hdu 5045(状态压缩dp)

  Contest

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

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

12 30.6 0.3 0.40.3 0.7 0.9

 

Sample Output

Case #1: 2.20000

 

思路：

因为限制条件，所以前ｎ道题，必须分给ｎ个人一人一道，那么每人只能做一次；

那么用状态压缩表示哪些人做过了；

比如５个人００１１０；那么下一道题就可以选择剩下的３个人做，并且比较出一个最优的选择；

当状态变成１１１１１时，就要重新置零；

所以用一个ｄｐ［ｉ］［ｊ］表示前ｉ道题，已经做的人的状态是ｊ；

AC：

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>using namespace std;int n,m;double p[15][1005];double dp[1005][1<<10];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 <= m; i++) for(int j = 0 ; j < (1 << n); j++)dp[i][j] = -1.0;dp[0][0] = 0;for(int i = 0; i < m; i++)for(int k = 0; k < (1<<n); k++){if(dp[i][k] == -1) continue;for(int j = 0; j < n; j++){int tmp = 1 << j;if(k & tmp) continue;int s = k | tmp;if(s == (1 << n) - 1) s = 0;dp[i+1][s] = max(dp[i+1][s],dp[i][k]+p[j][i]);}}double ans = 0;for(int i = 0; i < (1<<n); i++)ans = max(ans,dp[m][i]);printf("Case #%d: %.5f\n",cas++,ans);}return 0;}