hdu 5001 walk 概率dp入门题

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 

题目大意：在一幅图中，主人公随机选择一个点为起始点，然后每一步都是随机选择相邻点进行访问，一共走了d步。求各个点没有被访问的概率。

概率dp入门题，定义dp[u][i][j]：经过了j步之后落在i点且从来不经过u点的概率。最后对于各个点，从不经过它的概率就是dp[u][*][d]的和。

在实际操作过程中，可以把dp数组变成二维的，即变成dp[i][j]。

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
using namespace std;
const int maxn = 55;
int head[maxn],cur;
struct Edge
{
    int to;
    int next;
} edge[maxn*2];
void init()
{
    cur=0;
    memset(head,-1,sizeof(head));
}
void addedge(int u,int v)
{
    edge[cur].to = v;
    edge[cur].next = head[u];
    head[u] = cur++;
}
double dp[maxn][10005];
double ans[maxn];
int du[maxn],n;
void cal_dp(int x,int d)
{
    for(int u=1; u<=n; u++)
    {
        dp[u][d] = 0;
        if(u==x) continue;
        for(int e=head[u]; e!=-1; e=edge[e].next)
        {
            int v = edge[e].to;
            if(v==x) continue;
            dp[u][d] += dp[v][d-1]/du[v];
        }
    }
}
int main()
{
    int T,u,v,m,d;
    scanf("%d",&T);
    while(T--)
    {
        init();
        memset(du,0,sizeof(du));
        scanf("%d%d%d",&n,&m,&d);
        for(int i=0; i<m; i++)
        {
            scanf("%d%d",&u,&v);
            addedge(u,v);
            addedge(v,u);
            du[u]++;
            du[v]++;
        }
        for(int i=1; i<=n; i++)
        {
            dp[i][0] = 1.0/n;
            ans[i]=0;
        }
        for(int u=1; u<=n; u++)
        {
            for(int j=1; j<=d; j++)
                cal_dp(u,j);
            for(int v=1;v<=n;v++)
                ans[u] += dp[v][d];
        }
        for(int i=1; i<=n; i++)
        {
            printf("%.10lf
",ans[i]);
        }
    }
}