Pitcher Rotation

题意：

n个人m个对手给出每个人能战胜每个敌人的概率，现在有g个比赛，每个人赛完后要休息4天（可重复用），求能获得胜利的最大期望个数。

分析：

因为只有每个人5天就能用一次，所以对于每个人来说，只有得分前5的会被使用上，所以后4维状态只需要5^4,进行状态转移

dp[i][j][k][l][p]表示第i场比赛，前一天为j，两天为k，三天为l，四天为p，的最大得分,只和i-1状态有关用滚动数组。

#include <map>
#include <set>
#include <list>
#include <cmath>
#include <queue>
#include <stack>
#include <cstdio>
#include <vector>
#include <string>
#include <cctype>
#include <complex>
#include <cassert>
#include <utility>
#include <cstring>
#include <cstdlib>
#include <iostream>
#include <algorithm>
using namespace std;
typedef pair<int,int> PII;
typedef long long ll;
#define lson l,m,rt<<1
#define pi acos(-1.0)
#define rson m+1,r,rt<<11
#define All 1,N,1
#define read freopen("in.txt", "r", stdin)
const ll  INFll = 0x3f3f3f3f3f3f3f3fLL;
const int INF= 0x7ffffff;
const int mod =  1000000007;
struct node{
    int id,v;
}win[110][110];
int dp[2][6][6][6][6],n,m,g,d[255];
bool cmp(node x,node y){
    return x.v>y.v;
}
int cal(int i,int j){
    if(i==0||j==0)return 0;
    return win[i][j].id;
}
int t;
 int main() {
        scanf("%d", &t);
        while (t--) {
           int ans = 0;
            scanf("%d%d%d", &n, &m, &g);
            for (int i = 1; i <= m; i++) {
                for (int j = 1; j <= n; j++) {
                    scanf("%d", &win[i][j].v);
                    win[i][j].id = j;
                }
                sort(win[i] + 1, win[i] + 1 + n, cmp);
            }
            g += 10;
            int i,j,k,y,l,x;
            for (i = 1; i <= g; i++)
                scanf("%d", &d[i]);
            memset(dp[0], 0, sizeof(dp[0]));
            int now=0;
            for (i = 1; i <= g; i++) {
                now^=1;
                memset(dp[now], 0, sizeof(dp[now]));
                if (d[i]) {
                    for (y = 1; y <= 5; y++) {
                        for (j = 0; j <= 5; j++) {
                            if (i > 1 && cal(d[i], y) == cal(d[i - 1], j)) continue;

                            for (k = 0; k <= 5; k++) {
                                if (i > 2 && cal(d[i], y) == cal(d[i - 2], k)) continue;

                                for (l = 0; l <= 5; l++) {
                                    if (i > 3 && cal(d[i], y) == cal(d[i - 3], l)) continue;
                                    for (x = 0; x <= 5; x++) {
                                        if (i > 4 && cal(d[i], y) == cal(d[i - 4], x)) continue;
                                        dp[now][y][j][k][l] = max(dp[now][y][j][k][l], dp[now^1][j][k][l][x] + win[d[i]][y].v);
                                        ans = max(ans, dp[now][y][j][k][l]);
                                    }
                                }
                            }
                        }
                    }
                }
                else {
                    for (j = 0; j <= 5; j++) {
                        for (k = 0; k <= 5; k++) {
                            for (l = 0; l <= 5; l++) {
                                for (x = 0; x <= 5; x++) {
                                    dp[now][0][j][k][l] = max(dp[now][0][j][k][l], dp[now^1][j][k][l][x]);
                                    ans = max(ans, dp[now][0][j][k][l]);
                                }
                            }
                        }
                    }
                }
            }
            printf("%.2lf\n", ans * 1.0 / 100);
        }
        return 0;
}