312. 乌龟棋 - AcWing题库
/* 状态表示：f[b1,b2,b3,b4]表示所有第 i种卡片使用了 bi张的走法的最大分值。状态计算：将 f[b1,b2,b3,b4]表示的所有走法按最后一步选择哪张卡片分成四类：第 i类为最后一步选择第 i种卡片。比如 i=2，则这一类的最大分值是 f[b1,b2−1,b3,b4]+score[b1+2b2+3b3+4b4]
*/
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>using namespace std;const int N = 360, M = 41;int n, m;
int score[N];
int f[M][M][M][M];int main()
{int b[5] = {0};scanf("%d%d", &n, &m);for (int i = 0; i < n; i ++ ) scanf("%d", &score[i]);for (int i = 0; i < m; i ++ ){int t;scanf("%d", &t);b[t] ++ ;}for (int A = 0; A <= b[1]; A ++ )for (int B = 0; B <= b[2]; B ++ )for (int C = 0; C <= b[3]; C ++ )for (int D = 0; D <= b[4]; D ++ ){int t = score[A + B * 2 + C * 3 + D * 4];int &v = f[A][B][C][D];v = t;if (A) v = max(v, f[A - 1][B][C][D] + t);if (B) v = max(v, f[A][B - 1][C][D] + t);if (C) v = max(v, f[A][B][C - 1][D] + t);if (D) v = max(v, f[A][B][C][D - 1] + t);}printf("%d\n", f[b[1]][b[2]][b[3]][b[4]]);return 0;
}
