题目:http://www.lydsy.com/JudgeOnline/problem.php?id=1059
YY一下可以发现,假如某个黑色格子用掉了,那么与其同行同列的所有格子都不能用了,而且与其他行列的格子无关,那么就二分图匹配啦~
代码:
#include <cstdio>
#include <algorithm>
#include <cstring>
using namespace std ;
#define MAXN 100010
#define AddEdge( s , t , f ) Add( s , t , f ) , Add( t , s , 0 )
#define inf 0x7fffffff
struct edge {
int next , t , f ;
} E[ MAXN ] ;
int head[ MAXN ] , Index ;
void Add( int s , int t , int f ) {
E[ Index ].t = t , E[ Index ].f = f , E[ Index ].next = head[ s ] ;
head[ s ] = Index ++ ;
}
int node[ MAXN ][ 2 ] , V , S , T ;
int gap[ MAXN ] , h[ MAXN ] , d[ MAXN ] ;
int sap( int v , int flow ) {
if ( v == T ) return flow ;
int rec = 0 ;
for ( int p = d[ v ] ; p != - 1 ; p = E[ p ].next ) {
if ( E[ p ].f && h[ v ] == h[ E[ p ].t ] + 1 ) {
int ret = sap( E[ p ].t , min( flow - rec , E[ p ].f ) ) ;
E[ p ].f -= ret , E[ p ^ 1 ].f += ret , d[ v ] = p ;
if ( ( rec += ret ) == flow ) return flow ;
}
}
if ( ! ( -- gap[ h[ v ] ] ) ) h[ S ] = T ;
gap[ ++ h[ v ] ] ++ , d[ v ] = head[ v ] ;
return rec ;
}
int maxflow( ) {
memset( gap , 0 , sizeof( gap ) ) ;
memset( h , 0 , sizeof( h ) ) ;
for ( int i = 0 ; i ++ < T ; ) d[ i ] = head[ i ] ;
gap[ 0 ] = T ;
int flow = 0 ;
while ( h[ S ] < T ) flow += sap( S , inf ) ;
return flow ;
}
int tot , n ;
int main( ) {
scanf( "%d" , &tot ) ;
while ( tot -- ) {
Index = V = 0 ;
scanf( "%d" , &n ) ;
for ( int i = 0 ; i ++ < n ; ) {
node[ i ][ 0 ] = ++ V , node[ i ][ 1 ] = ++ V ;
}
S = ++ V ; T = ++ V ;
for ( int i = 0 ; i ++ < V ; ) head[ i ] = - 1 ;
for ( int i = 0 ; i ++ < n ; ) {
AddEdge( S , node[ i ][ 0 ] , 1 ) , AddEdge( node[ i ][ 1 ] , T , 1 ) ;
for ( int j = 0 ; j ++ < n ; ) {
int x ; scanf( "%d" , &x ) ;
if ( x ) AddEdge( node[ i ][ 0 ] , node[ j ][ 1 ] , 1 ) ;
}
}
printf( maxflow( ) == n ? "Yes\n" : "No\n" ) ;
}
return 0 ;
}