HDU 4336 Card Collector（動態規劃-機率DP）

標籤：des   style   class   blog   code   http   

Card CollectorProblem DescriptionIn 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. 
InputThe 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. 
OutputOutput 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. 
Sample Input

10.120.1 0.4

 
Sample Output

10.00010.500

 
Source2012 Multi-University Training Contest 4 
Recommendzhoujiaqi2010 
題目大意：

有n個卡片，你現在買一包即食麵，沒包即食麵出現其中一個卡片的機率為 p[i] ，問你集齊一套卡片需要的張數的數學期望。

解題思路：

機率DP，用位進位0表示這個卡片有了，1表示這個卡片還沒有，那麼 例如 “3” 用二進位表示 “1 1” 那麼 數組 dp[3] 記錄的就是 1號卡片和2號卡片都有的情況集齊一套卡片需要的張數的數學期望。

dp[sum]= ( 1+sum { dp[ sum + (1<<j )] *p[j] }   ) /sum{p[j] }

其中 ( i&(1<<j) )==0

解題代碼：

#include <iostream>#include <cstdio>using namespace std;const int maxn=(1<<20)+10;int n;double dp[maxn];double p[30];void solve(){    int sum=(1<<n)-1;    dp[sum]=0;    for(int i=sum-1;i>=0;i--){        double tmp=0;        dp[i]=1;        for(int j=0;j<n;j++){            if( ( i&(1<<j) )==0 ){                dp[i]+=dp[i+(1<<j)]*p[j];                tmp+=p[j];            }        }        dp[i]/=tmp;    }    printf("%lf\n",dp[0]);}int main(){    while(scanf("%d",&n)!=EOF){        for(int i=0;i<n;i++) scanf("%lf",&p[i]);        solve();    }    return 0;}