既然等概率又收敛，多跑一百次就是答案了。
#include <bits/stdc++.h> using namespace std; int n,m,k; const int N=100+10; double f[N][N]; vector<int> g[N]; void solve() { int n,m; while(~scanf("%d%d",&n,&m)) { for (int i = 1; i <= n; i++) g[i].clear(),g[i].push_back(i); for (int i = 1; i <= m; i++) { int a, b; scanf("%d%d", &a, &b); g[a].push_back(b); } memset(f,0,sizeof f); f[0][1] = 1; for (int i = 1; i <= 100; i++) { for (int j = 1; j <= n; j++) { for (auto k:g[j]) { f[i][k] += f[i - 1][j] * (1.0 / (g[j].size())); } } } double res = 0; for (int i = 1; i <= n; i++) res = max(f[100][i], res); cout << res << '\n'; } } signed main() { solve(); }