Hdu 5001 Walk 機率dp

Walk

Time Limit: 1 Sec  

Memory Limit: 256 MB

題目鏈接

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

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

HINT

 

題意

隨機走d步，請問沒有通過i點的機率是多少

題解：

數據範圍很小咯，點不多，那就直接暴力DP就行了

dp[i][j][k]表示沒有通過k點，走了i步，如今在j點的位置

代碼：

//qscqesze
#include <cstdio>
#include <cmath>
#include <cstring>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <set>
#include <bitset>
#include <vector>
#include <sstream>
#include <queue>
#include <typeinfo>
#include <fstream>
#include <map>
#include <stack>
typedef long long ll;
using namespace std;
//freopen("D.in","r",stdin);
//freopen("D.out","w",stdout);
#define sspeed ios_base::sync_with_stdio(0);cin.tie(0)
#define maxn 200500
#define mod 1001
#define eps 1e-9
#define pi 3.1415926
int Num;
//const int inf=0x7fffffff;
const ll inf=999999999;
inline ll read()
{
    ll x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
//*************************************************************************************

vector<int> E[maxn];
double dp[10100][101];
int main()
{
    int t = read();
    while(t--)
    {
        int n=read(),m=read(),d=read();
        for(int i=0;i<=n;i++)
            E[i].clear();
        for(int i=0;i<m;i++)
        {
            int x=read(),y=read();
            E[x].push_back(y);
            E[y].push_back(x);
        }
        for(int p=1;p<=n;p++)
        {
            memset(dp,0,sizeof(dp));
            for(int i=1;i<=n;i++)
                dp[0][i]=1.0/n;
            for(int i=1;i<=d;i++)
            {
                for(int j=1;j<=n;j++)
                {
                    if(j==p)continue;
                    int N = E[j].size();
                    for(int k=0;k<E[j].size();k++)
                        dp[i][E[j][k]] += dp[i-1][j] * 1.0 / N;
                }
            }
            double ans = 0;
            for(int i=1;i<=n;i++)
            {
                if(i==p)continue;
                ans += dp[d][i];
            }
            printf("%.6lf\n",ans);
        }
    }
}