洛谷 #1541. 乌龟棋

题意

n个格子中有一些分数，到达一个格子就可以拿到。初始在1，n为终点

有m张牌，保证用完刚好到达终点。牌有向前移动1，2，3，4格四种，求最高分数

题解

dp[j1][j2][j3][j4]表示用j1张‘1’牌，j2张‘2’牌，j3张‘3’牌，j4张‘4’牌能达到的最高分数

调试记录

判断j1 != 0要用if (j1),不是if (!j1)

#include 
#include  
#include 

using namespace std;

int dp[41][41][41][41], n, m, a[355], b[5];

int main(){
	memset(b, 0, sizeof b);
	scanf("%d%d", &n, &m);
	
	for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
	for (int t, i = 1; i <= m; i++){
		scanf("%d", &t); b[t]++;
	}
	
	memset(dp, 0, sizeof(dp));
	dp[0][0][0][0] = a[1];
	
	for (int j1 = 0; j1 <= b[1]; j1++){
		for (int j2 = 0; j2 <= b[2]; j2++){
			for (int j3 = 0; j3 <= b[3]; j3++){
				for (int j4 = 0; j4 <= b[4]; j4++){
					int tmp = 1 + j1 + j2 * 2 + j3 * 3 + j4 * 4;
					if (j1) dp[j1][j2][j3][j4] = max(dp[j1][j2][j3][j4], dp[j1 - 1][j2][j3][j4] + a[tmp]);
					if (j2) dp[j1][j2][j3][j4] = max(dp[j1][j2][j3][j4], dp[j1][j2 - 1][j3][j4] + a[tmp]);
					if (j3) dp[j1][j2][j3][j4] = max(dp[j1][j2][j3][j4], dp[j1][j2][j3 - 1][j4] + a[tmp]);
					if (j4) dp[j1][j2][j3][j4] = max(dp[j1][j2][j3][j4], dp[j1][j2][j3][j4 - 1] + a[tmp]);
				}
			}
		}
	}
	
	printf("%d\n", dp[b[1]][b[2]][b[3]][b[4]]);
	
	return 0;
}