BZOJ-3232: 圈地游戏（分数规划----二分+最小割）

题目：http://www.lydsy.com/JudgeOnline/problem.php?id=3232

分数规划的常用做法，二分答案，然后最小割判定，题解传送门：http://hi.baidu.com/strongoier/item/0425f0e5814e010265db0095

代码：

#include 

#include 

 

#define rep( i , x ) for ( int i = 0 ; i ++ < x ; )

#define REP( i , l , r ) for ( int i = l ; i <= r ; ++ i )

 

const double inf = 0x7fffffff , esp = 0.000001 ;

 

const int maxn = 55 , maxm = 21000 , maxv = 5010 ;

 

struct edge {

    edge *next , *pair ;

    int t ;

    double f ;

} E[ maxm ] ;

 

edge *pt , *head[ maxv ] , *d[ maxv ] ;

 

void Init(  ) {

    memset( head , 0 , sizeof( head ) ) ;

    pt = E ;

}

 

void add( int s , int t , double f ) {

    edge *p = pt ++ ;

    p -> t = t , p -> next = head[ s ] , p -> f = f ;

    head[ s ] = p ;

}

 

void addedge( int s , int t , double a , double b ) {

    add( s , t , a ) , add( t , s , b ) ;

    head[ s ] -> pair = head[ t ] , head[ t ] -> pair = head[ s ] ;

}

 

int gap[ maxv ] , h[ maxv ] , S , T ;

 

double min( double x , double y ) {

    return x < y ? x : y ;

}

 

double sap( int v , double flow ) {

    if ( v == T ) return flow ;

    double rec = 0 , ret ;

    for ( edge *p = d[ v ] ; p ; p = p -> next ) if ( p -> f > esp && h[ p -> t ] + 1 == h[ v ] ) {

        ret = sap( p -> t , min( flow - rec , p -> f ) ) ;

        p -> f -= ret , p -> pair -> f += ret , d[ v ] = p ;

        if ( ( rec += ret ) > flow - esp ) return flow ;

    }

    if ( ! ( -- gap[ h[ v ] ] ) ) h[ S ] = T ;

    gap[ ++ h[ v ] ] ++ , d[ v ] = head[ v ] ;

    return rec ;

}

 

double maxflow(  ) {

    memset( gap , 0 , sizeof( gap ) ) ;

    memset( h , 0 , sizeof( h ) ) ;

    for ( int i = 0 ; i ++ < T ; ) {

        d[ i ] = head[ i ] ;

    }

    gap[ 0 ] = T ;

    double flow = 0 ;

    for ( ; h[ S ] < T ; flow += sap( S , inf ) ) ;

    return flow ;

}

 

int a[ maxn ][ maxn ] , b[ maxn ][ maxn ] , c[ maxn ][ maxn ] , n , m , node[ maxn ][ maxn ] , V = 0 ;

double sum ;

 

bool check( double x ) {

    Init(  ) ;

    rep( i , n ) addedge( node[ i ][ 0 ] , T , inf , 0 ) , addedge( node[ i ][ m + 1 ] , T , inf , 0 ) ;

    rep( i , m ) addedge( node[ 0 ][ i ] , T , inf , 0 ) , addedge( node[ n + 1 ][ i ] , T , inf , 0 ) ;

    rep( i , n ) rep( j , m ) addedge( S , node[ i ][ j ] , a[ i ][ j ] , 0 ) ;

    double cost ;

    REP( i , 0 , n ) rep( j , m ) {

        cost = double( b[ i ][ j ] ) * x ;

        addedge( node[ i ][ j ] , node[ i + 1 ][ j ] , cost , cost ) ;

    }

    rep( i , n ) REP( j , 0 , m ) {

        cost = double( c[ i ][ j ] ) * x ;

        addedge( node[ i ][ j ] , node[ i ][ j + 1 ] , cost , cost ) ;

    }

    return maxflow(  ) < sum - esp ;

}

 

int main(  ) {

    scanf( "%d%d" , &n , &m ) ;

    rep( i , n ) rep( j , m ) scanf( "%d" , &a[ i ][ j ] ) , sum += double( a[ i ][ j ] ) ;

    REP( i , 0 , n ) rep( j , m ) scanf( "%d" , &b[ i ][ j ] ) ;

    rep( i , n ) REP( j , 0 , m ) scanf( "%d" , &c[ i ][ j ] ) ;

    REP( i , 0 , ( n + 1 ) ) REP( j , 0 , ( m + 1 ) ) {

        node[ i ][ j ] = ++ V ;

    }

    S = ++ V ; T = ++ V ;

    double l = 0 , r = inf , mid ;

    check( 3 ) ;

    while ( r - l > esp ) {

        mid = ( l + r ) / 2.0 ;

        if ( check( mid ) ) l = mid ; else r = mid ;

    }

    printf( "%.3f\n" , l ) ;

    return 0 ;

}