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hdu5001－－－Walk

2024-10-19 02:16:39   212人阅读

Walk

Time Limit: 30000/15000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 651    Accepted Submission(s): 402
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
2014 ACM/ICPC Asia Regional Anshan Online
 

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此题的状态是想到了，但是含义没搞对，本来想把到达某个点的概率求出来，然后用１去减就是答案，但是是不对的，因为那部分概率是有牵连的

设ｄｐ［ｉ］［ｊ］　表示走了ｊ步，目前在节点ｉ，且路径中不含节点ｕ的概率
最后只要累计出节点ｕ以外其他点的概率和就行

/*************************************************************************
    > File Name: hdu4945.cpp
    > Author: ALex
    > Mail: 405045132@qq.com 
    > Created Time: 2014年12月30日 星期二 19时34分57秒
 ************************************************************************/

#include <map>
#include <set>
#include <queue>
#include <stack>
#include <vector>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>

using namespace std;

const int D = 10010;
const int N = 55;
double dp[N][D];
int n, m, d;
vector <int> edge[N];

void DP(int u)
{
	for (int i = 1; i <= n; ++i)
	{
		for (int j = 0; j <= d; ++j)
		{
			dp[i][j] = 0;
		}
	}
	for (int i = 1; i <= n; ++i)
	{
		dp[i][0] = (1.0 / (double)n);
	}
	for (int i = 0; i < d; ++i)
	{
		for (int j = 1; j <= n; ++j)
		{
			if (j == u)
			{
				continue;
			}
			int cnt = edge[j].size();
			for (int k = 0; k < cnt; ++k)
			{
				int v = edge[j][k];
				if (v == u)
				{
					continue;
				}
				dp[j][i + 1] += dp[v][i] * (1.0 / cnt);
			}
		}
	}
	double ans = 0;
	for (int i = 1; i <= n; ++i)
	{
		if (i == u)
		{
			continue;
		}
		ans += dp[i][d];
	}
	printf("%.10f\n", ans);
}

int main()
{
	int t;
	int u, v;
	scanf("%d", &t);
	while (t--)
	{
		scanf("%d%d%d", &n, &m, &d);
		for (int i = 1; i <= n; ++i)
		{
			edge[i].clear();
		}
		for (int i = 1; i <= m; ++i)
		{
			scanf("%d%d", &u, &v);
			edge[u].push_back(v);
			edge[v].push_back(u);
		}
		for (int i = 1; i <= n; ++i)
		{
			DP(i);
		}
	}
	return 0;
}



hdu5001－－－Walk


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