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;
vector<long long> day;
void dfs(long long src, vector<vector<pair<long long, long long>>> &adj,
vector<bool> &vis, long long dont = -1, long long par = -1) {
vis[src] = true;
long long cur = 1;
for (auto &x : adj[src]) {
long long dest = x.first, roadInde... |
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 = 1e9 + 7;
long long dx[] = {-1, 0, 1, 0};
long long dy[] = {0, -1, 0, 1};
vector<pair<long long, long long> > g[200005];
vector<long long> ans[200005];
long long res;
void dfs(long long x, long long p, long long t) {
long long cnt = 0;
for (auto 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 N = 200200;
vector<int> d[N];
vector<pair<int, int>> g[N];
int n, k;
void go(int cur, int u = -1, int par = -1) {
int day = 0;
for (auto it : g[cur])
if (it.first != par) {
if (day == u) ++day;
d[day].push_back(it.second);
if (day > k) k ... |
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 MODD = 1000000007LL;
long long poww(long long x, long long y, long long MODD) {
if (x == 0LL && y == 0LL) {
return 1LL;
}
long long ret = 1;
while (y) {
if (y & 1LL) {
ret *= x;
ret %= MODD;
}
x *= x;
x %= MODD;
y ... |
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;
const int INF = 1e9 + 23;
const int MOD = 1e9 + 7;
const int SZ = 2e5 + 100;
vector<pair<int, int> > G[SZ];
vector<int> ans[SZ];
int nr[SZ];
void dfs(int v, int par) {
int p = 0;
for (auto& it : G[v]) {
int u = it.first, id = it.second;
if (u == par) continue;
... |
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 = 212345;
vector<pair<int, int> > v[N];
bool used[N];
set<int> st[N];
vector<int> ans[N];
int tt = 0;
void dfs(int x) {
used[x] = 1;
int cnt = 1;
for (int i = 0; i < v[x].size(); i++) {
int to = v[x][i].first;
int ti = v[x][i].second;
if (!used... |
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 par {
int ver, id;
par(){};
par(int _ver, int _id) : ver(_ver), id(_id){};
};
int i, j, n, m, maxi, x, y, id, b[300000];
vector<par> a[300000];
vector<int> ans[300000];
void DFS(int x, int k) {
b[x] = 1;
int cnt = 0;
for (int i = 0; i < (int)a[x].size(); ... |
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 comment(linker, "/STACK:66777216")
using namespace std;
long long n, m, t, l, r, x, y;
vector<long long> a[200005];
bool vis[200005];
bool maked[200005];
vector<long long> ans[200005];
map<pair<long long, long long>, long long> ind;
int main() {
cin >> n;
for (int i = 0; i < n - 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;
bool cmp(const pair<int, int> &a, const pair<int, int> &b) {
if (a.second != b.second) return a.second > b.second;
return a.first < b.first;
};
class graphal {
public:
int n, mx = 0;
vector<pair<int, int> > *ed;
vector<int> *ans;
int *taken;
graphal(int 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;
const int N = 200000 + 10;
struct node {
vector<int> children;
vector<int> road;
int color;
int parentDay;
};
node lst[N];
int main() {
int n, i, j;
scanf("%d", &n);
for (i = 1; i <= n; i++) {
lst[i].color = lst[i].parentDay = 0;
}
int ans = 0;
for (... |
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 Node {
vector<int> neib;
vector<int> ednum;
bool done;
};
Node* nodes;
vector<int>* days;
void dfs(int nodeNum, int parentDay) {
int day = 0;
nodes[nodeNum].done = true;
for (int i = 0; i < (int)nodes[nodeNum].neib.size(); i++) {
if (day == parentDay)... |
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 C {
BufferedReader br;
PrintWriter out;
StringTokenizer st;
boolean eof;
static class Edge {
int to, id;
public Edge(int to, int id) {
this.to = to;
this.id = id;
}
}
List<Edge>[] g;
int[] col;
void dfs(int v, int p, int forbColor) {
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 MAXN = 2e5 + 10;
const int INF = 1e9 + 7;
bool used[MAXN];
vector<pair<int, int>> G[MAXN];
vector<int> roads[MAXN];
int dfs(int v, int day) {
used[v] = 1;
int ans = G[v].size();
int firstday = day;
int toadd = -1;
for (int i = 0; i < G[v].size(); ++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>
#pragma comment(linker, "/STACK:128777216")
using namespace std;
const long long LINF = 1000000000000000000LL;
const int INF = 1000000000;
const long double eps = 1e-9;
const long double PI = 3.1415926535897932384626433832795l;
void prepare(string s) {
if (s.length() != 0) {
freopen((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;
const string nameFiles = "";
void error(vector<string>::iterator) {}
template <typename T, typename... Args>
void error(vector<string>::iterator cur_var, T a, Args... args) {
