<vector<int>> ans;
void dfs(int v, int p = -1) {
int t = -1;
for (auto i : g[v]) {
if (i.first == p) continue;
t++;
... |
Problem: In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to wor... | #include <bits/stdc++.h>
using namespace std;
vector<int> mpp[200010];
map<pair<int, int>, int> finde;
map<int, vector<int>> ans;
int vis[200010];
void dfs(int u, int color) {
int cnt = 1;
vis[u] = 1;
for (auto v : mpp[u])
if (!vis[v]) {
if (cnt == color) cnt++;
ans[cnt].push_back(finde[{u, v}]);
... |
Problem: In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to wor... | #include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 10;
vector<int> graph[N];
struct Node {
Node* parent;
vector<Node*> children;
int n;
};
int n;
map<pair<int, int>, int> numeral;
int maxColor = 0;
int maxDay = 0;
vector<int> days[N];
pair<int, int> mpm(int a, int b) { return make_pair(min(a, b), m... |
Problem: In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to wor... | #include <bits/stdc++.h>
using namespace std;
inline int getint() {
int x = 0, f = 1, c = getchar();
for (; c < 48 || c > 57; f = c == '-' ? -1 : f, c = getchar())
;
for (; c > 47 && c < 58; x = x * 10 + c - 48, c = getchar())
;
return x * f;
}
const int N = 200005;
int mx, ihead[N], cnt;
struct E {
i... |
Problem: In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to wor... | #include <bits/stdc++.h>
using namespace std;
vector<pair<int, int> > sp[200000];
vector<int> days[200000];
int maxday = 0;
void construct(int t, int d, int c) {
int day = 0;
for (int i = 0; i < sp[t].size(); i++) {
if (sp[t][i].second != c) {
day++;
if (day == d) day++;
days[day].push_back(sp... |
Problem: In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to wor... | #include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 5;
int n;
vector<pair<int, int> > v[N];
int a, b;
vector<int> ans[N];
int mx;
void dfs(int node, int parent, int time) {
int cur = 0;
for (auto it : v[node]) {
int next = it.first;
if (next != parent) {
++cur;
cur += (cur == time)... |
Problem: In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to wor... | #include <bits/stdc++.h>
using namespace std;
long long int node, edge, vis[100010], dist[100010], col[110000], in[110000],
out[110000], timer = 1;
vector<long long int> gh[200010], ind[200010], ans[200010];
void dfs(long long int node, long long int p, long long int blocked) {
long long int cnt = 0;
for (long ... |
Problem: In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to wor... | import sys
import threading
from collections import defaultdict
def put():
return map(int, input().split())
def dfs(i, p, m):
cnt = 1
z = 0
for j in tree[i]:
if j==p: continue
if cnt==m: cnt+=1
index = edge_index[(i,j)]
ans[cnt].append(index)
z = max(dfs(j,i,cnt... |
Problem: In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to wor... | #include <bits/stdc++.h>
using namespace std;
const int maxn = 200000 + 5;
int n;
vector<pair<int, int> > edge[maxn];
vector<int> a[maxn];
int res;
void dfs(int u, int past, int day) {
int now = 0;
for (int i = 0; i < edge[u].size(); i++) {
int v = edge[u][i].first;
if (v == past) continue;
int road = e... |
Problem: In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to wor... | #include <bits/stdc++.h>
using namespace std;
signed main() {
ios_base::sync_with_stdio(false);
cin.tie(nullptr);
int n;
cin >> n;
vector<vector<pair<int, int> > > graph(n + 1);
for (int e = 1; e < n; e++) {
int u, v;
cin >> u >> v;
graph[u].emplace_back(v, e);
graph[v].emplace_back(u, e);
... |
Problem: In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to wor... | #include <bits/stdc++.h>
using namespace std;
void bfs(vector<vector<uint64_t>>& g,
