网络流24题 洛谷 3355 骑士共存
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#include <bits/stdc++.h> using namespace std ; const int mx [ 9 ] = { 2 , 2 , -2 , -2 , -1 , 1 , -1 , 1 } ; const int my [ 9 ] = { -1 , 1 , -1 , 1 , 2 , 2 , -2 , -2 } ; const int N = 100000 + 10 , inf = 1e8 + 7 ; queue < int > q ; int hh [ N << 2 ] , head [ N << 2 ] , nxt [ N << 2 ] , to [ N << 2 ] , cn = 1 ; int flt [ N << 2 ] , dis [ N ] ; int src , sink , x , y , n , m ; bool vis [ N ] , tag [ 500 ] [ 500 ] ; int minx ( int a , int b ) { return a > b ? b : a ; } void create ( int u , int v , int f ) { cn ++ ; to [ cn ] = v ; flt [ cn ] = f ; nxt [ cn ] = head [ u ] ; head [ u ] = cn ; cn ++ ; to [ cn ] = u ; flt [ cn ] = 0 ; nxt [ cn ] = head [ v ] ; head [ v ] = cn ; } bool bfs ( ) { memset ( vis , 0 , sizeof ( vis ) ) ; memset ( dis , 0 , sizeof ( dis ) ) ; q . push ( src ) ; vis [ src ] = 1 ; while ( ! q . empty ( ) ) { int tmp = q . front ( ) ; q . pop ( ) ; for ( int i = head [ tmp ] ; i ; i = nxt [ i ] ) { int v = to [ i ] ; if ( flt [ i ] && ! vis [ v ] ) { dis [ v ] = dis [ tmp ] + 1 ; q . push ( v ) ; vis [ v ] = true ; } } } return vis [ sink ] ; } int dinic ( int u , int delta ) { if ( u == sink ) return delta ; int res = 0 , v ; for ( int i = hh [ u ] ; i && delta ; i = nxt [ i ] ) { v = to [ i ] ; if ( flt [ i ] && dis [ v ] == dis [ u ] + 1 ) { int dd = dinic ( v , minx ( delta , flt [ i ] ) ) ; flt [ i ] -= dd ; flt [ i ^ 1 ] += dd ; res += dd ; delta -= dd ; hh [ u ] = i ; } } return res ; } int main ( ) { scanf ( "%d%d" , & n , & m ) ; src = 0 ; sink = n * n + 1 ; for ( int i = 1 ; i <= m ; i ++ ) { scanf ( "%d%d" , & x , & y ) ; tag [ x ] [ y ] = 1 ; } for ( int i = 1 ; i <= n ; i ++ ) for ( int j = 1 ; j <= n ; j ++ ) { if( tag [ i ] [ j ] ) continue ; int id = ( i - 1 ) * n + j ; if( ( i + j ) & 1 ) { create ( src , id , 1 ) ; for( int k = 0 ; k <= 7 ; k ++ ) { int t1 = i + mx [ k ] , t2 = j + my [ k ] ; if ( t1 <= 0 || t1 > n || t2 <= 0 || t2 > n || tag [ t1 ] [ t2 ] ) continue ; create ( id , ( t1 - 1 ) * n + t2 , inf ) ; } } else create ( id , sink , 1 ); } int ans = 0 ; while ( bfs ( ) ) { ans += dinic ( src , inf ) ; for ( int i = 0 ; i <= n * n + 1 ; i ++ ) hh [ i ] = head [ i ] ; } printf ( "%d" , n * n - m - ans ) ; return 0 ; }
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