题目链接：hdu 5045 Contest

题目大意：一个队伍有N个人，比赛一共有M道题目，给定一个矩阵，表示每个人答对相应题目的正确率。现在对于每道题，可以派出一名学生参加答题，但是在任意时刻，任意两个学生答题数量不能相差2题以上。

解题思路：dp[i][s]，表示在第i道题，s表示一个二进制状态，表示哪些人答过题（相应的），2N−1=0

#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;
const int maxn = 15;
const int maxm = 2005;
const int maxs = (1<<10) + 5;
const double INF = 0x3f3f3f3f;

int N, M;
double p[maxn][maxm], dp[maxm][maxm];

void init () {
    scanf("%d%d", &N, &M);
    for (int i = 0; i < N; i++) {
        for (int j = 1; j <= M; j++)
            scanf("%lf", &p[i][j]);
    }
}

double solve () {
    int T = 1<<N;
    for (int i = 0; i <= M; i++)
        for (int j = 0; j < T; j++)
            dp[i][j] = -INF;
    dp[0][0] = 0;

    for (int i = 1; i <= M; i++) {
        for (int s = 0; s < T; s++) {
            for (int k = 0; k < N; k++) {
                if (s & (1<<k))
                    continue;
                dp[i][s | (1<<k)] = max(dp[i][s | (1<<k)], dp[i-1][s] + p[k][i]);
            }
        }
        dp[i][0] = dp[i][T-1];
    }

    double ret = 0;
    for (int i = 0; i < T; i++)
        ret = max(ret, dp[M][i]);
    return ret;
}

int main () {
    int cas;
    scanf("%d", &cas);
    for (int kcas = 1; kcas <= cas; kcas++) {
        init();
        printf("Case #%d: %.5lf\n", kcas, solve());
    }
    return 0;
}