Codeforces 16E Fish (状压dp+概率)

题意

有n条鱼，他们相遇时会吃掉对方，给出他们相遇时双方获胜的概率，求这n条鱼最后剩下自己的概率。

思路

看范围就要考虑状压DP，dp[s]表示当前剩下的鱼的状态为s时的概率。
那么P(i吃掉j) = P(i和j同时存在) *P(ij相遇)* P(i战胜j)
即dp[s ^ (1 << j)] += dp[s] * p[i][j] * 1 / (num * (num - 1) / 2)。num是当前状态下鱼的总数。

代码

#include <stdio.h>#include <string.h>#include <iostream>#include <algorithm>#include <vector>#include <queue>#include <stack>#include <set>#include <map>#include <list>#include <string>#include <math.h>#include <stdlib.h>#include <time.h>using namespace std;#define LL long long#define lowbit(x) ((x)&(-x))#define lson l, mid, rt << 1#define rson mid + 1, r, rt << 1|1#define MP(a, b) make_pair(a, b)const int INF = 0x3f3f3f3f;const int MOD = 1000000007;const int maxn = 1e5 + 10;const double eps = 1e-8;const double PI = acos(-1.0);typedef pair<int, int> pii;double dp[1<<19];double p[20][20];int main(){    //freopen("in.txt","r",stdin);    //freopen("out.txt","w",stdout);    int n;    cin >> n;    for (int i = 0; i < n; i++)        for (int j = 0; j < n; j++)            cin >> p[i][j];    dp[(1 << n) - 1] = 1.0;    for (int s = (1 << n) - 1; s > 0; s--)        for (int i = 0; i < n; i++)            for (int j = 0; j < n; j++)                if ((s & (1 << i)) && (s & (1 << j)))                {                    if (i == j) continue;                    int num = __builtin_popcount(s);                    dp[s ^ (1 << j)] += dp[s] * p[i][j] / (num * (num - 1) / 2);                }    for (int i = 0; i < n; i++)        printf("%.6f ", dp[1<<i]);    return 0;}