hdu4336 Card Collector 状态压缩dp

最新推荐文章于 2019-09-11 15:23:28 发布

转载最新推荐文章于 2019-09-11 15:23:28 发布 · 70 阅读

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原文链接：http://www.cnblogs.com/jiangu66/p/3246683.html

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#java

本文介绍了一个经典的概率问题：为了集齐一系列卡片所需的预期零食包数量。通过使用容斥原理和状态压缩DP两种方法来解决该问题，并给出了详细的代码实现。

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Card Collector

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1708 Accepted Submission(s): 780
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

当时看到，这一题，一点思路都没有啊，后来，发现是用容斥原理，这个表示没用过，补上这个知识点！

#include <iostream>
#include <stdio.h>
#include <string.h>

using namespace std;
double p[25];
int main()
{
    int n,i,flag,j,k,all,cnt;
    double sum,s,temp;
    while(scanf("%d",&n)!=EOF)
    {
        flag=1;
        for(i=0;i<n;i++)
        {
            scanf("%lf",&p[i]);
        }
        sum=0;
        flag=1;
        all=1<<n;
        for(i=1;i<all;i++)
        {
            cnt=0;
            temp=0;
            for(j=0;j<n;j++)
            {
               if(i&(1<<j))//相交
               {
                   temp+=p[j];cnt++;
               }
            }
            if(cnt&1)//奇加偶减
            sum+=1.0/temp;
            else
            sum-=1.0/temp;
        }
        printf("%.4f\n",sum);
    }
    return 0;
}

再来一个状态压缩dp,这里我们可以推出dp[n]=1+all dp[n](空的)+all dp[n](重复的)+all dp[k](没有但是加一个就有了)；这样就好做了，好像这一题还非要写成输出４位，输出３位还是错的，样例应该是有问题的！很坑啊，有木有！

#include <stdio.h>
#include <string.h>
#include <iostream>
using namespace std;
double p[25],dp[1<<20];
int main ()
{
    int n,i,j,all;
    double allp,tempallp,sum;
    while(scanf("%d",&n)!=EOF)
    {
        allp=0;
        for(i=0;i<n;i++)
        {
            scanf("%lf",&p[i]);
            allp+=p[i];
        }
        all=1<<n;
        dp[0]=0;

        for(i=1;i<all;i++)
        {
            tempallp=allp;
            sum=1;//买了一袋
            for(j=0;j<n;j++)
            {
                if(i&(1<<j))
                {
                    sum+=p[j]*dp[i^(1<<j)];


                }
                else
                {
                  tempallp-=p[j];

                }
            }
            dp[i]=sum/tempallp;

        }
        printf("%.4f\n",dp[all-1]);
    }

    return 0;
}

转载于:https://www.cnblogs.com/jiangu66/p/3246683.html