map<tuple<uint64_t, uint64_t>, uint64_t>& i2r) {
vector<bool> visit(g.size());
deque<tuple<uint64_t, uint64_t>> ev;
unordered_map<int64_t, vector<uint64_t>> roads_by_day;
visit[1] = true;
int day = 0;
for (auto e : ... |
Problem: In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to wor... | #include <bits/stdc++.h>
using namespace std;
const int M = 200200;
vector<int> h[M], g[M], ans[M];
int n, m;
bool used[M];
void read() {
cin >> n;
for (int i = 0; i < n - 1; ++i) {
int x, y;
cin >> x >> y;
g[x].push_back(y);
g[y].push_back(x);
h[x].push_back(i + 1);
h[y].push_back(i + 1);
... |
Problem: In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to wor... | #include <bits/stdc++.h>
using namespace std;
long long mod = 1e9 + 7;
long long min(long long a, long long b) { return (a < b) ? a : b; }
long long max(long long a, long long b) { return (a > b) ? a : b; }
long long fp(long long a, long long b) {
if (b == 0) return 1;
long long x = fp(a, b / 2);
x = (x * x) % mo... |
Problem: In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to wor... | #include <bits/stdc++.h>
using namespace std;
const int MAXN = 2e5 + 7;
vector<pair<int, int> > adj[MAXN];
vector<int> ans[MAXN];
int mx = 1;
void DFS(int v, int par = -1, int col = 0) {
int lst = 1;
for (auto p : adj[v]) {
int u = p.first, id = p.second;
if (u == par) {
continue;
}
if (lst ==... |
Problem: In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to wor... | #include <bits/stdc++.h>
using namespace std;
vector<pair<long long, long long> > g[200003];
vector<int> ans[200003];
bool f(pair<long long, long long> a, pair<long long, long long> b) {
return ((long long)(g[a.first]).size()) < ((long long)(g[b.first]).size());
}
int mx_day = -2;
void dfs(int v, int p, int day) {
... |
Problem: In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to wor... | #include <bits/stdc++.h>
using namespace std;
template <typename... Args>
void dbg(Args... args) {
((cout << args << " "), ...);
}
const int INF = 1e9, N = 1e6;
vector<pair<int, int>> g[N];
vector<int> ans[N];
bitset<N> used;
int main() {
int n;
cin >> n;
int maxi = 0;
for (int i = 1; i < n; ++i) {
int u,... |
Problem: In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to wor... | #include <bits/stdc++.h>
using namespace std;
constexpr int maxn = 2e5 + 5, inf = 0x3f3f3f3f;
int n, m, cnt = 1;
int head[maxn], col[maxn], du[maxn];
int k;
struct edge {
int u, v, nex;
} e[maxn * 2];
vector<int> d[maxn];
inline void add(int u, int v) {
e[++cnt].v = v;
e[cnt].u = u;
e[cnt].nex = head[u];
head... |
Problem: In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to wor... | #include <bits/stdc++.h>
using namespace std;
const int size = 200005;
int n, visit[size], tmax;
vector<vector<int> > cnt, vertices;
map<tuple<int, int>, int> roads;
int dfs(int v, int p, int t) {
tmax = max(t, tmax);
visit[v] = 1;
if (not cnt[t].size()) {
if (roads.count(make_tuple(p, v)))
cnt[t].push_... |
Problem: In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to wor... | import java.io.*;
import java.math.BigInteger;
import java.util.*;
public class C {
class Edge{
int to;
int ind;
int day;
public Edge(int to, int ind) {
this.to = to;
this.ind = ind;
}
}
ArrayList<Edge>[] g;
ArrayList<Edge> a=new Array... |
Problem: In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to wor... | #include <bits/stdc++.h>
using namespace std;
const long long MAX = 1e18;
const long long MOD = 1e9 + 7;
const double eps = 1e-8;
long long t, m, n, k;
vector<pair<int, int>> a[200005];
set<int> st;
vector<pair<pair<int, int>, int>> p1, p2;
vector<int> ans[200005];
void dfs(int u, int p, int color) {
int last = 0;
... |
