直接反圖+拓排然後跑一遍概率DP

#include <queue>
#include <cmath>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#define db double
#define ll long long
#define rep(i,x,y) for(ll i=(x);i<=(y);i++)
#define red(i,x,y) for(ll i=(x);i>=(y);i--)
using namespace std;

const ll N=1e5+5;

db f[N],edge[N<<1];
ll n,m,a[N],b[N];
ll cnt,to[N<<1],nxt[N<<1],head[N];

inline ll read() {
	ll x=0;char ch=getchar();bool f=0;
	while(ch>'9'||ch<'0'){if(ch=='-')f=1;ch=getchar();}
	while(ch>='0'&&ch<='9'){x=(x<<1)+(x<<3)+ch-'0';ch=getchar();}
	return f?-x:x;
}

void ins(ll x,ll y,db z) {
	to[++cnt]=y;edge[cnt]=z;nxt[cnt]=head[x];head[x]=cnt;
}

void dp() {
	queue<ll>q;
	rep(i,1,n) if(!a[i]) q.push(i);
	
	while(q.size()) {
		ll x=q.front();q.pop();
		
		for(ll i=head[x];i;i=nxt[i]) {
			ll y=to[i];db z=edge[i];
			
			f[y]+=(f[x]+z)/b[y];
			
			--a[y];
			if(!a[y]) q.push(y);
		}
	}
}

int main() {
	n=read(),m=read();
	
	rep(i,1,m) {
		ll x=read(),y=read(),z=read();
		ins(y,x,z*1.0);a[x]++,b[x]++;
	}
	
	dp();
	
	printf("%.2lf",f[1]);

	return 0;
}