【BZOJ3036】绿豆蛙的归宿 概率与期望

 1 #include <iostream>
 2 #include <cstdio>
 3 using namespace std;
 4 #define N 100010
 5 #define M 200020
 6 struct E
 7 {
 8     int next,to,v;
 9 }e[M];
10 double f[N];
11 int head[N],vis[N],oute[N];
12 int n,m,cnt;
13 inline int read()
14 {
15     int ans=0,f=1;
16     char c;
17     while (!isdigit(c=getchar())) {if (c==‘-‘) f=-1;}
18     ans=c-‘0‘;
19     while (isdigit(c=getchar())) ans=ans*10+c-‘0‘;
20     return ans*f;
21 }
22 inline void insert(int x,int y,int w)
23 {cnt++; e[cnt].next=head[x]; head[x]=cnt; e[cnt].to=y; e[cnt].v=w;}
24 void dfs(int x)
25 {
26     if (!vis[x]) vis[x]=1;
27     else return;
28     for (int i=head[x];i;i=e[i].next)
29     {
30         dfs(e[i].to);
31         f[x]+=e[i].v+f[e[i].to];
32     }
33     if (oute[x]) f[x]/=oute[x];
34 }
35 int main()
36 {
37     n=read(); m=read();
38     for (int i=1;i<=m;i++)
39     {
40         int x,y,w;
41         x=read(); y=read(); w=read();
42     //    printf("%d %d %d ",x,y,w); 
43         insert(x,y,w);
44         oute[x]++;
45     }
46     dfs(1);
47     printf("%.2lf\n",f[1]);
48     return 0;
49 }