ZOJ 3209

精确覆盖

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
using namespace std;

const int maxn=920;
const int maxnode=920*550;
const int maxr=550;
int ans;
struct DLX
{
  int n , sz;                                                 // 行数，节点总数
  int S[maxn];                                                // 各列节点总数
  int row[maxnode],col[maxnode];                              // 各节点行列编号
  int L[maxnode],R[maxnode],U[maxnode],D[maxnode];            // 十字链表
 
  int ansd;                                         // 解
 
  void init(int n )
  {
    this->n = n ;
    for(int i = 0 ; i <= n; i++ )
    {
      U[i] = i ;
      D[i] = i ;
      L[i] = i - 1;
      R[i] = i + 1;
    }
    R[n] = 0 ;
    L[0] = n;
    sz = n + 1 ;
    memset(S,0,sizeof(S));
  }
  void addRow(int r,vector<int> c1)
  {
    int first = sz;
    for(int i = 0 ; i < c1.size(); i++ ){
      int c = c1[i];
      L[sz] = sz - 1 ; R[sz] = sz + 1 ; D[sz] = c ; U[sz] = U[c];
      D[U[c]] = sz; U[c] = sz;
      row[sz] = r; col[sz] = c;
      S[c] ++ ; sz ++ ;
    }
    R[sz - 1] = first ; L[first] = sz - 1;
  }
  // 顺着链表A，遍历除s外的其他元素
  #define FOR(i,A,s) for(int i = A[s]; i != s ; i = A[i])
 
  void remove(int c){
    L[R[c]] = L[c];
    R[L[c]] = R[c];
    FOR(i,D,c)
      FOR(j,R,i) {U[D[j]] = U[j];D[U[j]] = D[j];--S[col[j]];}
  }
  void restore(int c){
    FOR(i,U,c)
      FOR(j,L,i) {++S[col[j]];U[D[j]] = j;D[U[j]] = j; }
    L[R[c]] = c;
    R[L[c]] = c;
  }
  void dfs(int d){
  	if(d>=ans) return ;
    if(R[0] == 0 ){
      ansd = d;
      ans=min(ans,ansd);
    }
    // 找S最小的列c
    int c = R[0] ;
    FOR(i,R,0) if(S[i] < S[c]) c = i;
 
    remove(c);
    FOR(i,D,c){
      FOR(j,R,i) remove(col[j]);
      dfs(d + 1);
      FOR(j,L,i) restore(col[j]);
    }
    restore(c);
  }
  bool solve(){
    dfs(0);
  }
};

DLX solver;

int main(){
	int T; int n,m ,p; int xx1,xx2,yy1,yy2;
	vector<int>colmuns;
	scanf("%d",&T);
	while(T--){
		ans=(1<<30);
		scanf("%d%d%d",&n,&m,&p);
		solver.init(n*m);
		for(int k=1;k<=p;k++){
			colmuns.clear();
			scanf("%d%d%d%d",&xx1,&yy1,&xx2,&yy2);
			for(int i=xx1+1;i<=xx2;i++){
				for(int j=yy1+1;j<=yy2;j++){
					colmuns.push_back((i-1)*m+j);
				}
			}
			solver.addRow(k,colmuns);
		}
		solver.solve();
		if(ans==(1<<30)) printf("-1\n");
		else printf("%d\n",ans);
	}
	return 0;
}