Problem: In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to wor... | #include <bits/stdc++.h>
using namespace std;
const int INF = int(1e9);
const long long INFll = 1ll * INF * INF;
const long double ldINF = 1e+018;
const long double EPS = 0.000000001;
const int N = 100001;
template <typename T>
ostream& operator<<(ostream& out, pair<T, T>& a) {
out << a.first << " " << a.second;
re... |
Problem: In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to wor... | #include <bits/stdc++.h>
using namespace std;
const int maxn = 200100;
vector<pair<int, int> > g[maxn];
vector<int> solution[maxn];
int n;
void dfs(int node, int p, int opentime) {
int br = 1;
for (auto i : g[node]) {
if (i.first != p) {
if (br == opentime) br++;
solution[br].push_back(i.second);
... |
Problem: In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to wor... | #include <bits/stdc++.h>
using namespace std;
int n, i, j, a, b, mx, gag, son, parent[200005], mark[200005];
vector<pair<int, int> > g[200005];
vector<int> q[200005];
void build(int v) {
for (int i = 0; i < g[v].size(); i++) {
int to = g[v][i].first;
if (to == parent[v]) continue;
parent[to] = v;
buil... |
Problem: In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to wor... | #include <bits/stdc++.h>
using namespace std;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
void rset();
void init_test();
void solve();
signed main() {
ios::sync_with_stdio(false);
cin.tie(0);
cout << fixed;
cout.precision(20);
init_test();
return 0;
}
template <typename T>
void chmi... |
Problem: In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to wor... | #include <bits/stdc++.h>
using namespace std;
const int MAX = 2 * 1e5 + 10;
vector<int> a[MAX], l[MAX];
int d[MAX];
map<pair<int, int>, int> mp;
void dfs(int v, int c = 0, int p = 0) {
int k = 0;
for (auto u : a[v]) {
if (u == p) continue;
k++;
if (k == c) k++;
l[k].push_back(mp[{v, u}]);
dfs(u,... |
Problem: In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to wor... | import java.io.*;
import java.util.*;
public class c2 {
BufferedReader br;
StringTokenizer in;
PrintWriter out;
public String nextToken() throws IOException {
while (in == null || !in.hasMoreTokens()) {
in = new StringTokenizer(br.readLine());
}
return in.nextToken();
}
public int nextInt() throws IOE... |
Problem: In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to wor... | #include <bits/stdc++.h>
using namespace std;
vector<int> answer[300000];
bool used[300000];
vector<pair<int, int> > adj[300000];
int n;
void dfs(int v = 1, int forb = -1) {
used[v] = true;
int step = 0;
for (int i = 0; i < adj[v].size(); ++i) {
int u, w;
u = adj[v][i].first;
w = adj[v][i].second;
... |
Problem: In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to wor... | #include <bits/stdc++.h>
using namespace std;
const int N = 200000;
vector<pair<int, int> > g[N];
vector<int> days[N];
int ans;
void dfs(int v, int p, int used) {
int nxt = 1;
for (int i = 0; i < g[v].size(); i++) {
int u = g[v][i].first;
int w = g[v][i].second;
if (u != p) {
if (nxt == used) nxt+... |
Problem: In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to wor... | #include <bits/stdc++.h>
using namespace std;
long long int INF = 1e9 + 5;
long long int mod = 998244353;
long long int nax = 2e5;
vector<vector<long long int>> days(nax + 1);
vector<vector<pair<long long int, long long int>>> g(nax + 1);
void dfs(long long int node, long long int pa, long long int k,
long lon... |
Problem: In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to wor... | #include <bits/stdc++.h>
using namespace std;
vector<int> ans[300001];
vector<pair<int, int> > v[300001];
void dfs(int cur, int par, int prev_day) {
