Cod sursa(job #2392807)

Utilizator Andrei-27Arhire Andrei Andrei-27 Data 30 martie 2019 14:16:34
Problema Flux maxim de cost minim Scor 70
Compilator cpp-64 Status done
Runda Arhiva educationala Marime 2.94 kb
/*
    ** Code by Andrei Arhire
    ** Tecuci , Galati
*/
#include <bits/stdc++.h>
#define pb push_back
#define s second
#define f first
#define it vector < int > :: iterator
using namespace std ;
const int NR = 355 , oo = ( 1 << 30 ) ;
ifstream in ("fmcm.in") ;
ofstream out ("fmcm.out") ;
int n , m , source , sink , capacity , z , x , y ;
int64_t ans ;
int cost [ NR ][ NR ] , cap [ NR ][ NR ] , t [ NR ] ;
vector < int > v [ NR ] , d_old ( NR , oo ) , d_dij ( NR , oo ) , d_new ( NR , oo ) ;
struct cmp  {
inline bool operator ()( int i , int j )    {
return d_dij [ i ] > d_dij [ j ] ;
}
};
priority_queue < int , vector < int > , cmp > q ;
bitset < NR > inq ;
bool dijkstra (  )  {
    int nod , i , flow ;
    for ( i = 1 ; i <= n ; ++ i )   d_dij [ i ] = oo ;
    d_dij [ source ] = 0 ;
    d_new [ source ] = 0 ;
    q.push( source ) ;
    while ( !q.empty() )    {
        nod = q.top() ;
        q.pop() ;
        inq [ nod ] = 0 ;
        for ( it i = v [ nod ].begin() ; i < v [ nod ].end() ; ++ i )   {
            if ( cap [ nod ][ *i ] )    {
                if ( d_dij [ nod ] + ( d_old [ nod ] + cost [ nod ][ *i ] - d_old [ *i ] ) < d_dij [ *i ] ) {
                    d_dij [ *i ] = d_dij [ nod ] + ( d_old [ nod ] + cost [ nod ][ *i ] - d_old [ *i ] ) ;
                    d_new [ *i ] = d_new [ nod ] + cost [ nod ][ *i ] ;
                    t [ *i ] = nod ;
                    if ( !inq [ *i ] )
                        q.push( *i ) ,
                        inq [ *i ] = 1 ;
                }
            }
        }
    }
    if ( d_dij [ sink ] == oo ) return 0 ;
    flow = oo ;
    for ( i = sink ; i != source ; i = t [ i ] )
        flow = min ( flow , cap [ t [ i ] ][ i ] ) ;
    for ( i = sink ; i != source ; i = t [ i ] )
        cap [ t [ i ] ][ i ] -= flow ,
        cap [ i ][ t [ i ] ] += flow ;
    for ( i = 1 ; i <= n ; ++ i )   d_old [ i ] = d_new [ i ] ;
    ans += flow * d_new [ sink ] ;
    return 1 ;
}
void bellman ( )    {
    int nod ;
        d_old [ source ] = 0 ;
        queue < int > q ;
        q.push( source ) ;
    while ( !q.empty() ) {
        nod = q.front() ;
        q.pop() ;
        inq [ nod ] = 0 ;
        for ( it i = v [ nod ].begin() ; i != v [ nod ].end() ; ++ i )  {
            if ( cap [ nod ][ *i ] )
                if ( d_old [ nod ] + cost [ nod ][ *i ] < d_old [ *i ] )  {
                    d_old [ *i ] = d_old [ nod ] + cost [ nod ][ *i ] ;
                    if ( !inq [ *i ] )
                        q.push( *i ) ,
                        inq [ *i ] = 1 ;
                }
            }
    }
}
int main () {
    in >> n >> m >> source >> sink ;
    while ( m -- )  {
        in >> x >> y >> capacity >> z ;
        cap [ x ][ y ] = capacity ;
        cost [ x ][ y ] = z ;
        cost [ y ][ x ] = -z ;
        v [ x ].pb ( y ) ;
        v [ y ].pb ( x ) ;
    }
    bellman() ;
    while ( dijkstra() ) ;
    out << ans ;
}