이 제출은 이전 버전의 oj.uz에서 채점하였습니다. 현재는 제출 당시와는 다른 서버에서 채점을 하기 때문에, 다시 제출하면 결과가 달라질 수도 있습니다.
#include<bits/stdc++.h>
#include<fstream>
//#pragma GCC optimize("O3")
//#pragma GCC target("avx2")
using namespace std;
#define sz(a) (int)a.size()
#define ALL(v) v.begin(), v.end()
#define ALLR(v) v.rbegin(), v.rend()
#define ll long long
#define pb push_back
#define forr(i, a, b) for(int i = a; i < b; i++)
#define dorr(i, a, b) for(int i = a; i >= b; i--)
#define ld long double
#define vt vector
#include<fstream>
#define fi first
#define se second
#define pll pair<ll, ll>
#define pii pair<int, int>
#define mpp make_pair
const ld PI = 3.14159265359, prec = 1e-9;;
//using u128 = __uint128_t;
const int dx[4] = {1, 0, -1, 0};
const int dy[4] = {0, -1, 0, 1};
const ll mod = 1e9 + 7, pr = 31;
const int mxn = 2e5 + 5, mxd = 250 * 250, sq = 100, mxv = 5e4 + 1, mxq = 2e6 + 5, T = 1000;
//const int base = (1 <<18);
const ll inf = 2e9 + 5, neg = -69420, inf2 = 1e14;
mt19937_64 rng(chrono::steady_clock::now().time_since_epoch().count());
#include <vector>
#include <cassert>
#include <cstdio>
//#include "september.h"
#include <vector>
int n, m;
bool vis[mxn + 1];
vt<int>adj[mxn + 1];
int cnt = 0;
void dfs(int s){
cnt += m; vis[s] = 1;
for(auto i: adj[s]){
if(!vis[i]){
dfs(i);
}
}
}
int solve(int N, int M, std::vector<int> F, std::vector<std::vector<int>> S) {
n = N; m = M;
cnt = 0;
for(int i = 0; i < N; i++){
adj[i].clear(); vis[i] = 0;
}
for(int i = 1; i < N; i++){
adj[F[i]].pb(i);
}
int ans = 0;
for(int i = 0; i < N - 1; i++){
for(int j = 0; j < M; j++){
if(!vis[S[j][i]])dfs(S[j][i]);
}
cnt -= M;
if(cnt == 0)ans++;
}
return(ans);
}
/*
#include <vector>
void taskcase() {
int N, M;
assert(2 == scanf("%d%d", &N, &M));
std::vector<int> F(N);
F[0] = -1;
for (int i = 1; i < N; ++i)
assert(1 == scanf("%d", &F[i]));
std::vector<std::vector<int>> S(M, std::vector<int>(N - 1));
for (int i = 0; i < M; ++i)
for (int j = 0; j < N - 1; ++j)
assert(1 == scanf("%d", &S[i][j]));
printf("%d\n", solve(N, M, F, S));
}
int main() {
int T;
assert(1 == scanf("%d", &T));
while(T--) taskcase();
return 0;
}
*/
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