題目鏈接：https://nanti.jisuanke.com/t/41301
設day[i]表示以i爲起點到終點的期望天數。
設ans[i]表示以i爲起點到終點的期望得分。
顯然day特別好求，然後考慮如何借用day去得到ans，這也沒什麼難的，按求day的那種感覺就能寫出方程來，然後驗證一下樣例就可以了。
#include <bits/stdc++.h>
#define rep(i ,a, b) for(int i = (a); i <= (b); i++)
#define per(i, a, b) for(int i = (a); i >= (b); i--)
#define ll long long
#define pb push_back
using namespace std;
const int N = 1e5+1000;
int n,m;
double day[N],ans[N];
bool vis[N];
vector<int>nxt[N];
void dfs(int u) { //求day
if(vis[u]) return;
vis[u] = 1;
int deg = nxt[u].size()+1;
if(deg==1) {
day[u] = 0;
return;
}
double sum = 0;
for(auto v:nxt[u]) {
dfs(v);
sum += day[v]/deg;
}
day[u] = (sum+1)*deg/(deg-1);
}
void dfss(int u) { //求ans
if(vis[u]) return;
vis[u] = 1;
int deg = nxt[u].size()+1;
if(deg==1) {
ans[u] = 0;
return;
}
double sum = (day[u]+1)/deg;
for(auto v:nxt[u]) {
dfss(v);
sum += (ans[v]+day[v]+1)/deg;
}
ans[u] = sum*deg/(deg-1);
}
int main() {
int T;
scanf("%d",&T);
while(T--) {
scanf("%d%d",&n,&m);
rep(i, 1, n) {
nxt[i].clear();
vis[i] = 0;
}
rep(i, 1, m) {
int u, v;
scanf("%d%d",&u,&v);
nxt[u].pb(v);
}
dfs(1);
rep(i, 1, n) vis[i] = 0;
dfss(1);
printf("%.2f\n",ans[1]);
}
return 0;
}