P4316 绿豆蛙的归宿 期望DP
P4316 绿豆蛙的归宿 期望DP
DAG上,每条边有边权,走向相连每条路的概率相等,问从起点到终点所经过的路径总长度期望
因为发现终点走到终点期望为0,定义\(f[i]\)从终点走到\(i\)所经过的路径总长度期望,所以\(f[n]=0\)。于是建反图,拓扑转移
\[f[v]+=\frac{w+f[u]}{edg[v]}
\]
\(edg[v]\)表示从节点\(v\)有\(edg[v]\)条路可走
#include <cstdio>
#include <queue>
#define MAXN 100010
using namespace std;
int head[MAXN],nxt[MAXN*2],vv[MAXN*2],ww[MAXN*2],tot;
inline void add_edge(const int &u, const int &v, const int &w){
vv[++tot]=v;
ww[tot]=w;
nxt[tot]=head[u];
head[u]=tot;
}
int read(){
char ch;int s=0;
ch = getchar();
while(ch<'0'||ch>'9') ch=getchar();
while(ch>='0'&&ch<='9') s=s*10+ch-'0',ch=getchar();
return s;
}
int rdu[MAXN],edg[MAXN],n,m;
double f[MAXN];
queue <int> q;
int main(){
n=read(),m=read();
for(int i=1;i<=m;++i){
int u,v,w;
u=read(),v=read(),w=read();
add_edge(v, u, w);
++edg[u];++rdu[u];
}
q.push(n);
while(!q.empty()){
int u=q.front();q.pop();
for(int i=head[u];i;i=nxt[i]){
int v=vv[i],w=ww[i];
f[v]+=(w+f[u])/edg[v];
--rdu[v];
if(rdu[v]==0) q.push(v);
}
}
printf("%.2f\n", f[1]);
return 0;
}