cerr << cur_var->substr((*cur_var)[0] == ' ') << " = " << a << endl;
error(++cur_var, args...);... |
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> adj[200009], num[200009], solution[200009];
int degree[200009];
void dfs(int u, int p, int upcolor) {
vector<int> v;
int timer = 0;
for (int i = 0; i < adj[u].size(); i++)
if (adj[u][i] != p) {
timer++;
if (timer == upcolor) {
timer... |
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> day;
void dfs(int src, vector<vector<pair<int, int>>> &adj, vector<bool> &vis,
int dont = -1, int par = -1) {
vis[src] = true;
int cur = 1;
for (auto &x : adj[src]) {
int dest = x.first, roadIndex = x.second;
if (dest == par) continue;
... |
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 x;
vector<int> g[200010];
map<pair<int, int>, int> ans;
map<pair<int, int>, int> pos;
bool cmp(int x, int y) {
if (g[x] >= g[y]) return false;
return true;
}
vector<vector<int> > res;
bool u[200010];
int main() {
cin >> x;
for (int i = 0; i < x - 1; i++) {
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>> adj[300001];
vector<int> parent(200001);
int cma = 0;
vector<int> make[200001];
vector<bool> visited(200001);
void dfs(int v, int p = -1) {
visited[v] = 1;
int cid = 0;
for (auto it : adj[v]) {
if (!visited[it.first]) {
if (cid == p) 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 N = (int)3e5;
int n;
vector<int> g[N];
vector<int> day[N];
struct edge {
int a, b;
} E[N];
void dfs(int v, int e, int bad) {
int curDay = 0;
for (int id : g[v]) {
if (id == e) continue;
if (++curDay == bad) curDay++;
day[curDay].push_back(id);
... |
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>
int N, head[200001], next[399999], to[399999], id[399999], E, O, q[200001],
fa[200001], d[200001];
std::vector<int> V[200001];
int main() {
scanf("%d", &N);
for (int i = 1, u, v; i < N; i++) {
scanf("%d%d", &u, &v);
d[u]++;
d[v]++;
next[++E] = head[u], to[E] = v, id[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;
typedef struct lnod {
int nod, nr;
lnod *next;
} * nod;
int i, n, x, y, timer[200005], nr[200005], ans;
nod lda[200005], rs[200005];
void add(int x, int z, nod &y) {
nod p = new lnod;
p->nod = x;
p->nr = z;
p->next = y;
y = p;
}
void dfs(int x, int tata) {
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 N = 1e6 + 10;
int du[N];
vector<int> g[N];
struct Edge {
int from, to, nex;
} edge[N];
int h[N], idx, ans;
void add(int u, int v) {
Edge E = {u, v, h[u]};
edge[idx] = E;
h[u] = idx++;
}
int n;
void dfs(int u, int fa, int last) {
int j = 1;
for (int 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;
void traverse(int index, int parent_index, int parent_color,
vector<vector<int> > &G_to, vector<vector<int> > &G_idx,
vector<vector<int> > &ans) {
int c = 0;
for (int j = 0; j < G_to[index].size(); j++) {
if (parent_index == G_to[index][j... |
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 int N = 1e5 + 7;
long long int ord[3 * N] = {0}, visit[3 * N] = {0}, indeg[3 * N] = {0}, f = 0;
vector<long long int> ans[2 * N];
map<pair<long long int, long long int>, long long int> mp;
void dfs(vector<long long int> vec[], long long int a) {
long 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;
vector<int> day;
void dfs(int src, vector<vector<pair<int, int>>> &adj, int dont = 0,
int par = -1) {
int cur = 1;
for (auto &x : adj[src]) {
if (x.first == par) continue;
if (cur == dont) cur++;
day[x.second] = cur;
dfs(x.first, adj, cur, src);... |
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;
struct node {
int v, next, num;
} e[maxn << 1];
int n;
int cnt, head[maxn];
inline void add(int x, int y, int num) {
e[++cnt].v = y;
e[cnt].next = head[x];
head[x] = cnt;
e[cnt].num = num;
}
int du[maxn], ans;
vector<int> pth[maxn];
void... |
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;
map<pair<int, int>, int> edge_no;
int max_day = 0;
vector<int> build_schedule[200001];
void addEdge(vector<int> adj[], int u, int v) {
adj[u].push_back(v);
adj[v].push_back(u);
}
void DFSUtil(int u, vector<int> adj[], vector<bool> &visited, int day) {
visited[u] = tru... |
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;
std::vector<pair<int, int>> adj[200005];
std::vector<int> ans[200005];
void dfs(int n, int u, int v) {
int j = 1;
for (auto i : adj[n]) {
if (j == v) j++;
if (i.first != u) {
ans[j].push_back(i.second);
dfs(i.first, n, j);
j++;
}
}
retu... |
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> day;
void dfs(int v, int edge) {
for (int i = 0, d = 0; i < g[v].size(); ++i, ++d) {
if (g[v][i].second == edge) {
--d;