int cur_day = 1;
for (int i = 0; i < v[cur].size(); i++) {
int x = v[cur][i].first;
if (x == par) continue;
if (cur_day == prev_day) cur_day++;
ans[cur... |
Problem: In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to wor... | import java.util.*;
import java.io.*;
public class C {
FastScanner in;
PrintWriter out;
class Edge {
int x, num;
public Edge(int x, int num) {
this.x = x;
this.num = num;
}
}
List<Edge>[] edges;
List<Integer>[] answer;
int ans;
void dfs... |
Problem: In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to wor... | #include <bits/stdc++.h>
using namespace std;
int n;
vector<pair<int, int> > adj[200013];
vector<int> has[200013];
void dfs(int x, int p = 0, int use = 0) {
int on = 0;
for (auto i : adj[x])
if (i.first != p) {
while (++on == use)
;
has[on].push_back(i.second);
dfs(i.first, x, on);
... |
Problem: In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to wor... | #include <bits/stdc++.h>
using namespace std;
const long long N = 2e5;
const long long mod = 998244353;
vector<vector<long long> > day(N + 1);
vector<vector<pair<long long, long long> > > adj(N + 1);
long long vis[N + 1], col[N + 1];
int32_t main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
long long n, m... |
Problem: In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to wor... | #include <bits/stdc++.h>
using namespace std;
vector<vector<pair<int, int> > > g;
vector<vector<int> > arr;
bool *used;
void dfs(int v, int day) {
used[v] = true;
int d = 0;
if (d == day) d++;
for (int i = 0; i < (int)g[v].size(); i++) {
if (used[g[v][i].first]) continue;
arr[d].push_back(g[v][i].second... |
Problem: In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to wor... | #include <bits/stdc++.h>
using namespace std;
struct Point {
double x, y;
Point(double x = 0, double y = 0) : x(x), y(y) {}
Point operator+(const Point& p) const { return Point(p.x + x, p.y + y); }
Point operator-() const { return Point(-x, -y); }
Point operator-(const Point& p) const { return *this + (-p); }... |
Problem: In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to wor... | #include <bits/stdc++.h>
using namespace std;
struct edge {
int to, next, id;
} e[200010 << 1];
int box[200010], cnt;
void init() {
memset(box, -1, sizeof(box));
cnt = 0;
}
int ma;
void add(int from, int to, int id) {
e[cnt].id = id;
e[cnt].to = to;
e[cnt].next = box[from];
box[from] = cnt++;
}
vector<int... |
Problem: In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to wor... | #include <bits/stdc++.h>
using namespace std;
const int N = 200005;
int n;
vector<pair<int, int> > v[N];
int a, b;
vector<int> ans[N];
int mx = 0;
void dfs(int node, int parent, int time) {
int cur = 0;
for (auto it = v[node].begin(); it != v[node].end(); ++it) {
int next = (*it).first;
if (next != parent) ... |
Problem: In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to wor... | #include <bits/stdc++.h>
using namespace std;
const int MOD = 1e9 + 7;
const int SIZE = 1e6 + 10;
vector<pair<int, int> > e[SIZE];
int an[SIZE];
vector<int> res[SIZE];
void dfs(int x, int lt, int col) {
set<int> use;
use.insert(col);
int now = 1;
for (int i = 0; i < (((int)(e[x]).size())); ++i) {
int y = e[... |
Problem: In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to wor... | #include <bits/stdc++.h>
using namespace std;
const int MN = 200005;
map<pair<int, int>, int> num;
vector<int> G[MN], wyn[MN];
int dp[MN], used[MN];
int res;
void dfs(int x) {
int c = 0;
used[x] = 1;
int s = G[x].size();
for (int i = 0; i < s; ++i) {
if (!used[G[x][i]]) {
++c;
if (c == dp[x]) ++... |
Problem: In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to wor... | /**
* Created by Anna on 26.03.2016.
