FZU 1686 神龙的难题（DLX反复覆盖）

FZU 1686 神龙的难题

题目链接

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
", gao.ansd);
	}
	return 0;
}