HDOJ 5001 Walk

概率DP

dp[j][d] 表示不经过i点走d步到j的概率, dp[j][d]=sigma ( dp[k][d-1] * Probability )

ans = sigma ( dp[j][D] )

Walk

Time Limit: 30000/15000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)

Total Submission(s): 401    Accepted Submission(s): 261

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

 

Source

field=problem&key=2014+ACM%2FICPC+Asia+Regional+Anshan+Online&source=1&searchmode=source" style="color:rgb(26,92,200); text-decoration:none">2014 ACM/ICPC Asia Regional Anshan Online

 

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

using namespace std;

const int maxn=10010;

int n,m,D;
vector<int> g[maxn];
double dp[55][maxn];

int main()
{
	int T_T;
	scanf("%d",&T_T);
	while(T_T--)
	{
		scanf("%d%d%d",&n,&m,&D);
		for(int i=0;i<=n+1;i++) g[i].clear();
		while(m--)
		{
			int a,b;
			scanf("%d%d",&a,&b);
			g[a].push_back(b);
			g[b].push_back(a);
		}
		for(int i=1;i<=n;i++)
		{
			memset(dp,0,sizeof(dp));
			for(int j=1;j<=n;j++)
			{
				if(i!=j) dp[j][0]=1.0/n;
			}

			for(int d=1;d<=D;d++)
			{
				for(int j=1;j<=n;j++)
				{
					if(j==i) continue;
					for(int k=0,sz=g[j].size();k<sz;k++)
					{
						int v=g[j][k];
						if(v!=i) dp[j][d]+=dp[v][d-1]*(1./g[v].size());
					}
				}
			}

			double ans=0.0;
			for(int j=1;j<=n;j++)
			{
				if(i!=j) ans+=dp[j][D];
			}
			printf("%.10lf\n",ans);
		}
	}
	return 0;
}