HDU Walk （概率DP）

HDU Walk （概率DP）：http://acm.hdu.edu.cn/showproblem.php?pid=5001

题面描述：

Time Limit: 30000/15000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 996    Accepted Submission(s): 634
Special Judge

Problem Description

I used to think I could be anything, but now I know that I couldn't do anything. So I started traveling.

The nation looks like a connected bidirectional graph, and I am randomly walking on it. It means when I am at node i, I will travel to an adjacent node with the same probability in the next step. I will pick up the start node randomly (each node in the graph has the same probability.), and travel for d steps, noting that I may go through some nodes multiple times.

If I miss some sights at a node, it will make me unhappy. So I wonder for each node, what is the probability that my path doesn't contain it.

 

Input

The first line contains an integer T, denoting the number of the test cases.

For each test case, the first line contains 3 integers n, m and d, denoting the number of vertices, the number of edges and the number of steps respectively. Then m lines follows, each containing two integers a and b, denoting there is an edge between node a and node b.

T<=20, n<=50, n-1<=m<=n*(n-1)/2, 1<=d<=10000. There is no self-loops or multiple edges in the graph, and the graph is connected. The nodes are indexed from 1.

 

Output

For each test cases, output n lines, the i-th line containing the desired probability for the i-th node.

Your answer will be accepted if its absolute error doesn't exceed 1e-5.

 

Sample Input

   
   
   
   

    
    
    
    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
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    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
   
   
   
   
   
   
   
   


   
   
   
   

题目大意：

在一个n给节点，m条边的无向图中，以每个顶点为起点的概率相等，求经过d步之后，不经过该点的概率为多少。

题目解析：

不经过这个节点的概率等于去掉该节点的图中走了d步到其他顶点的和。

特殊的枚举方式，对于每个i，因为不让他到达这个点i，所以不达到点i的概率就是到达其他顶点的概率和，不求该点即可。

代码实现：

#include <iostream>
#include <cstdio>
#include <vector>
#include <cstring>

using namespace std;
#define maxn 55
#define maxstep 10005
vector<int> V[maxn];
double dp[maxstep][maxn];

int main()
{
    int T;
    int n,m,d;
    scanf("%d",&T);
    while(T--)
    {
        int a,b;
        scanf("%d%d%d",&n,&m,&d);
        for( int i=1; i<=n; i++)
        {
            V[i].clear();
        }
        for( int i=1; i<=m; i++)
        {
            scanf("%d%d",&a,&b);
            V[a].push_back(b);
            V[b].push_back(a);
        }
        for(int i=1; i<=n; i++)
        {
            dp[0][i]=1.0/n;
        }
        for(int i=1; i<=n; i++)
        {
            //memset(dp,0,sizeof(dp));
            for(int j=1; j<=d; j++)
            {
                for(int k=1; k<=n; k++)
                {
                    if(k==i)
                        continue;
                    dp[j][k]=0;
                    for(int q=0; q<V[k].size(); q++)
                    {
                        if(V[k][q]==i) continue;
                        else dp[j][k]+=dp[j-1][V[k][q]]*(1.0/V[k].size());
                    }
                }
            }
            double ans=0;
            for(int j=1; j<=n; j++)
            {
                if(j==i)
                    continue;
                else ans+=dp[d][j];
            }
                printf("%.10lf\n",ans);
        }
    }
    return 0;
}