HDU 4336 – Card Collector [状压概率DP]

Link

http://acm.hdu.edu.cn/showproblem.php?pid=4336

Problem

In your childhood, do you crazy for collecting the beautiful cards in the snacks? They said that, for example, if you collect all the 108 people in the famous novel Water Margin, you will win an amazing award.
As a smart boy, you notice that to win the award, you must buy much more snacks than it seems to be. To convince your friends not to waste money any more, you should find the expected number of snacks one should buy to collect a full suit of cards.

The first line of each test case contains one integer N (1 <= N <= 20), indicating the number of different cards you need the collect. The second line contains N numbers p1, p2, …, pN, (p1 + p2 + … + pN <= 1), indicating the possibility of each card to appear in a bag of snacks.
Note there is at most one card in a bag of snacks. And it is possible that there is nothing in the bag.

Output one number for each test case, indicating the expected number of bags to buy to collect all the N different cards.
You will get accepted if the difference between your answer and the standard answer is no more that 10^-4.

Mean

有N种卡片和好多包,一个包中可能含有某种卡片或者空。收集若干个包后可以集齐所有种类的卡片,求包个数的期望。

Analysis

定义dp[i]为当前状态下期望收集几个包后集齐。假设i为“0110”,表示第14种还没有。则i==0表示集齐,所以dp[0]=0。

根据题意,dp[0110] = p[1] * dp[1110] + p[4] * dp[0111] + (1 – p[1] – p[4]) * dp[0110] + 1。

整理后可得:dp[0110] = (p[1] * dp[1110] + p[4] * dp[0111] + 1) / (p[1] + p[4])。

据此用二进制枚举从后往前dp即可。

Code

#include <bits/stdc++.h>
using namespace std;

int N;
double p[25], dp[(1<<20)+5];

int main() {
    while(~scanf("%d", &N)) {
        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 num = 1, den = 0;
            for(int j = 0; j < N; ++j) {
                if(!(i&(1<<j))) {
                    num += p[j] * dp[i|(1<<j)];
                    den += p[j];
                }
            }
            dp[i] = num / den;
        }
        printf("%.7f\n", dp[0]);
    }
    return 0;
}

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