P4316 绿豆蛙的归宿 期望DP

Posted 2022-09-13 santiego

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
\[ f[v]+=\fracw+f[u]edg[v] \]
\(edg[v]\)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;