图片说明
图片说明

需要使用记忆话，一个点的期望只去计算一次。

#include <stdio.h>
#include <cstring>
#include <algorithm>
#include <vector>
#include <stack>
#include <queue>
#include <iostream>
#include <map>

#define go(i, l, r) for(int i = (l), i##end = (int)(r); i <= i##end; ++i)
#define god(i, r, l) for(int i = (r), i##end = (int)(l); i >= i##end; --i)
#define ios ios_base::sync_with_stdio(0),cin.tie(0),cout.tie(0)
#define debug_in  freopen("in.txt","r",stdin)
#define debug_out freopen("out.txt","w",stdout);
#define pb push_back
#define all(x) x.begin(),x.end()
#define fs first
#define sc second
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<ll,ll> pii;
const ll maxn = 1e5+10;
const ll maxM = 1e6+10;
const ll inf_int = 1e8;
const ll inf_ll = 1e17;

template<class T>void read(T &x){
    T s=0,w=1;
    char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')w=-1;ch=getchar();}
    while(ch>='0'&&ch<='9') s=s*10+ch-'0',ch=getchar();
    x = s*w;
}
template<class H, class... T> void read(H& h, T&... t) {
    read(h);
    read(t...);
}

void pt(){ cout<<'\n';}
template<class H, class ... T> void pt(H h,T... t){ cout<<" "<<h; pt(t...);}

//--------------------------------------------
int N,M;
int h[maxn],w[maxn*2],e[maxn*2],ne[maxn*2],idx = 1;
int sz[maxn];
double rec[maxn];
void add(int a,int b,int c){
    e[++idx] = b;
    w[idx] = c;
    ne[idx] = h[a];
    h[a] = idx;
}
double E[maxn];
void dfs(int u){
   if(rec[u]>=0) return ;//记录一下，算过的期望就不要再算，不然算重，多加了。
   for(int i = h[u];i;i = ne[i]){
       int v = e[i];
       dfs(v);
       E[u] += 1.0 * (E[v] + w[i])/(double)sz[u];
   }
   rec[u] = E[u];
}
int main() {
//    debug_in;
//    debug_out;

    read(N,M);
    for(int i = 1;i<=M;i++){
        int x,y,z;read(x,y,z);
        add(x,y,z);
        sz[x]++;
    }
    memset(rec,-1,sizeof rec);
    dfs(1);
    printf("%.2f\n",E[1]);



    return 0;
}