continue;
}
if (edge >= 0 && d == day[edge]) ++d;
day[g[v][i].second] = 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;
const int MAXN = 2e5;
int n;
vector<pair<int, int> > edge[MAXN + 5];
int ind[MAXN + 5];
vector<int> ans[MAXN + 5];
inline void OPEN(string s) {
freopen((s + ".in").c_str(), "r", stdin);
freopen((s + ".out").c_str(), "w", stdout);
}
void dfs(int u, int fa, int No) {
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;
const int MAXN = 200000;
int n, degree[MAXN];
vector<pair<int, int> > edges[MAXN];
vector<vector<int> > answers;
void dfs(int u, int parent, int day) {
int dayIdx = 0;
for (auto &edge : edges[u])
if (edge.first != parent) {
dayIdx += dayIdx == day;
if (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... | import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.util.HashMap;
import java.util.HashSet;
import java.util.ArrayList;
import java.util.List;
public class RoadImp{
private static void dfs(Node[] nodes, int n, HashMap<Integer, List<Integer>> res, int cur, int parentLabel, HashSet<Integer>... |
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>
using namespace std;
void boost() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
}
const int N = 2e5 + 5;
vector<int> day[N];
int l[N], r[N];
vector<pair<int, int> > adj[N];
void dfs(int u, int p) {
for (auto v : adj[u]) {
if (v.first != p) {
if (l[u] - 1 > 0) {
l[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;
const int lim = (int)(2e5 + 5);
int n, ans;
vector<pair<int, int> > v[lim];
vector<int> vec[lim];
void dfs(int x, int back, int y) {
int cn = 0;
for (int i = 0; i < (int)v[x].size(); i++) {
if (v[x][i].first == back) continue;
cn++;
if (cn == y) cn++;
an... |
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;
class graphal {
public:
int n, mx = 0;
vector<pair<int, int> > *ed;
vector<int> *ans;
graphal(int n) {
this->n = n;
ed = new vector<pair<int, int> >[n];
ans = new vector<int>[n + 1]();
}
~graphal() {
delete[] ed;
delete[] ans;
}
void add... |
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.*;
import java.math.BigInteger;
import java.util.Map.Entry;
import static java.lang.Math.*;
public class C extends PrintWriter {
class Road {
final int id, dst;
Road ret;
public Road(int id, int dst) {
this.id = id;
this.dst = dst;
}
}
void run() {
int n = next... |
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.util.stream.IntStream;
import java.util.Arrays;
import java.util.InputMismatchException;
import java.util.OptionalInt;
import java.util.ArrayList;
import java.io.OutputStreamWriter;
import java.util.NoSuchElementException;
... |
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>> G[200005];
vector<int> ans[200005];
void dfs(int n, int p, int dayp) {
int k = 1;
for (auto z : G[n]) {
if (z.first != p) {
if (k == dayp) k++;
ans[k].push_back(z.second);
dfs(z.first, n, k);
k++;
}
}
}
int main()... |
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) {
int cnt = 0;
for (int i = 0; i < adj[u].size(); i++) {
int v = adj[u][i].first, num = adj[u][i].second;
if (v == par) 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.*;
import java.util.*;
public class tests {
Fs scn = new Fs(System.in);
int n = scn.nextInt();
ArrayList<Pair>[] graph = new ArrayList[n];
boolean[] used = new boolean[n];
ArrayList<Pair2> pars = new ArrayList<>();
PrintWriter out = new PrintWriter(System.out);
public stat... |
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 = 9e5 + 6;
int ans[N];
vector<pair<int, int> > G[N];
vector<int> aa[N];
int n;
void dfs(int x, int f, int c) {
int st = 1;
for (auto k : G[x]) {
if (k.first != f) {
if (st == c) st++;
ans[k.second] = st;
dfs(k.first, x, st);
++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;
vector<int> v[200005];
vector<pair<int, int> > adj[200005];
int n, a, b, tot;
void dfs(int pos, int par, int used) {
int start = 0, baby, path, j;
for (int i = 0; i < adj[pos].size(); i++) {
baby = adj[pos][i].first;
path = adj[pos][i].second;
if (baby != pa... |
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 = 200010;
vector<int> g[MAX];
vector<int> ans[MAX];
int rootId = 0;
int rootPow = 0;
map<pair<int, int>, int> m;
int bad[MAX];
bool used[MAX];
void dfs(int v, int parent) {
if (used[v]) {
return;
}
used[v] = true;
int sz = (g[v].size());
int 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... | import java.io.*;
import java.math.BigInteger;
import java.util.*;
public class Main {
static int MOD = 1000000007;
// After writing solution, quick scan for:
// array out of bounds
// special cases e.g. n=1?
