美文网首页信息学竞赛题解(IO题解)
BZOJ-1565: [NOI2009]植物大战僵尸(Tarja

BZOJ-1565: [NOI2009]植物大战僵尸(Tarja

作者: AmadeusChan | 来源:发表于2018-10-16 20:43 被阅读0次

    题目:http://www.lydsy.com/JudgeOnline/problem.php?id=1565

    2014刷的第一道题~~

    裸的最小割模型,但是我们发现有些点相互依赖,这个在直接建图之后是无法割去的,那么求一次SCC,对于每个大小大于1的SCC的点,连到汇点,连一无穷大的边,表示强制割掉,然后跑最大流。

    代码:

    #include<cstdio>
    
    #include<algorithm>
    
    #include<cstring>
    
    #include<stack>
    
     
    
    using namespace std ;
    
     
    
    #define MAXN 40
    
    #define inf 0x7fffffff
    
    #define MAXV 2010
    
     
    
    struct edge{
    
                edge *next ,*pair ;
    
                int t , f ;
    
    };
    
     
    
    struct network{
    
               
    
                edge *head[ MAXV ];
    
               
    
                network(  ){
    
                            memset( head ,0,sizeof( head ));
    
                }
    
               
    
                void Add(int s ,int t ,int f ){
    
                            edge *p =new( edge );
    
                            p -> t = t , p -> f = f , p -> next = head[ s ];
    
                            head[ s ]= p ;
    
                }
    
               
    
                void AddEdge(int s ,int t ,int f ){
    
                            Add( s , t , f ),Add( t , s ,0);
    
                            head[ s ]-> pair = head[ t ], head[ t ]-> pair = head[ s ];
    
                }
    
               
    
                int h[ MAXV ], gap[ MAXV ], S , T ;
    
                edge *d[ MAXV ];
    
               
    
                int sap(int v ,int flow ){
    
                            if( v == T )return flow ;
    
                            int rec =0;
    
                            for( edge *p = d[ v ]; p ; p = p -> next )if( p -> f && h[ v ]== h[ p -> t ]+1){
    
                                         int ret =sap( p -> t ,min( flow - rec , p -> f ));
    
                                         p -> f -= ret , p -> pair -> 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(  ){
    
                            int rec =0;
    
                            for(int i =0; i ++< T ;) h[ i ]= gap[ i ]=0, d[ i ]= head[ i ];
    
                            gap[ S ]= T ;
    
                            while( h[ S ]< T ) rec +=sap( S , inf );
    
                            return rec ;
    
                }
    
               
    
    } net ;
    
     
    
    int node[ MAXN ][ MAXN ], V =0, n , m , sum =0, ws[ MAXV ];
    
     
    
    stack <int> S ;
    
    int dfn[ MAXV ], low[ MAXV ], scc[ MAXV ], size[ MAXV ], cnt =0;
    
    bool f[ MAXV ];
    
     
    
    void Tarjan(int v ){
    
                dfn[ v ]= low[ v ]=++ cnt ;
    
                S.push( v ), f[ v ]=true;
    
                for( edge *p = net.head[ v ]; p ; p = p -> next )if( p -> f ){
    
                            if(! dfn[ p -> t ]){
    
                                         Tarjan( p -> t );
    
                                         low[ v ]=min( low[ v ], low[ p -> t ]);
    
                            }else if( f[ p -> t ]) low[ v ]=min( low[ v ], low[ p -> t ]);
    
                }
    
                if( low[ v ]== dfn[ v ]){
    
                            int u ;
    
                            do{
    
                                         u = S.top(  ); S.pop(  );
    
                                         f[ u ]=false, scc[ u ]= v , size[ v ]++;
    
                            }while( u != v );
    
                }
    
    }
    
     
    
    int main(  ){
    
                scanf("%d%d",&n ,&m );
    
                for(int i =0; i < n ; i ++)for(int j =0; j < m ; j ++) node[ i ][ j ]=++ V ;
    
                net.S =++ V ; net.T =++ V ;
    
                for(int i =0; i < n ; i ++)for(int j =0; j < m ; j ++){
    
                            int x , r , c ;scanf("%d",&x ); ws[ node[ i ][ j ]]= x ;
    
                            if( x >0) net.AddEdge( net.S , node[ i ][ j ], x ), sum += x
    
                            ;else net.AddEdge( node[ i ][ j ], net.T ,- x );
    
                            scanf("%d",&x );
    
                            while( x --){
    
                                         scanf("%d%d",&r ,&c );
    
                                         net.AddEdge( node[ r ][ c ], node[ i ][ j ], inf );
    
                            }
    
                }
    
                for(int i =0; i < n ; i ++)for(int j =0; j < m -1; j ++){
    
                            net.AddEdge( node[ i ][ j ], node[ i ][ j +1], inf );
    
                }
    
                memset( dfn ,0,sizeof( dfn ));
    
                memset( f ,false,sizeof( f ));
    
                memset( size ,0,sizeof( size ));
    
                for(int i =0; i ++< net.S -1;)if(! dfn[ i ]){
    
                            Tarjan( i );
    
                }
    
                for(int i =0; i ++< net.S -1;)if( size[ scc[ i ]]>1){
    
                            net.AddEdge( i , net.T , inf );
    
                }
    
                printf("%d\n", sum - net.maxflow(  ));
    
                return 0;
    
    }
    

    相关文章

      网友评论

        本文标题:BZOJ-1565: [NOI2009]植物大战僵尸(Tarja

        本文链接:https://www.haomeiwen.com/subject/unlkzftx.html