*/
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.lang.reflect.Array;
import java.util.*;
public class TaskC {
StringTokenizer st;
PrintWriter out;
BufferedReader in;
public st... |
Problem: In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to wor... | #include <bits/stdc++.h>
using namespace std;
const int N = 200000 + 1000;
vector<pair<int, int> > g[N];
vector<int> ans[N];
int n, sum, tot, u, v;
void dfs(int x, int fa, int son_t) {
int l = g[x].size();
int sum = 0;
for (int i = 0; i < l; i++) {
int v = g[x][i].first;
if (v == fa) continue;
sum++;
... |
Problem: In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to wor... | #include <bits/stdc++.h>
using namespace std;
const long long mod = 998244353;
int n, rs = 0;
vector<pair<int, int> > adj[200005];
vector<int> ans[200005];
void dfs(int nw, int pr, int dy) {
int cnt = 0;
for (auto i : adj[nw]) {
int nx = i.first, nm = i.second;
if (nx == pr) continue;
cnt++;
if (cnt... |
Problem: In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to wor... | #include <bits/stdc++.h>
using namespace std;
template <class T>
void show(T a, int n) {
for (int i = 0; i < n; ++i) cerr << a[i] << ' ';
cerr << endl;
}
template <class T>
void show(T a, int r, int l) {
for (int i = 0; i < r; ++i) show(a[i], l);
cerr << endl;
}
const int INF = 1000 * 1000 * 1000 + 1000 * 1000;... |
Problem: In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to wor... | #include <bits/stdc++.h>
using namespace std;
vector<vector<pair<int, int> > > g;
vector<int> edge_color;
void dfs(int cur, int forb) {
int cur_color = 0;
for (int i = 0; i < g[cur].size(); i++) {
if (edge_color[g[cur][i].second] < 0) {
if (cur_color == forb) {
cur_color++;
}
edge_colo... |
Problem: In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to wor... | #include <bits/stdc++.h>
using namespace std;
inline int max3(int a, int b, int c) { return max(a, max(b, c)); }
inline int min3(int a, int b, int c) { return min(a, min(b, c)); }
const int INF = 1e9;
const long long LINF = 1e18;
const double EPS = 1e-6;
const double PI = acos(-1.0);
const int maxn = 200020;
int n;
vec... |
Problem: In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to wor... | #include <bits/stdc++.h>
#pragma GCC optimize("Ofast")
#pragma GCC optimization("unroll-loops")
const int N = 2e5 + 5;
const long long int mod = 1e9 + 7;
const long long int Mod = 998244353;
const long double Pi = acos(-1);
const long long int Inf = 4e18;
using namespace std;
vector<int> mark(N);
vector<int> g[N], ans[... |
Problem: In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to wor... | #include <bits/stdc++.h>
using namespace std;
int dx[] = {-1, -1, -1, 0, 0, 1, 1, 1};
int dy[] = {-1, 0, 1, -1, 1, -1, 0, 1};
int xChange[] = {0, 1, 0, -1};
int yChange[] = {1, 0, -1, 0};
const int N = 200005;
int n, mx, cnt[N], d, ans;
vector<int> adj[N];
map<pair<int, int>, int> mp;
vector<vector<int>> vv(N);
bool vi... |
Problem: In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to wor... | import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.Comparator;
import java.util.HashSet;
import java.util.Scanner;
public class VKCupQualC {
private static class pair {
public int index, v, day = 0;
public pair(int i, int v) {
this.... |
Problem: In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to wor... | #include <bits/stdc++.h>
using namespace std;
vector<pair<int, int>> adj[200005];
vector<vector<int>> edges;
void dfs(int u, int p, int idx) {
int i = 0;
for (auto v : adj[u])
if (v.first != p) {
if (i == idx) i++;
if (i == edges.size()) edges.push_back(vector<int>());
edges[i].push_back(v.sec... |
Problem: In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to wor... | #include <bits/stdc++.h>
using namespace std;
void ga(int N, int *A) {
for (int i(0); i < N; i++) scanf("%d", A + i);
}
vector<int> O[(200005)];
vector<pair<int, int> > g[(200005)];
int N, X, a, b;
