codeforces 16E Fish

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概述

http://www.elijahqi.win/archives/1302
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 n real 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 i should 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

概率dp

我们每次从现在存活的小鱼中枚举两两相遇的情况那么每对相遇的概率就是C（2/剩余）所以产生下一种状态的概率就是吃与被吃的概率*相遇的概率

#include<cstdio>
double dp[1<<19],map[20][20];int n;
int main(){
    freopen("cf.in","r",stdin);
    scanf("%d",&n);
    for (int i=0;i<n;++i)
        for (int j=0;j<n;++j) scanf("%lf",&map[i][j]);
    int bin=(1<<n)-1;dp[bin]=1;
    for (int s=bin;s;s--){
        double ans=0;for (int i=0;i<n;++i) if (s&(1<<i)) ans++;
        if (ans==1) continue;ans=2.0/(ans*(ans-1));
        for (int i=0;i<n;++i){
            if (s&(1<<i)){
                for (int j=i+1;j<n;++j){
                    if (s&(1<<j)){
                        dp[s-(1<<i)]+=dp[s]*map[j][i]*ans;
                        dp[s-(1<<j)]+=dp[s]*map[i][j]*ans;
                    }
                }
            }
        }
    }
    for (int i=0;i<n;++i)printf("%.6f ",dp[1<<i]);

    return 0;
}