居然不知道acyclic是没有环的意思GG。
如果是拓扑图的话, 用bitset暴力更新就完事了。
#include<bits/stdc++.h> #define LL long long #define LD long double #define ull unsigned long long #define fi first #define se second #define mk make_pair #define PLL pair<LL, LL> #define PLI pair<LL, int> #define PII pair<int, int> #define SZ(x) ((int)x.size()) #define ALL(x) (x).begin(), (x).end() #define fio ios::sync_with_stdio(false); cin.tie(0); using namespace std; const int N = 2e4 + 2; const int inf = 0x3f3f3f3f; const LL INF = 0x3f3f3f3f3f3f3f3f; const int mod = 1e9 + 7; const double eps = 1e-8; const double PI = acos(-1); template<class T, class S> inline void add(T &a, S b) {a += b; if(a >= mod) a -= mod;} template<class T, class S> inline void sub(T &a, S b) {a -= b; if(a < 0) a += mod;} template<class T, class S> inline bool chkmax(T &a, S b) {return a < b ? a = b, true : false;} template<class T, class S> inline bool chkmin(T &a, S b) {return a > b ? a = b, true : false;} mt19937 rng(chrono::steady_clock::now().time_since_epoch().count()); int n, m, ans, deg[N]; vector<int> G[N]; bitset<20000> bit[20000]; bool vis[N]; void dfs(int u) { if(vis[u]) return; vis[u] = true; for(auto &v : G[u]) { dfs(v); bit[u] |= bit[v]; } for(auto &v : G[u]) { if(bit[u][v]) { ans++; } else { bit[u][v] = 1; } } } int main() { int T; scanf("%d", &T); while(T--) { scanf("%d%d", &n, &m); for(int i = 0; i < n; i++) { G[i].clear(); bit[i].reset(); deg[i] = 0; vis[i] = 0; } for(int i = 1; i <= m; i++) { int u, v; scanf("%d%d", &u, &v); u--; v--; G[u].push_back(v); deg[v]++; } ans = 0; for(int i = 0; i < n; i++) { if(!deg[i]) { dfs(i); } } printf("%d ", ans); } return 0; } /* */