cogs 1065 绿豆蛙的归宿 图上概率dp

Posted 2020-09-28 Loser Of Life

http://cogs.pro/cogs/problem/problem.php?pid=1065

题目大意就是，给一个DAG，求出期望路径长度。

那么我们计算一下每个点到终点步数的期望值，公式为f[i]=sigma{(f[j]+weight[j][i])/exit[j]},其中j为所有与i相连的点。

建反图倒推即可。f[1]即为所求。

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<algorithm>
 4 #include<cstring>
 5 #include<queue>
 6 using namespace std;
 7 const int maxn=100005;
 8 struct node
 9 {
10     int from,to,next,weight;
11 }edge[maxn<<1];
12 int head[maxn],chudu[maxn],n,m,tot,tmp[maxn];
13 double f[maxn];
14 void addedge(int u,int v,int w)
15 {
16     edge[++tot]=(node){u,v,head[u],w};head[u]=tot;
17 }
18 int haha()
19 {
20     freopen("ldfrog.in","r",stdin);
21     freopen("ldfrog.out","w",stdout);
22     scanf("%d%d",&n,&m);
23     for(int i=1;i<=m;i++)
24     {
25         int x,y,z;scanf("%d%d%d",&x,&y,&z);
26         addedge(y,x,z);
27         tmp[x]++;chudu[x]++;
28     }
29     queue<int>q;q.push(n);
30     while(!q.empty())
31     {
32         int t=q.front();q.pop();
33         for(int i=head[t];i;i=edge[i].next)
34         {
35             int v=edge[i].to;
36             tmp[v]--;
37             f[v]+=(f[t]+edge[i].weight)/chudu[v];
38             if(!tmp[v])q.push(v);
39         }
40     }
41     printf("%0.2lf",f[1]);
42 }
43 int sb=haha();
44 int main(){;}