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题意：

给定n个点m条边的无向图

问：

从任意点出发任意走d步，从不经过某个点的概率

dp[i][j]表示从不经过i点的前提下，走了d步到达j点的概率。

#include <iostream>
#include <cstdio>
#include <string.h>
#include <queue>
#include <vector>
#include <algorithm>
#include <set>

using namespace std;
#define N 55
#define eps (1e-8)
struct Edge{
    int from, to, nex;
}edge[N*N*N];
int head[N], edgenum;
void init(){memset(head, -1, sizeof head); edgenum = 0;}
void add(int u, int v){
    Edge E = {u,v,head[u]};
    edge[edgenum] = E;
    head[u] = edgenum++;
}

double dp[2][N][N], siz[N];
int n, m, d;
void solve(){
    scanf("%d %d %d", &n, &m, &d);
    init();
    int u, v;
    for(int i = 1; i <= n; i++)siz[i] = 0.0;
    while(m--){
        scanf("%d %d",&u,&v);
        add(u,v); add(v,u);
        siz[u] += 1.0; siz[v] += 1.0;
    }
    int cur = 0;
    for(int i = 1; i <= n; i++)
        for(int j = 1; j <= n; j++)
            if(i == j)
                dp[cur][i][j] = 0;
            else 
                dp[cur][i][j] = 1.0/(double)n;
    while(d--) {
        cur ^= 1;
        for(int i = 1; i <= n; i++)
            for(int j = 1; j <= n; j++)
                dp[cur][i][j] = 0;
        for(int i = 1; i <= n; i++)
            for(int j = 1; j <= n; j++)
                if(i!=j)
                {
                    double Size = siz[j];
                    if(Size < eps)continue;
                    for(int k = head[j]; ~k; k = edge[k].nex) if(edge[k].to != i)
                    {
                        dp[cur][i][edge[k].to] += dp[cur^1][i][j] / Size;
                    }
                }
    }    
    for(int i = 1; i <= n; i++) {
        double ans = 0;
        for(int j = 1; j <= n; j++)
            ans += dp[cur][i][j];
        printf("%.10f\n", ans);
    }
}
int main(){
    int T;scanf("%d",&T);
    while(T--){
        solve();
    }
    return 0;
}
/*
3 3 2  
1 2 
2 3 
3 1 


*/

HDU 5001 Walk 求从任意点出发任意走不经过某个点的概率 概率dp 2014 ACM/ICPC Asia Regional Anshan Online

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原文地址：http://blog.csdn.net/qq574857122/article/details/39253169