//
// Big numbers arithmetic bugs:
// int overflow
// sorting, or tak... |
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... | 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 {
public static void main(String[] args) {
InputReader inputReader = new InputReader(System.in);
PrintWriter printWriter =... |
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 + 5;
int n;
vector<pair<int, int>> g[maxn];
vector<int> ans[maxn];
int tot;
void dfs(int u, int fa, int ti) {
int cur = 0;
for (int i = 0; i < g[u].size(); i++) {
int v = g[u][i].first;
if (v == fa) continue;
if (++cur == ti) 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;
int n;
vector<int> res[200009];
vector<pair<int, int> > path[200009];
int last[200009];
queue<pair<int, int> > Q;
int main() {
cin >> n;
int a, b;
for (int i = 1; i < n; i++) {
scanf("%d%d", &a, &b);
path[a].push_back({b, i});
path[b].push_back({a, 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 N = 2e5 + 5;
vector<pair<long long, long long> > adj[N];
long long deg[N];
vector<long long> ans[N];
long long anss[N];
void dfs(long long u, long long p, long long prev) {
long long curr = 1;
for (auto k : adj[u]) {
if (k.first == p) continue;
if (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;
int gcd1(int a, int b) {
if (a == 0) return b;
return gcd1(b % a, a);
}
long long modx(long long base, long long ex) {
long long ans = 1LL, val = base;
while (ex > 0LL) {
if (ex & 1LL) ans = (ans * val) % 1000000007LL;
val = (val * val) % 1000000007LL;
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 MAXN = 200 * 1000 + 20;
bool mark[MAXN];
int n, x, y, g;
vector<pair<int, int>> gr[MAXN];
vector<int> ans[MAXN];
void dfs(int v, int u, int g) {
int h = 1;
for (pair<int, int> i : gr[v]) {
if (i.first != u) {
if (h == g) h++;
ans[h].push_back(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;
struct Edge {
int u, id;
Edge(int _u, int _id) {
u = _u;
id = _id;
}
};
vector<Edge> to[200005];
vector<int> ans[200005];
int skp[200005];
void dfs(int v, int p) {
int s = 1;
for (auto x : to[v]) {
if (x.u == p) continue;
s += (skp[v] == s);
an... |
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<int>> g;
vector<vector<pair<int, int>>> res;
void dfs(int node, int parent, int day_parent) {
int cnt = 0;
for (auto neighbor : g[node]) {
if (neighbor == parent) continue;
if (cnt == day_parent) cnt++;
if (cnt >= res.size()) res.push_back(vect... |
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> > > vec;
vector<vector<int> > ans;
void w(int v, int u, int k) {
int k2 = 1;
for (int i = 0; i < vec[v].size(); ++i) {
if (k2 == k) {
++k2;
}
if (vec[v][i].first == u) {
continue;
}
ans[k2].push_back(vec[v][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> > gr[200005];
vector<int> ans[200005];
int mx;
void dfs(int v, int pr, int day) {
int cd = 0;
for (auto &i : gr[v]) {
if (i.first == pr) continue;
while (++cd == day)
;
mx = max(mx, cd);
ans[cd].push_back(i.second);
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;
const int maxn = 3e5;
struct edge {
int to, id;
};
vector<edge> g[maxn];
vector<int> res[maxn];
void dfs(int v, int p, int f) {
int c = 0;
for (auto &e : g[v]) {
if (e.to == p) {
continue;
}
if (c == f) {
++c;
}
res[c].push_back(e.id);
... |
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 main() {
ios::sync_with_stdio(false);
cin.tie(0);
int n;
cin >> n;
vector<pair<int, int>> g[n + 1];
for (int i = 1; i < n; ++i) {
int u, v;
cin >> u >> v;
g[u].emplace_back(v, i);
g[v].emplace_back(u, i);
}
vector<vector<int>> ans(n + 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;
const int N = 2e5 + 5;
int n, res;
bool visited[N];
vector<pair<int, int> > adj[N];
vector<int> output[N];
void Input() {
cin >> n;
for (int i = 1; i <= n - 1; i++) {
int u, v;
cin >> u >> v;
adj[u].push_back(make_pair(v, i));
