2014鞍山網路賽 E題||hdu 5001 概率dp

life4711發表於2014-09-13

http://acm.hdu.edu.cn/showproblem.php?pid=5001

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





 




雖然我不會概率dp並且水平還遠遠沒到在比賽的時候可以A出這道題來。。不過在學長的指點之下還是在比賽之後把這個題個過了。。



解題思路：  dp[i][j][k]表示，i代表當前  第i步，j代表當前走到第j個節點，k是沒有路過k，整體是  走到第i步恰好在j節點，這個過程中沒有路過k的概率 




#include <stdio.h>
#include <string.h>
#include <iostream>
using namespace std;

double a[3][55][55],ans[55];
int mp[55][55],cnt[55];
int T,n,m,d;

int main()
{
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d%d%d",&n,&m,&d);
        memset(mp,0,sizeof(mp));
        memset(a,0,sizeof(a));
        memset(cnt,0,sizeof(cnt));
        for(int i=0; i<m; i++)
        {
            int u,v;
            scanf("%d%d",&u,&v);
            mp[u][v]=mp[v][u]=1;
            cnt[u]++;
            cnt[v]++;
        }
        for(int i=1; i<=n; i++)
            for(int j=1; j<=n; j++)
                if(i!=j)
                    a[1][i][j]=1.0/n;
                else
                    a[1][i][j]=0;
        for(int i=1; i<=d; i++)
        {
            memset(a[(i+1)%2],0,sizeof(a[(i+1)%2]));
            for(int j=1; j<=n; j++)
                for(int k=1; k<=n; k++)
                    if(mp[j][k])
                        for(int p=1; p<=n; p++)
                            if(k!=p)
                                a[(i+1)%2][k][p]+=a[i%2][j][p]*(1.0/cnt[j]);
                            else
                                a[(i+1)%2][k][p]=0;
        }
        for(int i=1;i<=n;i++)
        {
            ans[i]=0;
            for(int j=1;j<=n;j++)
                ans[i]+=a[(d+1)%2][j][i];
        }
        for(int i=1;i<=n;i++)
            printf("%.10lf\n",ans[i]);
    }
    return 0;
}









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