【BZOJ 1087】【SCOI 2005】互不侵犯King

http://www.lydsy.com/JudgeOnline/problem.php?id=1087
很简单的状压，需要预处理，我两个状态可不可以挨着的预处理出错WA了好几次。
这个位运算预处理好神奇啊

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
typedef long long ll;
const int N = 10;
int in() {
	int k = 0; char c = getchar();
	for (; c < '0' || c > '9'; c = getchar());
	for (; c >= '0' && c <= '9'; c = getchar())
		k = k * 10 + c - 48;
	return k;
}

bool can[1 << 9], c[1 << 9][1 << 9];
ll f[N][N * N][1 << 9];
int tot, n, K, num[1 << 9];
bool flag;

int main() {
	n = in(); K = in();
	
	tot = (1 << n) - 1;
	for (int i = 0; i <= tot; ++i)
		if (can[i] = !(i & (i >> 1)))
			for (int j = 0; j < n; ++j)
				if ((1 << j) & i)
					++num[i];
	for (int i = 0; i < tot; ++i)
		if (can[i])
			for (int j = i + 1; j <= tot; ++j)
				if (can[j])
					c[i][j] = c[j][i] = !(i & j) && !(i & (j >> 1)) && !(j & (i >> 1));
	c[0][0] = true;
	
	f[0][0][0] = 1;
	for (int i = 1; i <= n; ++i)
		for (int j = min(K, i * n); j >= 0; --j)
			for (int k = 0; k <= tot; ++k)
				if (can[k])
					for (int l = 0; l <= tot; ++l)
						if (can[l] && c[k][l] && num[l] <= j)
							f[i][j][l] += f[i - 1][j - num[l]][k];
	
	ll ans = 0;
	for(int i = 0; i <= tot; ++i)
		ans += f[n][K][i];
	printf("%lld\n", ans);
	return 0;
}