HDOJ 5045 Contest

状压DP。。

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Contest

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

Total Submission(s): 666    Accepted Submission(s): 300

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 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

1
2 3
0.6 0.3 0.4
0.3 0.7 0.9

 

Sample Output

Case #1: 2.20000

 

Source

2014 ACM/ICPC Asia Regional Shanghai Online

 

#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <queue>

using namespace std;

const int maxn=1100;

int N,M;
double P[15][maxn];

int num[1100];

int getONE(int x)
{
    int ans=0;
    while(x)
    {
        x=x&(x-1);
        ans++;
    }
    return ans;
}

void init()
{
    for(int i=0;i<=1024;i++)
        num[i]=getONE(i);
}

double pb[maxn];
bool vis[maxn];

double getMX(int i,int n)
{
    double ans=0;
    queue<int> q;
    memset(vis,0,sizeof(vis));
    memset(pb,0,sizeof(pb));
    q.push(0);
    for(int j=1;j<=n;j++)///第i组任务里的第j个
    {
        int task=i+j-1;
        while(q.empty()!=true)
        {
            int t=q.front();
            if(num[t]>=j) break;
            q.pop();
            if(num[t]==j-1)
            for(int k=0;k<N;k++)
            {
                if((t&(1<<k))==0)
                {
                    int id=t|(1<<k);
                    pb[id]=max(pb[id],pb[t]+P[k+1][task]);
                    ans=max(pb[id],ans);
                    if(vis[id]==false)
                    {
                        vis[id]=true;
                        q.push(id);
                    }
                }
            }
        }
    }
    return ans;
}

int main()
{
    int T_T,cas=1;
    init();
    scanf("%d",&T_T);
    while(T_T--)
    {
        scanf("%d%d",&N,&M);
        for(int i=1;i<=N;i++)
            for(int j=1;j<=M;j++)
                scanf("%lf",&P[i][j]);
        double ans=0;
        for(int i=1;i+N-1<=M;i+=N)///第i组任务
        {
            getMX(i,N);
            ans+=pb[(1<<N)-1];
        }
        int res=M%N;
        if(res) ans+=getMX(M-res+1,res);
        printf("Case #%d: %.5lf\n",cas++,ans);
    }
    return 0;
}