hdu5045

A B C D

A - Contest

Time Limit:1000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u

Submit Status Practice HDU 5045

Description

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

 

Input

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 .

 

Output

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.

 

Sample Input

12 30.6 0.3 0.40.3 0.7 0.9 

 

Sample Output

Case #1: 2.20000 
题意：有n个人做m道题目，每个人对于每题都有答对的概率，求最后答出所有题目概率的最大值。有要求就是每两个人之间答题的数目不能超过1。
学长用的网络流过得，但是这道题更适合状态dp啊，dp[i][j] 表示第i道题完成的人得状态 j,
dp[i+1]=  max（ dp[i][st] ，dp[i][j]+num[k][i]) ,st表示j状态加上帝k个人的状态，j表示j状态，num [k][i]表示第k个人回答i问题概率
ac代码有解释：
#include <stdio.h>#include <string.h>#include <algorithm>#include <iostream>using namespace std;const int maxn=1000;double  dp[1050][1050],num[15][maxn+10];///dp[i][j]  表示第i道题完成的人得状态 jint main(){    int n,m;    int T=0;    int t;    cin>>t;    while(t--)    {        scanf("%d%d",&n,&m);        for(int i=0;i<n;i++)            for(int j=0;j<m;j++)            scanf("%lf",&num[i][j]);        for(int i=0;i<1005;i++)            for(int j=0;j<1055;j++)            dp[i][j]=-1.0;        dp[0][0]=0.0;        int sm=(1<<n)-1;        for(int i=0;i<m;i++)        {            for(int j=0;j<(1<<n);j++)///n个人，2^n个状态            {                if(dp[i][j]<0.0)                    continue;                int st=0;                for(int k=0;k<n;k++)                {                    if(!((1<<k)&j))///第k个人要做第m道题                    {                         st=(1<<k)|j;///j状态加上第k个人的状态                         if(st==sm)///如果状态满了，全是1，说明n个人都答完一道题了                            st=0;///清零，这样每个人彼此之间都不会超过1题                        dp[i+1][st]=max(dp[i+1][st],dp[i][j]+num[k][i]);                    }                }            }        }        double ans=0;        for(int i=0;i<(1<<n);i++)            ans=max(ans,dp[m][i]);        printf("Case #%d: %.5lf\n",++T,ans);    }    return 0;}