hdu 5001 Walk（概率dp）

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

 

 

 题意：一张无向图，要你求出走d步之后，每个点无法到达的概率。

dp[i][j]表示走了i步时，到达点j的概率

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<math.h>
#include<algorithm>
#include<queue>
#include<set>
#include<bitset>
#include<map>
#include<vector>
#include<stdlib.h>
#include <stack>
using namespace std;
#define PI acos(-1.0)
#define max(a,b) (a) > (b) ? (a) : (b)  
#define min(a,b) (a) < (b) ? (a) : (b)
#define ll long long
#define eps 1e-10
#define MOD 1000000007
#define N 56
#define M 10006
#define inf 1e12
int n,m,d;
vector<int> g[N];
double dp[M][N];
int main()
{
    int t;
    scanf("%d",&t);
    while(t--){
        for(int i=0;i<N;i++){
            g[i].clear();
        }
        scanf("%d%d%d",&n,&m,&d);
        for(int i=0;i<m;i++){
            int u,v;
            scanf("%d%d",&u,&v);
            g[u].push_back(v);
            g[v].push_back(u);
        }
        
        for(int i=1;i<=n;i++){
            dp[0][i]=1.0/n;
        }
        
        for(int l=1;l<=n;l++){
            for(int i=1;i<=d;i++){
                for(int j=0;j<=n;j++){
                    dp[i][j]=0;
                }
            }
            for(int i=1;i<=d;i++){
                for(int j=1;j<=n;j++){
                    if(l==j){
                        continue;
                    }
                    int len=g[j].size();
                    for(int k=0;k<len;k++){
                        if(g[j][k]==l){
                            continue;
                        }
                        dp[i][g[j][k]]+=dp[i-1][j]/len;
                    }
                }
            }
            double ans=0;
            for(int i=1;i<=n;i++){
                ans+=dp[d][i];
            }
            printf("%.10lf
",ans);
        }
        
    }
    return 0;
}