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每次都去掉一个点求出到达 其他点的概率就是不能到达这个点的概率。

Walk

2
5 10 100
1 2
2 3
3 4
4 5
1 5
2 4
3 5
2 5
1 4
1 3
10 10 10
1 2
2 3
3 4
4 5
5 6
6 7
7 8
8 9
9 10
4 9

0.0000000000
0.0000000000
0.0000000000
0.0000000000
0.0000000000
0.6993317967
0.5864284952
0.4440860821
0.2275896991
0.4294074591
0.4851048742
0.4896018842
0.4525044250
0.3406567483
0.6421630037

#include <algorithm>
#include <iostream>
#include <stdlib.h>
#include <string.h>
#include <iomanip>
#include <stdio.h>
#include <string>
#include <queue>
#include <cmath>
#include <stack>
#include <map>
#include <set>
#define eps 1e-12
///#define M 1000100
#define LL __int64
///#define LL long long
///#define INF 0x7ffffff
#define INF 0x3f3f3f3f
#define PI 3.1415926535898
#define zero(x) ((fabs(x)<eps)?0:x)

using namespace std;

const int maxn = 10010;

double dp[55][maxn];
double ans[55];
vector<int>g[55];

int main()
{
    int T;
    cin >>T;
    while(T--)
    {
        int n, m, d;
        scanf("%d %d %d",&n, &m, &d);
        for(int i = 1; i <= n; i++) g[i].clear();
        int x, y;
        for(int i = 1; i <= m; i++)
        {
            scanf("%d %d",&x, &y);
            g[x].push_back(y);
            g[y].push_back(x);
        }
        for(int p = 1; p <= n; p++)
        {
            memset(dp, 0, sizeof(dp));
            for(int i = 1; i <= n; i++) dp[i][0] = 1.0/n;
            for(int i = 0; i < d; i++)
            {
                for(int j = 1; j <= n; j++)
                {
                    if(j == p) continue;
                    int N = g[j].size();
                    for(int k = 0; k < N; k++)
                    {
                        int u = g[j][k];
                        dp[u][i+1] += dp[j][i]*(1.0/N);
                    }
                }
            }
            ans[p] = 0.0;
            for(int i = 1; i <= n; i++)
            {
                if(i == p) continue;
                ans[p] += dp[i][d];
            }
            printf("%.9lf\n",ans[p]);
        }
    }
    return 0;
}

HDU 5001 Walk (暴力)

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原文地址：http://blog.csdn.net/xu12110501127/article/details/39254403