【HDU】3209 Treasure Map 精确覆盖

传送门：【HDU】3209 Treasure Map

题目分析：精确覆盖模板题。。。将矩形的行*列转化为列。

代码如下：

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

#define REP( i , a , b ) for ( int i = a ; i < b ; ++ i )
#define REV( i , a , b ) for ( int i = a - 1 ; i >= b ; -- i )
#define FOR( i , a , b ) for ( int i = a ; i <= b ; ++ i )
#define FOV( i , a , b ) for ( int i = a ; i >= b ; -- i )
#define REC( i , A , o ) for ( int i = A[o] ; i != o ; i = A[i] )
#define CLR( a , x ) memset ( a , x , sizeof a )

const int MAXN = 1000 ;
const int MAXNODE = 460000 ;
const int INF = 0x3f3f3f3f ;

struct DLX {
	int U[MAXNODE] , D[MAXNODE] , L[MAXNODE] , R[MAXNODE] ;
	int row[MAXNODE] , col[MAXNODE] ;
	int S[MAXN] , H[MAXN] ;
	int deep , ans[MAXN] ;
	int n , m , q ;
	int size ;
	
	void init () {
		CLR ( H , -1 ) ;
		FOR ( i , 0 , n ) {
			S[i] = 0 ;
			U[i] = i ;
			D[i] = i ;
			L[i] = i - 1 ;
			R[i] = i + 1 ;
		}
		L[0] = n ;
		R[n] = 0 ;
		size = n ;
		deep = INF ;
	}
	
	void link ( int r , int c ) {
		++ size ;
		++ S[c] ;
		row[size] = r ;
		col[size] = c ;
		U[size] = U[c] ;
		D[size] = c ;
		D[U[c]] = size ;
		U[c] = size ;
		if ( ~H[r] ) {
			L[size] = L[H[r]] ;
			R[size] = H[r] ;
			L[R[size]] = size ;
			R[L[size]] = size ;
		}
		else
			H[r] = L[size] = R[size] = size ;
	}
	
	void remove ( int c ) {
		L[R[c]] = L[c] ;
		R[L[c]] = R[c] ;
		REC ( i , D , c )
			REC ( j , R , i ) {
				U[D[j]] = U[j] ;
				D[U[j]] = D[j] ;
				-- S[col[j]] ;
			}
	}
	
	void resume ( int c ) {
		REC ( i , U , c )
			REC ( j , L , i ) {
				++ S[col[j]] ;
				D[U[j]] = j ;
				U[D[j]] = j ;
			}
		R[L[c]] = c ;
		L[R[c]] = c ;
	}
	
	void dance ( int d ) {
		if ( R[0] == 0 ) {
			deep = min ( deep , d ) ;
			return ;
		}
		int c = R[0] ;
		REC ( i , R , 0 )
			if ( S[c] > S[i] )
				c = i ;
		remove ( c ) ;
		REC ( i , D , c ) {
			//ans[d] = row[i] ;
			REC ( j , R , i )
				remove ( col[j] ) ;
			dance ( d + 1 ) ;
			REC ( j , L , i )
				resume ( col[j] ) ;
		}
		resume ( c ) ;
	}
	
	void solve () {
		int x1 , x2 , y1 , y2 ;
		scanf ( "%d%d%d" , &n , &m , &q ) ;
		n *= m ;
		init () ;
		FOR ( r , 1 , q ) {
			scanf ( "%d%d%d%d" , &x1 , &y1 , &x2 , &y2 ) ;
			FOR ( i , x1 + 1 , x2 )
				FOR ( j , y1 + 1 , y2 )
					link ( r , ( i - 1 ) * m + j ) ;
		}
		dance ( 0 ) ;
		printf ( "%d\n" , deep == INF ? -1 : deep ) ;
	}
} dlx ;

int main () {
	int T ;
	scanf ( "%d" , &T ) ;
	while ( T -- )
		dlx.solve () ;
	return 0 ;
}