zoj3209 精确覆盖

题目：有一个n*m的藏宝图，已经碎了，现在你有p个矩形，你知道p个矩形的左下角坐标x1,y1，右上角坐标x2,y2，让你从中选择最少的矩形重新构成这个藏宝图

思路：将p个矩形看成行，n*m个坐标看成列

代码：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<string>
#include<vector>
#include<map>
#include<set>
#include<queue>
#include<stack>
#include<list>
#include<numeric>
using namespace std;
#define PI acos(-1.0)
#define LL long long
#define ULL unsigned long long
#define INF 0x3f3f3f3f
#define mm(a,b) memset(a,b,sizeof(a))
#define PP puts("*********************");
template<class T> T f_abs(T a){ return a > 0 ? a : -a; }
template<class T> T gcd(T a, T b){ return b ? gcd(b, a%b) : a; }
template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;}
// 0x3f3f3f3f3f3f3f3f
// 0x3f3f3f3f

int limit;//重复覆盖时使用，行数选择的最大数目
struct DLX {
	const static int ROW = 1003, COL = 1003, SIZE = ROW * COL;
	int L[SIZE], R[SIZE], U[SIZE], D[SIZE];  //模拟指针
	int col[SIZE], row[SIZE]; //所在列 所在行
	int visn, visited[COL]; //用于估价函数
	int sel[ROW], seln; //选择的行
	int sz[COL]; //列元素数
	int total/*节点编号*/, H[ROW];
	void init(int clen) { //初始化列头指针
		for(int i = 0; i <= clen; ++i) {
			L[i] = i - 1; R[i] = i + 1;
			U[i] = D[i] = i; sz[i] = 0;
		}
		for(int i=0;i<ROW;i++) H[i]=-1;
		for(int i=0;i<COL;i++) visited[i]=0; visn = 0; //用于重复覆盖的A*剪枝
		L[0] = clen; R[clen] = 0; total = clen + 1;
	}
	void link(int r, int c) {
		U[total] = c; D[total] = D[c];
		U[D[c]] = total; D[c] = total;
		if(H[r] < 0) H[r] = L[total] = R[total] = total;
		else {
			L[total] = H[r]; R[total] = R[H[r]];
			L[R[H[r]]] = total; R[H[r]] = total;
		}
		sz[c]++; col[total] = c; row[total++] = r;
	}
//-------------------- 精确覆盖 -------------------------
	void remove(const int &c) { //删除c列上所有1元素所在的行
		L[R[c]] = L[c]; R[L[c]] = R[c];
		for(int i = D[c]; i != c; i = D[i])
			for(int j = R[i]; j != i; j = R[j])
				U[D[j]] = U[j], D[U[j]] = D[j], --sz[col[j]];
	}
	void resume(const int &c) { //恢复c列上所有1元素所在的行
		R[L[c]] = L[R[c]] = c;
		for(int i = U[c]; i != c; i = U[i])
			for(int j = L[i]; j != i; j = L[j])
				++sz[col[U[D[j]] = D[U[j]] = j]];
	}
	void dance(int now) {
		if(R[0] == 0) {
            if(seln==-1) seln=now;
			seln = min(seln,now);
			return;
		}
		int c=R[0], i, j;
		for(i = R[0]; i != 0; i = R[i]) //选择元素最少的列c
			if(sz[c] > sz[i]) c = i;
		remove(c);
		for(i = D[c]; i != c; i = D[i]) { //删除与c相连的行i
			for(j = R[i]; j != i; j = R[j]) //删除行元素所在的列j
				remove(col[j]);
			sel[now] = row[i]; //选择此行 保存行号
			dance(now+1);
			for(j = L[i]; j != i; j = L[j])
				resume(col[j]);
		}
		resume(c);
	}
//-------------------- 重复覆盖 -------------------------
	void _remove(const int &c) {
		for(int i = D[c]; i != c; i = D[i])
			L[R[i]] = L[i], R[L[i]] = R[i];
	}
	void _resume(const int &c) {
		for(int i = U[c]; i != c; i = U[i])
			L[R[i]] = R[L[i]] = i;
	}
	int Astar(){ //估价函数
		int res = 0; ++visn;
		for(int i = R[0]; i != 0; i = R[i]) {
			if(visited[i] != visn) {
				++res; visited[i] = visn;
				for(int j = D[i]; j != i; j = D[j])
					for(int k = R[j]; k != j; k = R[k])
						visited[col[k]] = visn;
			}
		} return res;
	}
	bool _dance(int now) {
	    if(now + Astar() > limit) return false;
		if(R[0] == 0) return now<=limit;
		int c=R[0], i, j;
		for(i = R[0]; i != 0; i = R[i]) //选择元素最少的列c
			if(sz[c] > sz[i]) c = i;
		for(i = D[c]; i != c; i = D[i]) {
			_remove(i);
			for(j = R[i]; j != i; j = R[j])
				_remove(j);
			if(_dance(now + 1)) { //对于不同的题 这个地方常常需要改动
				return true;
			}
			for(j = L[i]; j != i; j = L[j])
				_resume(j);
			_resume(i);
		}
		return false;
	}
}dl;
int main(){

    int T,n,m,p,x1,y1,x2,y2;
    scanf("%d",&T);
    while(T--){
        scanf("%d%d%d",&n,&m,&p);
        dl.init(n*m);
        for(int r=1;r<=p;r++){
            scanf("%d%d%d%d",&x1,&y1,&x2,&y2);
            for(int i=x1+1;i<=x2;i++)
                for(int j=y1+1;j<=y2;j++){
                    int c=(i-1)*m+j;
                    dl.link(r,c);
                }
        }
        dl.seln=-1;
        dl.dance(0);
        printf("%d\n",dl.seln);
    }
    return 0;
}