HDU4336 Card Collector

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Card Collector

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 2656 Accepted Submission(s): 1251

Special Judge

Problem Description
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.

Input
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
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.

Sample Input
1
0.1
2
0.1 0.4

Sample Output
10.000
10.500

Source

2012 Multi-University Training Contest 4

题意:有n中卡片，给出每种卡片出现的概率，每天可以买一张卡片，求集齐所有卡片的期望天数。

所有卡片出现的概率相加可能会小于1，所以可能会有没有卡片的情况。n只有20可以状压一下，dp[i]，i表示状态。(每一位的0和1表示取了还是没取)。

所以dp[i]=(1-sigma(p[j]))*dp[i]+sigma(p[k])*dp[i]+sigma(p[kk])*dp[i|(1<<k)];

j表示所有的种类，1-就是表示没有取到卡片，所以保持原状态，k表示取过了的，也表示原状态，kk表示还没有取过的，状态就转移了。。。

将上面的式子移项化简一下就等于dp[i]=sigma(p[kk]*dp[i|(1<<kk)])/sigma(p[kk]);

/*************************************************************************
	> File Name: 4336.cpp
	> Author: tjw
	> Mail: 
	> Created Time: 2014年10月21日 星期二 12时03分19秒
 ************************************************************************/

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<queue>
#include<vector>
#include<stack>
#include<map>
#define ll long long
#define ls k<<1
#define rs k<<1|1
using namespace std;
const int MAXN=20;
double dp[1<<MAXN];
double p[MAXN];
int main()
{
    int n,i,j;
    while(scanf("%d",&n)==1)
    {
        for(i=0;i<n;i++)
            scanf("%lf",&p[i]);
        dp[(1<<n)-1]=0.0;
        for(i=(1<<n)-2;i>=0;i--)
        {
            dp[i]=1;
            double temp=0;
            for(j=0;j<n;j++)
            {
                if((i&(1<<j))==0)
                {
                    dp[i]+=p[j]*dp[i|(1<<j)];
                    temp+=p[j];
                }
            }
            dp[i]/=temp;
        }
        printf("%.4f\n",dp[0]);
    }
    return 0;
}