Codeforces 16E Fish dp(概率)

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概述

题目链接：点击打开链接

题意：池中有n条鱼,每天会有两条鱼相遇，任意两条鱼两两相遇的概率相等，如果两条鱼相遇，其中一条一定会被吃掉，给出所有两两相遇时吃与被吃的概率，求每条鱼存活的概率。

状态转移方程： dp[s2]=∑(win[j][i]*dp[s]/C[num][2]);

i∈s

s2=s^i;

j∈s2,

起点： dp[1<<n]=1

因为没有被吃的固定顺序，所以应该用集合表示状态。

考虑到存活的鱼两两相遇的概率相等，要用组合公式。

n fish, numbered from 1 to n, live in a lake. Every day right one pair of fish meet, and the probability of each other pair meeting is the same. If two fish with indexes i and j meet, the first will eat up the second with the probability aij, and the second will eat up the first with the probability aji = 1 - aij. The described process goes on until there are at least two fish in the lake. For each fish find out the probability that it will survive to be the last in the lake.

Input

The first line contains integer n (1 ≤ n ≤ 18) — the amount of fish in the lake. Then there follow n lines with nreal numbers each — matrix a. aij (0 ≤ aij ≤ 1) — the probability that fish with index i eats up fish with index j. It's guaranteed that the main diagonal contains zeros only, and for other elements the following is true:aij = 1 - aji. All real numbers are given with not more than 6 characters after the decimal point.

Output

Output n space-separated real numbers accurate to not less than 6 decimal places. Number with index ishould be equal to the probability that fish with index i will survive to be the last in the lake.

        Examples 
      
          input 
        
2
0 0.5
0.5 0

          output 
        
0.500000 0.500000 

          input 
        
5
0 1 1 1 1
0 0 0.5 0.5 0.5
0 0.5 0 0.5 0.5
0 0.5 0.5 0 0.5
0 0.5 0.5 0.5 0

          output 
        
1.000000 0.000000 0.000000 0.000000 0.000000

#include<cstdio>
#include<string>
#include<cstring>
#include<iostream>
#include<cmath>
#include<algorithm>
#include<vector>
using namespace std;

#define all(x) (x).begin(), (x).end()
#define for0(a, n) for (int (a) = 0; (a) < (n); (a)++)
#define for1(a, n) for (int (a) = 1; (a) <= (n); (a)++)
#define ysk(x)  (1<<(x))
typedef long long ll;
typedef pair<int, int> pii;
const int INF =0x3f3f3f3f;
const int maxn=18    ;
int ed,n;
double win[maxn+10][maxn+10];
double dp[262144+10];


int C[maxn+5][maxn+5];
void pre()
{
    C[0][0]=1;
    for(int i=1;i<=maxn;i++)
    {
        C[i][0]=C[i][i]=1;
        for(int j=1;j<i; j++)
        {
            C[i][j]=C[i-1][j-1]+C[i-1][j];
        }
    }
}

int getfishnum(int s)//统计存活的鱼的数量
{
    int ans=0;
    for0(i,n) if(s&ysk(i))
    {
        ans++;
    }
    return ans;
}
int main()
{
    pre();
    while(~scanf("%d",&n))
    {
        for0(i,n) for0(j,n) scanf("%lf",&win[i][j]);

         ed=ysk(n)-1;

         memset(dp,0,(ed+1)*sizeof dp[0]);
         dp[ed]=1;
         for(int s=ed;s>=0;s--)
         {
             int num=getfishnum(s);
             for0(i,n) if(ysk(i)&s)//枚举第一个被吃的鱼
             {
                 int s2=s^ysk(i);
                 for0(j,n) if(s2&ysk(j))//枚举吃它的鱼
                 {
                     dp[s2]+=win[j][i]*dp[s]/C[num][2];//因为任意两条鱼两两相遇的概率相等
                 }
             }

         }
         for0(i,n)
         {
             if(i)  putchar(' ');
             printf("%f",dp[ysk(i)]);
         }
         putchar('n');



    }

   return 0;
}