标签：signed ++ set -- ble def html ret 反向
题意：每包干脆面可能开出卡或者什么都没有，一共n种卡，每种卡每包爆率pi，问收齐n种卡的期望
思路：期望求解公式为：$E(x) = \sum_{i=1}^{k}pi * xi + (1 - \sum_{i = 1}^{k}pi) * [1 + E(x)]$，即能转换到x情况的期望+x情况原地踏步的期望。
因为n比较小，我们可以直接状压来表示dp[x]为x状态时集齐的期望。那么显然dp[111111111] = 0。然后我们状态反向求解。最终答案为dp[0]。
然后来看期望的求解：$E(x) = \sum_{i = 1}^{k}pi * [1 + E(xi)] + (1 - \sum_{i = 1}^{k}pi) * [1 + E(x)]$，E(xi)是E(x)某一位0变成1后的期望。
化简后：$E(x) = (\sum_{i = 1}^{k}pi * E(xi) + 1) / \sum_{i = 1}^{k}pi$
代码：
#include<set> #include<map> #include<cmath> #include<queue> #include<cstdio> #include<vector> #include<cstring> #include <iostream> #include<algorithm> using namespace std; typedef long long ll; typedef unsigned long long ull; const int maxn = 20 + 5; const int M = maxn * 30; const ull seed = 131; const int INF = 0x3f3f3f3f; const int MOD = 1e4 + 7; double dp[1 << maxn]; double p[maxn]; int main(){ int n; while(~scanf("%d", &n)){ // for(int i = 0; i < (1 << n); i++) dp[i] = 0; for(int i = 0; i < n; i++){ scanf("%lf", &p[i]); } dp[(1 << n) - 1] = 0; for(int i = (1 << n) - 2; i >= 0; i--){ double sump = 0, sumpe = 0; for(int j = 0; j < n; j++){ if(!(i & (1 << j))){ sump += p[j]; sumpe += p[j] * dp[i | (1 << j)]; } } dp[i] = (sumpe + 1) / sump; } printf("%.6f\n", dp[0]); } return 0; }
HDU 4336 Card Collector（状压 + 概率DP 期望）题解
标签：signed ++ set -- ble def html ret 反向
原文地址：https://www.cnblogs.com/KirinSB/p/10987317.html