void dfs(int u, int p, int F) {
int I = -1;
for (auto &h : g[u])
if (h.first ^ p) {
if (++I == F) ++I;
... |
Problem: In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to wor... | #include <bits/stdc++.h>
using namespace std;
const int MAXN = 2e5 + 1;
const int INF = 1e9 + 7;
const int inf = -1e9 - 7;
const long double PI = 3.1415926535897932384626433832795;
struct pt {
int x, y, id;
};
bool comp(pt a, pt b) { return a.x < b.x; }
long long iyuza(pt a, pt b, pt c);
long double yuza(pt a, pt b, ... |
Problem: In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to wor... | #include <bits/stdc++.h>
using namespace std;
long long int n;
vector<long long int> adj[200015], ans[200015];
long long int mx = 0;
map<pair<long long int, long long int>, long long int> ma;
void dfs(long long int cur, long long int par, long long int not_allowed) {
long long int c = 1;
for (auto x : adj[cur]) {
... |
Problem: In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to wor... | #include <bits/stdc++.h>
using namespace std;
const int inf = (int)2 * 1e9;
vector<vector<pair<int, int> > > graph;
vector<vector<int> > road;
vector<set<int> > busy;
void dfs(int x, int par) {
int curr = 0;
for (int i = 0; i < graph[x].size(); i++)
if (graph[x][i].first != par) {
while (busy[x].find(curr... |
Problem: In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to wor... | #include <bits/stdc++.h>
using namespace std;
const long long MAX = 1000000000000000000;
long long mod = 1000000000;
long double pi = 3.141592653589793238;
void pls() {
ios_base::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
}
const long long N = 2e5 + 2;
vector<pair<int, int>> g[N];
vector<int> res[N];
int mx;
vo... |
Problem: In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to wor... | #include <bits/stdc++.h>
using namespace std;
const int MAXN = 2e5 + 100;
vector<pair<int, int> > E[MAXN];
vector<int> answer[MAXN];
int seen[MAXN];
set<int> added[MAXN];
void dfs(int u) {
seen[u] = 1;
int pointer = 1;
for (pair<int, int> edge : E[u]) {
int v = edge.first;
int id = edge.second;
if (se... |
Problem: In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to wor... | import java.io.*;
import java.util.ArrayList;
public class ProblemC {
public static void main(String[] args) throws IOException {
new ProblemC().run();
}
public void run() throws IOException {
MyInput sc = new MyInput(System.in);
int n = sc.nextInt();
Graph g = new Graph(n)... |
Problem: In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to wor... | #include <bits/stdc++.h>
using namespace std;
int n;
vector<vector<int> > G;
vector<vector<int> > ID;
vector<vector<int> > ans;
void addp(int id, int t) {
while (ans.size() <= t) ans.push_back(vector<int>());
ans[t].push_back(id);
}
void solve(int from, int lab, int v) {
int ctr = 0;
for (int i = 0; i < G[v].si... |
Problem: In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to wor... | #include <bits/stdc++.h>
using namespace std;
const int MaxN = 2e5;
int n, all;
int pre[2 * MaxN + 5], last[MaxN + 5], other[2 * MaxN + 5];
bool vis[MaxN + 5];
int ans[2 * MaxN + 5], num[2 * MaxN + 5], seq[2 * MaxN + 5], used[MaxN + 5];
vector<int> v[MaxN + 5];
void build(int x, int y, int k) {
pre[++all] = last[x];
... |
Problem: In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to wor... | #include <bits/stdc++.h>
using namespace std;
const int maxn = 200000;
int n, x;
bool was[maxn];
vector<pair<int, int> > g[maxn];
vector<int> ans[maxn];
void dfs(int u, int c) {
was[u] = true;
int cur_c = 0;
for (int i = 0; i < int(g[u].size()); i++) {
int v = g[u][i].first;
int id = g[u][i].second;
i... |
Problem: In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to wor... | #include <bits/stdc++.h>
using namespace std;
int answ;
vector<vector<pair<int, int> > > v;
vector<vector<int> > q;
vector<bool> used;
void dfs(int vertex, int color) {
used[vertex] = true;
int c = 0;
