BZOJ4950 [Wf2017][Mission Improbable] 二分圖匹配

題目鏈接

---

題解：減去行列最大值，再通過二分圖匹配加上多減的邊。

#include <vector>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
#define dnt long long
const 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 ;
}