hihoCoder1159 扑克牌

一道记忆化搜索

原题链接

和着色方案很像，这里就不详细阐述，可以去我博客里的着色方案里看。
但要注意本题不一样的是同种面值的牌花色不同，所以在转移时还需要乘上同种面值的牌的个数。

#include<cstdio>
#include<cstring>
using namespace std;
typedef unsigned long long ull;
const int N = 14;
ull f[N][N][N][N][5];
int v[N], S[5];
int re()
{
	int x = 0;
	char c = getchar();
	bool p = 0;
	for (; c<'0' || c>'9'; c = getchar())
		p = (c == '-' || p) ? 1 : 0;
	for (; c >= '0'&&c <= '9'; c = getchar())
		x = x * 10 + (c - '0');
	return p ? -x : x;
}
int re_l()
{
	char c = getchar();
	for (; (c<'2' || c>'9') && c != 'T'&&c != 'J'&&c != 'Q'&&c != 'K'&&c != 'A'; c = getchar());
	if (c == 'T')
		return 10;
	if (c == 'J')
		return 11;
	if (c == 'Q')
		return 12;
	if (c == 'K')
		return 13;
	if (c == 'A')
		return 1;
	return c - '0';
}
ull dp(int a, int b, int c, int d, int la)
{
	ull s = 0, &k = f[a][b][c][d][la];
	if (k)
		return k;
	if (!(a | b | c | d))
		return 1;
	if (a)
		s += 1llu * (a - (la == 2))*dp(a - 1, b, c, d, 1);
	if (b)
		s += (1llu * (b - (la == 3))*dp(a + 1, b - 1, c, d, 2)) << 1;
	if (c)
		s += 1llu * (c - (la == 4))*dp(a, b + 1, c - 1, d, 3) * 3;
	if (d)
		s += (1llu * d*dp(a, b, c + 1, d - 1, 4)) << 2;
	return k = s;
}
int main()
{
	int i, j, n, t;
	t = re();
	for (j = 1; j <= t; j++)
	{
		n = re();
		memset(v, 0, sizeof(v));
		memset(S, 0, sizeof(S));
		for (i = 1; i <= n; i++)
			v[re_l()]++;
		for (i = 1; i <= 13; i++)
			S[v[i]]++;
		printf("Case #%d: %llu\n", j, dp(S[1], S[2], S[3], S[4], 0));
	}
	return 0;
}