FZU 1686 神龙的难题（DLX反复覆盖）

FZU 1686 神龙的难题

题意：中文题

思路：每个1看成列，每个位置作为左上角的矩阵看成行。dlx反复覆盖就可以

代码：

#include <cstdio>

#include <cstring>

using namespace std;

const int MAXNODE = 66666;

const int INF = 0x3f3f3f3f;

const int MAXM = 230;

const int MAXN = 230;

int K;

struct DLX {

	int n,m,size;

	int U[MAXNODE], D[MAXNODE], R[MAXNODE], L[MAXNODE], row[MAXNODE], col[MAXNODE];

	int H[MAXN], S[MAXM];

	int ansd, ans[MAXN];

	void init(int n,int m) {

		this->n = n;

		this->m = m;

		ansd = INF;

		for(int i = 0; i <= m; i++) {

			S[i] = 0;

			U[i] = D[i] = i;

			L[i] = i - 1;

			R[i] = i + 1;

		}

		R[m] = 0; L[0] = m;

		size = m;

		for(int i = 1; i <= n; i++)

			H[i] = -1;

	}

	void Link(int r,int c) {

		++S[col[++size] = c];

		row[size] = r;

		D[size] = D[c];

		U[D[c]] = size;

		U[size] = c;

		D[c] = size;

		if(H[r] < 0) H[r] = L[size] = R[size] = size;

		else {

			R[size] = R[H[r]];

			L[R[H[r]]] = size;

			L[size] = H[r];

			R[H[r]] = size;

		}

	}

	void remove(int c) {

		for(int i = D[c]; i != c; i = D[i]) {

			L[R[i]] = L[i];

			R[L[i]] = R[i];

		}

	}

	void resume(int c) {

		for(int i = U[c]; i != c; i = U[i])

			L[R[i]] = R[L[i]] = i;

	}

	bool v[MAXNODE];

	int f() {

		int ret = 0;

		for(int c = R[0]; c != 0; c = R[c]) v[c] = true;

		for(int c = R[0]; c != 0; c = R[c]) {

			if(v[c]) {

				ret++;

				v[c] = false;

				for(int i = D[c]; i != c; i = D[i])

					for(int j = R[i]; j != i; j = R[j])

						v[col[j]] = false;

			}

		}

		return ret;

	}

	void Dance(int d) {

		if(d + f() >= ansd) return;

		if(R[0] == 0) {

			if (d < ansd) ansd = d;

			return;

		}

		int c = R[0];

		for(int i = R[0]; i != 0; i = R[i]) {

			if(S[i] < S[c])

				c = i;

		}

		for(int i = D[c]; i != c; i = D[i]) {

			remove(i);

			for(int j = R[i]; j != i; j = R[j]) remove(j);

			ans[d] = row[i];

			Dance(d + 1);

			for(int j = L[i]; j != i; j = L[j]) resume(j);

			resume(i);

		}

	}

} gao;

const int N = 20;

int n, m, g[N][N], n1, m1, to[N][N];

int main() {

	while (~scanf("%d%d", &n, &m)) {

		int cnt = 0;

		for (int i = 0; i < n; i++)

			for (int j = 0; j < m; j++) {

				scanf("%d", &g[i][j]);

				if (g[i][j]) to[i][j] = ++cnt;

			}

		scanf("%d%d", &n1, &m1);

		gao.init(n * m, cnt);

		for (int i = 0; i < n; i++) {

			for (int j = 0; j < m; j++) {

				for (int x = 0; x < n1 && i + x < n; x++) {

					for (int y = 0; y < m1 && j + y < m; y++) {

						if (g[i + x][j + y]) gao.Link(i * m + j + 1, to[i + x][j + y]);

					}

				}

			}

		}

		gao.Dance(0);

		printf("%d\n", gao.ansd);

	}

	return 0;

}