HDU P5001 Walk

2014區域網絡賽鞍山校區上的第五題：

Walk

Time Limit: 30000/15000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 183    Accepted Submission(s): 127
Special Judge

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

 

題意大致爲一張雙向圖相同概率一個點進入，任意走，問走完d步所有點沒走過的概率。要是你欲先求出所有點能走過的概率，然後1-它輸出的話，你就走遠了。。。

正解應該是剛開始每個點1/n的初始概率，然後刪除I點，之後求出其餘點能走到的概率，能走到的概率總和即I點不能走到的概率，然後注意到任意兩點直接走到的概率可以形成一個矩陣，走了d步相當於初始概率形成的矩陣*概率矩陣^d。 故可以用矩陣快速冪來做。

由於時限較寬，其餘方法還有很多，若暫時看不懂本題含義，可以先去看這道題的概率DP題解，其實思想是一樣的。

代碼：

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<ctime>
#include<string>
#include<cstring>
#include<algorithm>
#include<fstream>
#include<queue>
#include<stack> 
#include<vector>
#include<cmath>
#include<iomanip>
#define rep(i,n) for(i=1;i<=n;i++)
#define MM(a,t) memset(a,t,sizeof(a))
#define INF 1e9
typedef long long ll;
#define mod 1000000007
using namespace std;
int n,m,d;
vector<int> eg[100];
struct matrix{
	double a[100][100];
}pan,st,npan;
matrix mul(matrix a1,matrix a2,int i1,int i2){
	int i,j,k;
	matrix c;
	
	rep(i,i1)
	  rep(j,i2){
  		c.a[i][j]=0;
  		rep(k,i2){
		  	 c.a[i][j]+=a1.a[i][k]*1.0*a2.a[k][j];
		  }
  	}
  	
  	return c;
}
void qmi(int nn){
	int j,k;
	while(nn){
	  if(nn%2==1) st=mul(st,npan,1,n);
	  npan=mul(npan,npan,n,n);
	  nn=(nn>>1);	
	}
}
int main()
{
	int i,j,k,T;

    scanf("%d",&T);
    while(T--){
    	scanf("%d%d%d",&n,&m,&d);
    	rep(i,n) eg[i].clear();
    	MM(pan.a,0);
    	rep(i,m){
	    	int s,e;
	    	scanf("%d%d",&s,&e);
	    	eg[s].push_back(e);
	    	eg[e].push_back(s);
	    }
        rep(i,n){
          int sz=eg[i].size();
		  for(j=0;j<sz;j++) pan.a[i][eg[i][j]]=1.0/sz;	
        }
        rep(i,n){
        	double res=0;
        	rep(j,n) st.a[1][j]=1.0/n;
        	MM(npan.a,0);
        	rep(j,n)
        	  rep(k,n)
        	  if(j!=i && k!=i) npan.a[j][k]=pan.a[j][k];
      	    qmi(d);
      	    rep(j,n) res+=st.a[1][j];
      	    printf("%.8lf\n",res);
        }
    }
    
	return 0;
}