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
题意：每个人做题是有概率的，每个做题的序列任意两个人做题的个数不超过1，求最大的概率。
思路：设dp[i][j]表示前i道题，j的状态，因为不能超过1，所以每个状态我们都能用01表示，1表示当前做的题目最大，0表示最小的。例如3232，就表示1010，所以我们就能用1<<10的状态表示了，每次从0的人开始转移，因为只有0转移才能是符合条件的。
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int maxn = 1100;
double dp[maxn][maxn];
double p[maxn][maxn];
int main() {
int n, m;
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 j = 0; j < (1<<n); j++) {
if (dp[i][j] < 0) continue;
int st;
for (int k = 0; k < n; k++)
if (((1<<k)&j) == 0) {
st = j | (1<<k);
if (st == (1<<n)-1) st = 0;
if (dp[i+1][st] < dp[i][j] + p[k][i])
dp[i+1][st] = dp[i][j] + p[k][i];
}
}
}
double ans = 0;
for (int i = 0; i < (1<<n); i++)
ans = max(ans, dp[m][i]);
printf("Case #%d: %.5lf
", cas++, ans);
}
return 0;
}