codeforces 16E 概率+状压dp

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思路：

n为18.每一为为0则是鱼死了，1则是活着。

初始状态：全活着的概率为1

当前状态的鱼j杀死鱼k的概率,相遇的概率乘以j杀k的概率，累加至新状态即可。

#include <iostream>#include <cstdio>#include <string>#include <cstring>#include <cmath>#include <stack>#include <queue>#include <algorithm>#include <vector>#include <map>#include <set>#include <stdlib.h>#include <iomanip>#include <fstream>using namespace std;#pragma comment(linker, "/STACK:102400000,102400000")#define maxn 105#define MOD 1000000007#define mem(a , b) memset(a , b , sizeof(a))#define LL long long#define ULL unsigned long long#define FOR(i , n) for(int i = 1 ;  i<= n ; i ++)typedef pair<int , int> pii;#define INF 100000000int n;double a[maxn][maxn];double dp[1<<18];int main(){    while(scanf("%d" , &n) != EOF)    {        for(int i = 0 ; i < n ; i ++)        {            for(int j = 0 ; j < n ; j ++)            {                scanf("%lf" , &a[i][j]);                //cout << a[i][j] << endl;            }        }        mem(dp , 0);        dp[(1<<n) - 1] = 1;        for(int i = (1<<n) - 1 ; i >= 1 ; i --)        {            int num = 0 , tmp = i;            while(tmp != 0)            {                if(tmp & 1) num ++;                tmp >>= 1;            }            if(num == 1) continue;            double ch = 2.0 * dp[i] / num / (num-1);            for(int j = 0 ; j < n ; j ++)            {                if(i & ( 1 << j))                {                    for(int k = 0 ; k < n ; k ++)                    {                        if(i & ( 1 << k))                        {                            dp[i^(1<<k)] += ch * a[j][k];                        }                    }                }            }        }        for(int i = 0 ; i < n ; i ++)        {            printf("%.6lf" , dp[(1 << i)]);            if(i != n-1) printf(" ");        }        cout <<endl;    }    return 0;}

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