BZOJ4950 [Wf2017][Mission Improbable] 二分图匹配

题目链接

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题解：减去行列最大值，再通过二分图匹配加上多减的边。

#include <vector>#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>using namespace std;#define dnt long longconst int N = 105;int mp [N] [N] , sl [N] , sr [N] , lk [N] , vis [N] ;int n , m ;vector <int> G [N] ;int find ( int u ){    for ( int i = 0 ; i < G [u].size () ; ++ i ){        int v = G [u] [i] ;        if ( ! vis [v] ) {            vis [v] = 1;            if ( lk [v] == -1 || find ( lk [v] ) ){                lk [v] = u ;                return 1 ;            }        }    }    return 0 ;}int main (){    scanf ( "%d%d" , &n , &m ) ;    dnt ans = 0;    for ( int i = 1 ; i <= n ; ++ i )        for ( int j = 1 ; j <= m ; ++ j ){            scanf ( "%d" , & mp [i] [j] );            sr [i] = max ( sr [i] , mp [i] [j] );            sl [j] = max ( sl [j] , mp [i] [j] );            ans += mp [i] [j] ;        }//  for ( int i = 1 ; i <= m ; ++ i ) printf ( "%d\n" , sl [i] );    for ( int i = 1 ; i <= n ; ++ i )        for ( int j = 1 ; j <= m ; ++ j ) if ( mp [i] [j] ){            ans -- ;            if ( sr[i] > 1 && sr [i] == sl [j] )                 G [i].push_back ( j );        }    for ( int i = 1 ; i <= n ; ++ i )         if ( sr [i] ) ans -= sr [i] - 1 ;    for ( int i = 1 ; i <= m ; ++ i )         if ( sl [i] ) ans -= sl [i] - 1 ;//  printf ( "%d\n" , ans );    memset ( lk , -1 , sizeof ( lk ) ) ;    for ( int i = 1 ; i <= n ; ++ i ) {        if ( sr [i] ) memset ( vis , 0 , sizeof ( vis ) ) , ans += find ( i ) * ( sr [i] - 1 ) ;    }           printf ( "%lld" , ans ) ;    return 0 ;}