洛谷P3886 [JLOI2009]神秘的生物(插头dp)

洛谷P3886 [JLOI2009]神秘的生物(插头dp)

题目大意

求带权正方形棋盘中最大权联通块

数据范围

[1 le n le 9 ]

解题思路

自己做出来的第一道插头 dp，不过很简单就是了

由于这道题维护的是连通性，所以要用最小表示法记录一下每个点所在的联通块编号，然后就是轮廓线保留 n 个格子即可，因为插头处只用看看左边有没有块

那么就分为两种情况了，一种是不放插头，一种是放一个插头

不放插头要看看是不是会导致图不连通，放插头要将左和上的两个格子的联通块合并

代码

const int N = 5050;
const int P = 3592;
int h[N], ne[N], st[2][N], tot[2], val[2][N], nw, pr;
void insert(int bit, int num) {
//	printf ("%d %d
", bit, num);
	int x = bit % P + 1;
	for (int i = h[x]; i; i = ne[i]) 
		if (st[nw][i] == bit) return Mx(val[nw][i], num);
	ne[++tot[nw]] = h[x], st[nw][h[x] = tot[nw]] = bit, val[nw][tot[nw]] = num;
}

int vis[9], ans = -1e6, n;
int reb(int bit) {
	int now = 0, mx = 0;
	memset(vis, 0, sizeof(vis));
	for (int i = 0;i < n; i++) {
		int p = bit >> (i * 3) & 7;
		if (!p) continue;
		if (!vis[p]) vis[p] = ++mx;
		now |= vis[p] << (i * 3);
	}
	return now;
}

void update(int now, int val) {
	for (int l = 0;l < n; l++)
		if ((now >> (l * 3) & 7) > 1) return; 
	Mx(ans, val);
}

int a[50][50];
int main() {
	read(n);
	for (int i = 1;i <= n; i++)
		for (int j = 1;j <= n; j++) 
			read(a[i][j]), Mx(ans, a[i][j]);
	if (ans <= 0) return write(ans), 0;
	insert(0, 0); 
	for (int i = 1;i <= n; i++) {
		for (int j = 1;j <= n; j++) {
			pr = nw, nw = nw ^ 1, tot[nw] = 0;
			memset(h, 0, sizeof(h));
			for (int k = 1;k <= tot[pr]; k++) {
				int now = st[pr][k], V = val[pr][k];
				int b1 = (j == 1) ? 0 : now >> ((j - 2) * 3) & 7; 
				int b2 = now >> ((j - 1) * 3) & 7;
				update(now, V);
				int cc = 0;
				for (int l = 0;l < n; l++) {
					int pp = now >> (l * 3) & 7;
					if (pp == b2) cc++;
				}
				if (cc > 1 || b2 == 0) insert(now ^ (b2 << ((j-1) * 3)), V);
				for (int l = 0;l < n; l++) {
					int pp = now >> (l * 3) & 7;
					if (pp == 0) continue;
					if (pp == b1 || pp == b2) 
						now ^= (7 ^ pp) << (l * 3);
				}
				if (b2 == 0) now |= 7 << ((j-1) * 3);
				insert(reb(now), V + a[i][j]);
			}
		}
	}
	for (int i = 1;i <= tot[nw]; i++)
		update(st[nw][i], val[nw][i]);
	write(ans);
	return 0;
}