[HDU5001]Walk

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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题意：n个点，m条边的无向图，你随机的从一个点开始，走k步，问你对于每一个点，它不被经过的概率是多少。

我们这样考虑，一个点不被经过的概率就是1-这个点经过的概率，所以就设f[i][j]为已经走了i步, 不经过x点，走到第j个点的概率。

$large f[i+1][to]+=frac{1}{deg[j]}f[i][j]$

于是对于每一个点我们都跑一遍dp。

然后每个点x的答案就是1-∑dp[i][x].

然后 就水过去了

 

 

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#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
using namespace std;
#define gc getchar()
inline int read(){
    int res=0;char ch=gc;
    while(!isdigit(ch))ch=gc;
    while(isdigit(ch)){res=(res<<3)+(res<<1)+(ch^48);ch=gc;}
    return res;
}
#undef gc

int T, n, m, K;
struct edge{
    int nxt, to;
}ed[2505];
int head[55], cnt;
inline void add(int x, int y)
{
    ed[++cnt] = (edge){head[x], y};
    head[x] = cnt;
}
int deg[55];
double f[10005][55];

inline double DP(int cur)
{
    memset(f, 0, sizeof f);
    double res = 0;
    for (int i = 1 ; i <= n ; i ++) f[0][i] = (double)(1.0/(double)n);
    for (int j = 0 ; j <= K ; j ++)
    {
        for (int x = 1 ; x <= n ; x ++)
        {
            if (x == cur) continue;
            for (int i = head[x] ; i ; i = ed[i].nxt)
            {
                int to = ed[i].to;
                f[j+1][to] += (double)(f[j][x] / (double)deg[x]);
            }
        }
        res += f[j][cur];
    }
    return res;
}

int main()
{
    T = read();
    while(T--)
    {
        memset(head, 0, sizeof head);
        memset(deg, 0, sizeof deg);
        cnt = 0;
        n = read(), m = read(), K = read();
        for (int i = 1 ; i <= m ; i ++)
        {
            int x = read(), y = read();
            add(x, y), add(y, x);
            deg[x]++, deg[y]++;
        }
        for (int i = 1 ; i <= n ; i ++)
            printf("%.10lf
", 1 - DP(i));
    }
    return 0;
}