for (int i = 0; i < v[vertex].size(); i++) {
if (!used[v[vertex][i].first]) {
if (c == color) c++;
... |
Problem: In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to wor... | #include <bits/stdc++.h>
using namespace std;
const long long MN = 200111;
long long deg[MN];
vector<long long> ke[MN];
vector<long long> ls[MN];
map<long long, long long> id[MN];
void dfs(long long u, long long fu, long long last) {
long long c = 0;
for (long long v : ke[u]) {
if (v == fu) continue;
++c;
... |
Problem: In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to wor... | import java.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.OutputStream;
import java.io.PrintWriter;
import java.util.Iterator;
import java.io.BufferedWriter;
import java.util.InputMismatchException;
import java.io.IOException;
import java.util.ArrayList;
import java.io.Writer;
... |
Problem: In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to wor... | #include <bits/stdc++.h>
using namespace std;
const long long N = 2e5 + 3;
vector<vector<pair<long long, long long>>> v(N);
vector<vector<long long>> day;
void dfs(long long x, long long d, long long p) {
long long temp = d;
long long m = day.size();
for (pair<long long, long long> i : v[x]) {
if (i.first != ... |
Problem: In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to wor... | #include <bits/stdc++.h>
using namespace std;
const int N = 200005;
struct Edge {
int id;
int from, to;
int day;
Edge *prev;
bool used;
Edge(int id, int from, int to) : id(id), from(from), to(to) { used = false; }
};
int n, m;
vector<Edge> g[N];
Edge *q[N];
vector<int> res[N * 10];
int main() {
int n;
c... |
Problem: In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to wor... | import java.io.*;
import java.math.BigInteger;
import java.util.*;
public class C {
static BufferedReader in;
static PrintWriter out;
static StringTokenizer tok;
public static void main(String[] args) throws IOException {
try {
in = new BufferedReader(new FileReader("quarth.in"));... |
Problem: In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to wor... | #include <bits/stdc++.h>
using namespace std;
inline int nxt() {
int first;
scanf("%d", &first);
return first;
}
vector<vector<int>> a;
vector<char> used;
vector<int> color;
vector<int> p;
void dfs(int v) {
used[v] = 1;
int i = 0;
for (auto first : a[v]) {
if (used[first]) continue;
while (i == colo... |
Problem: In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to wor... | #include <bits/stdc++.h>
unsigned long int size = 0;
struct point {
unsigned int visited;
std::vector<unsigned long int> neigbors;
};
std::vector<point> graph;
std::map<unsigned long int, std::vector<unsigned long int> > colors;
std::map<std::pair<unsigned long int, unsigned long int>, unsigned long int>
routes... |
Problem: In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to wor... | #include <bits/stdc++.h>
using namespace std;
template <typename A, typename B>
string to_string(pair<A, B> p);
template <typename A, typename B, typename C>
string to_string(tuple<A, B, C> p);
template <typename A, typename B, typename C, typename D>
string to_string(tuple<A, B, C, D> p);
string to_string(const string... |
Problem: In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to wor... | #include <bits/stdc++.h>
using namespace std;
int l[200005], r[200005], sz;
vector<int> g[200005], arr[200005];
void dfs(int v, int pr, int wt) {
int cnt = 0;
for (int i = 0, id, to; i < g[v].size(); ++i) {
id = g[v][i];
if (v == l[id]) {
to = r[id];
} else {
to = l[id];
}
if (to == ... |
Problem: In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to wor... | #include <bits/stdc++.h>
using namespace std;
vector<int> mat[200030];
map<pair<int, int>, int> mp;
int main() {
int n;
cin >> n;
vector<int> val(n);
for (int i = 0; i < n - 1; i++) {
int a, b;
cin >> a >> b;
a--;
b--;
if (a > b) {
swap(a, b);
}
pair<int, int> pq;
pq.first ... |
