hdu 5045 - Contest（2014 ACM/ICPC Asia Regional Shanghai Online ）概率dp

Contest

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)

Total Submission(s): 974 Accepted Submission(s): 416

Problem 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 ith student solve the jth problem, the probability of correct solve is Pij .

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 jth number in the ith line is Pij .

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

1

2 3

0.6 0.3 0.4

0.3 0.7 0.9

Sample Output

Case #1: 2.20000

题意：n个学生m道题，一个n*m的矩阵代表第n个学生解第m题AC的概率，任意两学生做题数差距不能大于1，问AC所有题目概率的最大值

思路：dp[i][j]表示前i个题，学生做题数的状态为j的最大概率，因为最大只能查一题，那么当j全1时，就清空j变成0

[code]#include<iostream>
#include<cstdio>
#include<string>
#include<cstring>
#include<vector>
#include<cmath>
#include<queue>
#include<stack>
#include<map>
#include<set>
#include<algorithm>
using namespace std;
const int maxn=11;
const int maxm=1050;
int N,M;
double P[maxn][maxm];
double dp[maxm][maxm];
int main()
{
    int T,cas=1;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d%d",&N,&M);
        for(int i=0;i<N;i++)
            for(int j=0;j<M;j++)scanf("%lf",&P[i][j]);
        for(int i=0;i<maxm;i++)
            for(int j=0;j<maxm;j++)dp[i][j]=-1.0;
        dp[0][0]=0;
        int S=(1<<N);
        for(int i=0;i<M;i++)
        {
            for(int j=0;j<S;j++)
            {
                if(dp[i][j]<0)continue;
                for(int k=0;k<N;k++)
                {
                    if(!(j&(1<<k)))
                    {
                        int tmp=(j|(1<<k));
                        if(tmp==(S-1))tmp=0;
                        dp[i+1][tmp]=max(dp[i+1][tmp],dp[i][j]+P[k][i]);
                    }
                }
            }
        }
        double ans=0;
        for(int i=0;i<S;i++)
            ans=max(ans,dp[M][i]);
        printf("Case #%d: %.5lf\n",cas++,ans);
    }
    return 0;
}