题解：

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();
}