hdu 5045 F - Contest

问题描述

In the ACM International Collegiate Programming Contest, each team consist of three students. And the teams are given 5 hours to solve between 8 and 12 programming problems.

On Mars, there is programming contest, too. Each team consist of N students. The teams are given M hours to solve M programming problems. Each team can use only one computer, but they can’t cooperate to solve a problem. At the beginning of the ith hour, they will get the ith programming problem. They must choose a student to solve this problem and others go out to have a rest. The chosen student will spend an hour time to program this problem. At the end of this hour, he must submit his program. This program is then run on test data and can’t modify any more.

Now, you have to help a team to find a strategy to maximize the expected number of correctly solved problems.

For each problem, each student has a certain probability that correct solve. If the i ^th student solve the j ^th problem, the probability of correct solve is P _ij .

At any time, the different between any two students’ programming time is not more than 1 hour. For example, if there are 3 students and there are 5 problems. The strategy {1,2,3,1,2}, {1,3,2,2,3} or {2,1,3,3,1} are all legal. But {1,1,3,2,3},{3,1,3,1,2} and {1,2,3,1,1} are all illegal.

You should find a strategy to maximize the expected number of correctly solved problems, if you have know all probability

 

输入

The first line of the input is T (1 ≤ T ≤ 20), which stands for the number of test cases you need to solve.

The first line of each case contains two integers N ,M (1 ≤ N ≤ 10,1 ≤ M ≤ 1000),denoting the number of students and programming problem, respectively.

The next N lines, each lines contains M real numbers between 0 and 1 , the j ^th number in the i ^th line is P _ij .

 

输出

For each test case, print a line “Case #t: ”(without quotes, t means the index of the test case) at the beginning. Then a single real number means the maximal expected number of correctly solved problems if this team follow the best strategy, to five digits after the decimal point. Look at the output for sample input for details.

 

样例输入

     
     
     
     

      
      
      
       1
2 3
0.6 0.3 0.4
0.3 0.7 0.9 
     
     
     
     

 

样例输出

   
   
   
   

    
    
    
     Case #1: 2.20000 
   
   
   
   

状态压缩dp；用二进制表示做题人数的状态，比如00101表示第一个和第三个人；

dp[i][j]表示 第i道题状态为j时的ac的题数的最大期望，其中j表示状态，即，由哪些人去做前i道题，所以j状态的人数不能超过当前题数；

那么状态的枚举就可以用两个for循环:

       for(int j=0;j<=l;j++)

           for(int k=0;k<n;k++)

其中j表示前面i-1的状态状态，k+1表示当前新加的人；比如（j=0101，k=1时就表示，前面i-1道是由第一个和第三个ac的，现在在这基础上解决第i题，新加了第二个人）所以当前状态就可以用 j ||（1<<k)来表示了，用这种方法可以枚举出所有的状态

int tem=j || (1<<k),所以状态转移方程为:

dp[i][tem]=max(dp[i][tem],dp[i-1][j]+a[i][k+1])；

最后在找出dp[m][l]中最大的就可以了；（l = 1<<n-1)

代码如下：

#include <iostream>
#include <stdio.h>
#include <algorithm>
#include <cstring>
#include <iomanip>
using namespace std;

const int maxx=1100;

int n,m;
double a[maxx][15];
double dp[maxx][1<<11];


int main()
{
   int t;
   scanf("%d",&t);
   for(int k=1;k<=t;k++)
   {
       int n;int m;
       cin>>n>>m;

       for(int i=1;i<=n;i++)
        for(int j=1;j<=m;j++)
       {
           cin>>a[j][i];
       }
         //先全部初始化为-1；
         for(int i=0;i<maxx;i++)
            for(int j=0;j<maxx;j++)
                dp[i][j]=-1;
        dp[0][0]=0;
        int l=(1<<n)-1;
            for(int i=1;i<=m;i++)
                for(int j=0;j<=l;j++)
                    {
                        if(dp[i-1][j]<0.0)continue;//达到前i道题只有i个人去做的目的；
                        for(int k=0;k<n;k++)
                        {
                            if(!(j&(1<<k)))
                            {
                                int tem=j|(1<<k);
                                if(tem==l)tem=0;//不能忽视的细节；如果不写这句，下一次就用不了人数满了的结果
                                dp[i][tem]=max(dp[i][tem],dp[i-1][j]+a[i][k+1]);

                            }
                        }
                    }
        double ans=0;
        for(int i=0;i<=l;i++)
        {
            ans=max(ans,dp[m][i]);
        }
       printf("Case #%d: %.5lf\n",k,ans) ;
   }
}