adj[v].push_back(make_pair(... |
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 read() {
int x = 0, fh = 1;
char ch = getchar();
while (!isdigit(ch)) {
if (ch == '-') fh = -1;
ch = getchar();
}
while (isdigit(ch)) {
x = (x << 1) + (x << 3) + ch - '0';
ch = getchar();
}
return x * fh;
}
const int maxn = 2e5 + 5;
... |
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;
int n;
vector<int> v[maxn];
vector<pair<int, int> > g[maxn];
set<int> s;
int color[maxn];
int ans;
void dfs(int v, int par = 0, int c = n) {
s.erase(c);
for (auto u : g[v])
if (u.first != par) {
color[u.second] = *s.begin();
s.er... |
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 = 1000 * 1000 * 2;
vector<pair<int, int>> gr[MAXN];
vector<int> ans[MAXN];
int check[MAXN];
void dfs(int, int);
int main() {
int n;
cin >> n;
for (int i = 1; i < n; i++) {
int u, v;
cin >> u >> v;
gr[u].push_back({v, i});
gr[v].push_back... |
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 = 2000 * 100 + 10;
int n, d;
vector<int> ans[N];
vector<pair<int, int>> adj[N];
void dfs(int root, int par = -1, int prv = 0) {
int nxt = 1;
for (auto u : adj[root])
if (u.first ^ par) {
if (nxt == prv) nxt++;
ans[nxt].push_back(u.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 maxn = 5e5 + 10;
int f[maxn], color[maxn];
vector<pair<int, int> > g[maxn];
vector<int> vmark[maxn];
int n;
void input() {
int z = 1;
cin >> n;
for (int i = 1; i < n; i++) {
int x, y;
cin >> x >> y;
g[x].push_back(make_pair(y, z));
g[y].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.util.*;
import java.io.*;
public class Main {
public static void main(String[] args) throws IOException {
FastScanner in = new FastScanner(System.in);
PrintWriter out = new PrintWriter(System.out);
new Main().run(in, out);
out.close();
}
// should be the max(ind... |
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>> g(nax + 1), days(nax + 1);
map<pair<long long int, long long int>, long long int> mp;
void dfs(long long int node, long long int pa, long long int k,
... |
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, id;
};
int n, a, b, r;
vector<edge> g[200009];
vector<int> ret[200009];
void dfs(int pos, int pre, int pid) {
int curid = 0;
for (edge e : g[pos]) {
if (e.to != pre) {
if (pid == curid) curid++;
ret[curid].push_back(e.id);
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... | 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.util.TreeMa... |
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, ans;
vector<int> v[200010], v_ans[200010];
map<pair<int, int>, int> m;
void dfs(int node, int par, int col) {
ans = max(ans, col);
int c = 1;
for (int i = 0; i < v[node].size(); i++) {
if (v[node][i] == par) continue;
if (c == col) c++;
dfs(v[node][... |
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 = 200009;
vector<pair<int, int> > g[N];
bool used[N];
vector<int> steps[N];
int u[N], v[N];
deque<int> q;
int ind[N];
bool usedTo[N];
int main() {
int n;
cin >> n;
for (int i = 1; i < n; ++i) {
scanf("%d%d", &u[i], &v[i]);
g[u[i]].push_back(make_pa... |
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>> G[200005];
vector<int> work[200005];
int mxday;
void dfs(int now, int pa, int pud) {
int nt = 1;
for (auto i : G[now]) {
if (i.first == pa) continue;
if (nt == pud) ++nt;
work[nt].push_back(i.second);
mxday = max(mxday, nt);
df... |
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 comment(linker, "/STACK:67108864")
using namespace std;
int n;
vector<int> vc[222222];
vector<int> cn[222222];
set<int, less<int> > vv[222222];
int dr[222222][2];
bool color[222222];
int mx = 0;
void dfs(int v) {
color[v] = true;
for (size_t i = 0; i < cn[v].size(); i++) {
int j... |
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, uint64_t>> ev;
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;
int n, k;
