hdu 5001 概率dp

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

25 10 1001 22 33 44 51 52 43 52 51 41 310 10 101 22 33 44 55 66 77 88 99 104 9

Sample Output

0.00000000000.00000000000.00000000000.00000000000.00000000000.69933179670.58642849520.44408608210.22758969910.42940745910.48510487420.48960188420.45250442500.3406567483
0.6421630037
起点随意，往哪走随意（概率相同，与面临选择的个数有关），问走d步走不到每个点的概率
思路：
概率dp，dp[i][j]表示走了i步第一次到达j的概率
ac代码：
#include <iostream>#include <cstdio>#include <cstring>#include <vector>using namespace std;vector<int> mp[55];double dp[10005][55];int main(){    int n,m,d;    int t;    cin>>t;    while(t--)    {        cin>>n>>m>>d;        for(int i=0;i<=n;i++)        mp[i].clear();        int a,b;        for(int i=1;i<=m;i++)        {            scanf("%d%d",&a,&b);            mp[a].push_back(b);            mp[b].push_back(a);        }        for(int i=1;i<=n;i++)        {            double ans=0;            memset(dp,0,sizeof(dp));            for(int j=1;j<=n;j++)            dp[0][j]=1.0/n;            for(int j=0;j<d;j++)            {                for(int k=1;k<=n;k++)                {                    if(k==i)//关键判断，如果走到i的话就不再往后扩展，保证后面的都是第一一次到i                    continue;                    for(int h=0;h<mp[k].size();h++)                    {                        dp[j+1][mp[k][h]]+=(dp[j][k]/mp[k].size());                    }                }            }            for(int j=0;j<=d;j++)//求一下和就可以了                ans+=dp[j][i];            printf("%.6lf\n",1.0-ans);        }    }    return 0;}