UVA 1379 - Pitcher Rotation(DP + 贪心)

题目链接：

option=com_onlinejudge&Itemid=8&page=show_problem&problem=4125" rel="nofollow">1379 - Pitcher Rotation

题意：n个人，m个敌人。去比赛，有得分。n个人能够反复比。可是每次比完要歇息4天，问最大得分

思路：dp[i][j][k][l][x] 表示第场比赛，前一天为j，两天为k，三天为l。四天为x，的最大得分，然后因为仅仅有每一个人5天就能用一次，所以对于每一个人来说，仅仅有得分前5的会被使用上。所以后4维状态仅仅须要5^4,进行状态转移。不用比赛的情况分开考虑。还有这题内存有限，要用滚动数组优化不然会RE

代码：

#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;
#define max(a,b) ((a)>(b)?(a):(b))
const int N = 105;
const int M = 205;
int t, n, m, G, d[M + 10], i, x, j, k, l, y, dp[2][6][6][6][6], ans;

struct Map {
	int v, id;
} g[N][N];

bool cmp(Map a, Map b) {
	return a.v > b.v;
}

int cal(int u, int v) {
	if (u == 0 || v == 0)
		return 0;
	return g[u][v].id;
}

int main() {
	scanf("%d", &t);
	while (t--) {
		ans = 0;
		scanf("%d%d%d", &n, &m, &G);
		for (i = 1; i <= m; i++) {
			for (j = 1; j <= n; j++) {
				scanf("%d", &g[i][j].v);
				g[i][j].id = j;
			}
			sort(g[i] + 1, g[i] + 1 + n, cmp);
		}
		G += 10;
		for (i = 1; i <= G; i++)
			scanf("%d", &d[i]);
		memset(dp[0], 0, sizeof(dp[0]));
		for (i = 1; i <= G; i++) {
			int now = i % 2;
			int pre = (!now);
			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[pre][j][k][l][x] + g[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[pre][j][k][l][x]);
								ans = max(ans, dp[now][0][j][k][l]);
							}
						}
					}
				}
			}
		}
		printf("%.2lf\n", ans * 1.0 / 100);
	}
	return 0;
}