【BZOJ1806】[Ioi2007]Miners 礦工配餐【DP】

【題目鏈接】

設dp[i][a1][a2][b1][b2]表示，前i個食物，第一個礦場最後兩次的食物分別爲a1、a2，b1、b2同理，所得到的最大煤炭數。

/* Think Thank Thunk */
#include <cstdio>
#include <cstring>
#include <algorithm>
#define rec(i, x, y) for(int i = x; i <= y; i++)

using namespace std;

int n, dp[2][4][4][4][4];

inline int cread() {
	char ch = getchar();
	for(; ch != 'M' && ch != 'F' && ch != 'B'; ch = getchar());
	if(ch == 'M') return 1;
	if(ch == 'F') return 2;
	if(ch == 'B') return 3;
}

inline int calc(int a, int b, int c) {
	int res = 1;
	if(a != 0 && a != b && a != c) res++;
	if(b != 0 && b != c) res++;
	return res;
}

int main() {
	scanf("%d", &n);
	memset(dp, -1, sizeof(dp));
	dp[0][0][0][0][0] = 0;

	rec(i, 0, n - 1) {
		int t = cread();
		rec(a1, 0, 3) rec(a2, 0, 3) rec(b1, 0, 3) rec(b2, 0, 3) {
			if(!~dp[i & 1][a1][a2][b1][b2]) continue;
			dp[~i & 1][a2][t][b1][b2] = max(dp[~i & 1][a2][t][b1][b2], dp[i & 1][a1][a2][b1][b2] + calc(a1, a2, t));
			dp[~i & 1][a1][a2][b2][t] = max(dp[~i & 1][a1][a2][b2][t], dp[i & 1][a1][a2][b1][b2] + calc(b1, b2, t));
		}
	}

	int ans = 0;
	rec(a1, 0, 3) rec(a2, 0, 3) rec(b1, 0, 3) rec(b2, 0, 3)
		ans = max(ans, dp[n & 1][a1][a2][b1][b2]);
	printf("%d\n", ans);
	return 0;
}