Problem: In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to wor... | #include <bits/stdc++.h>
using namespace std;
const int MX = 2e5 + 7;
int n;
vector<pair<int, int> > g[MX];
int id[MX];
vector<int> ans[MX];
int c[MX];
int sz;
void solve(int v, int p = -1) {
int d = 1;
for (auto u : g[v]) {
if (u.first == p) {
continue;
}
while (c[v] == d) {
d++;
}
... |
Problem: In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to wor... | #include <bits/stdc++.h>
using namespace std;
long long int gcd(long long int a, long long int b) {
return b ? gcd(b, a % b) : a;
}
long long int lcm(long long int a, long long int b) {
return a * b / gcd(a, b);
}
const int N = 2e5 + 10;
vector<pair<int, int>> ad[N];
vector<int> ans[N];
int n;
void dfs(int v, int l... |
Problem: In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to wor... | import java.io.BufferedReader;
// import java.io.FileInputStream;
// import java.io.FileOutputStream;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.IOException;
import java.io.PrintWriter;
import java.util.*;
import static java.lang.Math.*;
import static java.util.Arrays.copyOf;
import s... |
Problem: In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to wor... | #include <bits/stdc++.h>
using namespace std;
long long n, mx = 0;
vector<long long> g[200001], used(200001), ans[200001];
map<pair<long long, long long>, long long> mp;
void dfs(long long v, long long p, long long pr) {
long long cnt = 0, c = 0, o = 0, q = 0;
used[v] = 1;
for (auto k : g[v]) {
cnt++;
c++... |
Problem: In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to wor... | #include <bits/stdc++.h>
template <typename T>
bool domax(T &a, T b) {
return (b > a ? (a = b, true) : false);
}
template <typename T>
bool domin(T &a, T b) {
return (b < a ? (a = b, true) : false);
}
const int maxn = 200 * 1000 + 5;
int n;
std::vector<std::pair<int, int>> e[maxn];
std::vector<int> v[maxn];
void go... |
Problem: In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to wor... | import java.util.*;
import java.math.*;
import java.io.*;
import java.io.File;
import java.io.FileInputStream;
import java.io.FileNotFoundException;
public class Hello {
static ArrayList<Integer> a[] = new ArrayList[200005];
static ArrayList<Integer> e[] = new ArrayList[200005];
static ArrayList<Integer> ans[] = ... |
Problem: In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to wor... | #include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 5;
vector<pair<int, int> > adj[N];
vector<int> plan[N];
int mxsz;
void dfs(int nd, int pr, int fndn_day) {
int day = 1;
for (auto &&i : adj[nd]) {
if (i.first == pr) continue;
if (day == fndn_day) day++;
mxsz = max(mxsz, day);
plan[da... |
Problem: In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to wor... | #include <bits/stdc++.h>
using namespace std;
int n;
vector<int> ans[200005];
vector<pair<int, int> > pms[200005];
int used[200005];
int MX = 0;
void dfs(int cur, int denied) {
used[cur] = 1;
int cnt = 1;
for (int i = 0; i < pms[cur].size(); ++i) {
if (cnt == denied) cnt++;
if (!used[pms[cur][i].first]) {... |
Problem: In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to wor... | import java.io.*;
import java.math.BigInteger;
import java.util.ArrayDeque;
import java.util.ArrayList;
import java.util.List;
import java.util.StringTokenizer;
public class CF {
static int[] excludeDay;
static int[] nextFreeDay;
static boolean[] isVisited;
static ArrayList<Integer>[] schedule;
sta... |
Problem: In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to wor... | #include <bits/stdc++.h>
using namespace std;
struct aj {
int ct;
int r;
aj(int a, int b) {
ct = a;
r = b;
}
};
vector<aj> adj[200005];
vector<int> day[200005];
int road[200005] = {0}, maxd = 0, n, i, a, b;
void repai(int i, int rped) {
int k = 0;
for (vector<aj>::iterator it = adj[i].begin(); it !=... |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.