乌龟棋（洛谷P1541）

$乌龟棋$

传送门

题目大意

①：从起点走到终点，每次只会前进不会后退 ②：所有卡片一定用完且用完后正好到达终点 ③：卡片只有四种，且数量最多也只有40 ④：知道使用哪几张卡片后，我们就能推出现在到了哪一个位置

动态规划显然

关于卡片 $a$ 的转移方程

if(a!=0)
	F[a][b][c][d]=max(F[a][b][c][d],F[a-1][b][c][d]+num[r])

其中 $r=1+a+b\times2+c\times3+d\times4$

AC code

#include <bits/stdc++.h>
using namespace std;
int n, m, a[351], r[5], dp[41][41][41][41];
#define getchar() (p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1 << 21, stdin), p1 == p2) ? EOF : *p1++)
char buf[1 << 21], *p1 = buf, *p2 = buf;
inline int read()
{
	char c = getchar();
	int x = 0;
	bool f = 0;
	for (; !isdigit(c); c = getchar())
		f ^= !(c ^ 45);
	for (; isdigit(c); c = getchar())
		x = (x << 1) + (x << 3) + (c ^ 48);
	if (f)
		x = -x;
	return x;
}
int main()
{
	n = read();
	m = read();
	for (int i = 1; i <= n; i++)
	{
		a[i] = read();
	}
	for (int i = 1; i <= m; i++)
	{
		int b;
		b = read();
		r[b]++;
	}
	dp[0][0][0][0] = a[1];
	for (register int r1 = 0; r1 <= r[1]; r1++)
		for (register int r2 = 0; r2 <= r[2]; r2++)
			for (register int r3 = 0; r3 <= r[3]; r3++)
				for (register int r4 = 0; r4 <= r[4]; r4++)
				{
					int j = 1 + r1 * 1 + r2 * 2 + r3 * 3 + r4 * 4;
					int mmax = 0;
					if (r1 != 0)
						mmax = max(mmax, dp[r1 - 1][r2][r3][r4] + a[j]);
					if (r2 != 0)
						mmax = max(mmax, dp[r1][r2 - 1][r3][r4] + a[j]);
					if (r3 != 0)
						mmax = max(mmax, dp[r1][r2][r3 - 1][r4] + a[j]);
					if (r4 != 0)
						mmax = max(mmax, dp[r1][r2][r3][r4 - 1] + a[j]);
					dp[r1][r2][r3][r4] = max(dp[r1][r2][r3][r4], mmax);
				}
	cout << dp[r[1]][r[2]][r[3]][r[4]];
	return 0;
}