vector<pair<int, int> > g[200055];
vector<int> res[200055];
void dfs(int id, int par, int pre) {
int cnt = 0;
for (int i = 0; i < (int)g[id].size(); i++) {
int to = g[id][i].first, u = g[id][i].second;
if (to == par) 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;
vector<pair<long long, long long> > adj[200001];
bool visit[200001];
const int mod = 1000000007;
long long mod_mult(long long a, long long b) {
return ((a % mod) * (b % mod)) % mod;
}
long long mult(long long x, long long y) {
long long prod = 1;
while (y > 0) {
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 read() {
int ans = 0, f = 1;
char c = getchar();
while (c > '9' || c < '0') {
if (c == '-') f = -1;
c = getchar();
}
while (c >= '0' && c <= '9') {
ans = ans * 10 + c - '0';
c = getchar();
}
return ans * f;
}
const int N = 2e5 + 5;
int n, 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;
const int INF = 0x3f3f3f3f;
vector<pair<int, int>> v[200001];
vector<int> ans[200000];
int k = 0;
void dfs(int cur, int pre, int tm) {
int fixtime = 0;
for (auto i : v[cur]) {
int nxt = i.first;
if (nxt == pre) continue;
++fixtime;
if (fixtime == tm) ++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... | #include <bits/stdc++.h>
using namespace std;
long long gcd(long long a, long long b) { return b == 0 ? a : gcd(b, a % b); }
const int MAXN = 200000;
const int MAXM = MAXN - 1;
int n;
int deg[MAXN];
int ghead[MAXN], gprv[2 * MAXM], gnxt[2 * MAXM], gto[2 * MAXM];
int ihead[MAXN], iprv[2 * MAXM], inxt[2 * MAXM];
int qhea... |
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 double eps = 1e-7;
const double PI = acos(-1);
const long long INFF = (long long)1e18;
const int INF = (int)1e9;
const int mod = (int)1e9 + 7;
const int MAX = (int)2e5 + 7;
vector<pair<int, int> > edge[MAX];
vector<int> ans[MAX];
int ck[MAX];
void dfs(int u, int pa, 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... | import sys
mx=2*10**5+10
g=[[] for _ in range(mx)]
days=[[] for _ in range(mx)]
maxi=-1
n=int(raw_input())
t=sys.stdin.read().split("\n")
for i in range(n-1):
l=t[i]
u,v=map(int,l.split())
g[u].append((v,i+1))
g[v].append((u,i+1))
q=[(1,1,-1) for i in range(n)]
h,t=0,1
while h<t:
u,pu,c=q[h][0],q[h][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>
#pragma GCC optimize("Ofast")
using namespace std;
const int N = 200000;
vector<pair<int, int> > g[N];
int depth[N], pa[N];
void dfs(int u, int par, int d) {
pa[u] = par;
depth[u] = d;
for (pair<int, int>& i : g[u])
if (i.first != par) dfs(i.first, u, d + 1);
}
void solve() {
int 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;
const int N = 2e5 + 25;
int n, u, v, res;
vector<pair<int, int> > G[N];
vector<int> ans[N];
void dfs(int u, int p, int last) {
res = max(res, (int)G[u].size());
int cnt = 0;
for (int i = 0; i < G[u].size(); ++i) {
int v = G[u][i].first;
if (v == p) continue;
... |
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;
import java.util.StringTokenizer;
public class Main {
static MyScanner in;
static PrintWriter out;
// static Timer timer = new Timer();
public static void main(String[] args) throws IOException {
in = new MyScanner();
out = new PrintWrite... |
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 MyClass {
public static void main(String args[]) throws IOException{
Scan sc = new Scan();
int n = sc.scanInt();
Solution sol = new Solution(n);
for(int i = 1; i < n; ++i){
sol.addEdge(i, sc.scanInt(), sc.scanInt());
}
... |
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.HashMap;
import java.util.InputMismatchException;
import java.io.IOException;
import java.util.ArrayLis... |
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 + 7;
int n, cou;
int arr_p[N], color[N], a[N];
vector<int> ans[N];
pair<int, int> road[N];
vector<int> gr[N];
int f(int);
void dfs(int);
int main() {
ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
cin >> n;
int x, y;
for (int 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 Max = 2e5;
vector<pair<int, int> > G[Max + 5];
vector<int> ans[Max + 5];
int n;
int sum;
void dfs(int x, int fa, int t) {
int k = 0;
for (int i = 0; i < G[x].size(); i++) {
if (G[x][i].first == fa) continue;
k++;
if (k == t) k++;
dfs(G[x][i].fi... |
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 N = 1000001;
using namespace std;
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
cout.tie(NULL);
int n;
cin >> n;
vector<pair<int, int>> a[n + 1];
int ans = 0;
for (int i = 0; i < n - 1; i++) {
int u, v;
cin >> u >> v;
a[u].push_back(make_pai... |
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 dp[N];
int n, a, b;
vector<pair<int, int> > g[N];
vector<int> ans[N];
void dfs(int v, int p = -1, int d = -1) {
int cnt = 0;
for (int(i) = (0); i < (g[v].size()); ++(i)) {
int to = g[v][i].first;
if (to == p) continue;
if (cnt == 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;
const int maxn = 200010;
vector<pair<int, int> > G[maxn];
vector<int> res[maxn];
int cnt = 0;
void dfs(int v, int p, int x) {
int vl = 1;
if (vl == x) {
++vl;
}
for (int i = 0; i < (int)G[v].size(); ++i) {
int u = G[v][i].first;
if (u != p) {
res[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;
int n;
const int maxn = 2e5 + 10;
struct node {
int next, v, num;
} e[maxn * 2];
int tot, head[maxn];
vector<int> ans[maxn];
int du[maxn];
void add(int x, int y, int num) {
e[++tot].v = y;
e[tot].next = head[x];
e[tot].num = num;
head[x] = tot;
}
void dfs(int u, 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 N = 200010;
vector<pair<int, int> > g[N];
vector<int> r[N];
int d[N];
void dfs(int u, int fa, int x) {
int c = 0;
for (int i = 0; i < g[u].size(); i++) {
int v = g[u][i].first;
if (v == fa) continue;
c++;
if (c == x) c++;
r[c].push_back(g[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>
void quit();
using namespace std;
const long double PI = acos(-1);
const long double EPS = 1e-10;
double __t;
int n;
int d[200100];
vector<pair<int, int> > g[200100];
int day[200100];
vector<int> ans[200100];
void dfs(int v, int skip = -1) {
int cnt = 0;
for (auto &e : g[v]) {
if (day[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;
int gcd1(int a, int b) {
if (a == 0) return b;
return gcd1(b % a, a);
}
long long modx(long long base, long long ex) {
long long ans = 1LL, val = base;
while (ex > 0LL) {
if (ex & 1LL) ans = (ans * val) % 1000000007LL;
val = (val * val) % 1000000007LL;
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;
vector<vector<pair<int, int> > > graph;
vector<vector<int> > ans;
int maxDeg;
int root;
void dfs(int u, int p, int day) {
int add = 1;
for (int i = 0; i < graph[u].size(); i++) {
int v = graph[u][i].first;
int id = graph[u][i].second;
if (v == p) continue;
... |
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<int> > edge(200005, vector<int>());
vector<vector<int> > ans(200005, vector<int>());
vector<bool> visited(200005, false);
map<pair<int, int>, int> mp;
void dfs(int i, int num) {
visited[i] = true;
int it = 1;
for (auto ed : edge[i]) {
if (!visited[ed... |
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, in[200005], day[200005];
vector<int> ans[200005];
vector<pair<int, int> > adj[200005];
void dfs(int now, int pre) {
map<int, int> M;
for (int i = 0; i < adj[now].size(); i++) {
int road = adj[now][i].second;
if (day[road]) {
M[day[road]] = 1e9;
... |
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;
vector<pair<int, int> > adj[maxn];
vector<int> v[maxn];
bool visit[maxn];
int n, m, u, v1, ans;
long long modx(long long base, long long ex) {
long long ans = 1LL, val = base;
while (ex > 0LL) {
if (ex & 1LL) ans = (ans * val) % 1000000007... |
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