【ACDream】1355 Domino in Casino 费用流

传送门：【ACDream】1355 Domino in Casino

题目分析：费用流水题（前提是能看出来）。将矩阵黑白染色。黑色结点和源点建边，容量1，费用0。白色结点和汇点建边，容量1，费用0。黑色结点向相邻的白色结点建边，容量1，费用为两结点内权值的乘积的相反数。再建立超级源汇，超级源点向源点建边，容量k，费用0。汇点向超级汇点建边，容量k，费用0。最后跑一遍最小费用最大流即可。费用的相反数即答案。

就是二分图上的匹配问题，限制就是至多只能有K对匹配，所以用费用流求解。

代码如下：

#include <cmath>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std ;

#define rep( i , a , b ) for ( int i = a ; i < b ; ++ i )
#define rev( i , a , b ) for ( int i = a ; i >= b ; -- i )
#define For( i , a , b ) for ( int i = a ; i <= b ; ++ i )
#define clr( a , x ) memset ( a , x , sizeof a )
#define cpy( a , x ) memcpy ( a , x , sizeof a )

const int MAXN = 2005 ;
const int MAXQ = 1000005 ;
const int MAXE = 1000005 ;
const int INF = 0x3f3f3f3f ;

struct Edge {
	int v , c , w , n ;
	Edge () {}
	Edge ( int v , int c , int w , int n ) : v ( v ) , c ( c ) , w ( w ) , n ( n ) {}
} E[MAXE] ;

int H[MAXN] , cntE ;
int d[MAXN] , cur[MAXN] , cap[MAXN] ;
int vis[MAXN] , Time ;
int Q[MAXQ] , head , tail ;
int ss , tt ;
int s , t ;
int flow ;
int cost ;
int n , m , k ;
int a[17][105] ;

void clear () {
	cntE = 0 ;
	clr ( H , -1 ) ;
}

void addedge ( int u , int v , int c , int w ) {
	E[cntE] = Edge ( v , c , +w , H[u] ) ;
	H[u] = cntE ++ ;
	E[cntE] = Edge ( u , 0 , -w , H[v] ) ;
	H[v] = cntE ++ ;
}

int spfa () {
	head = tail = 0 ;
	clr ( d , INF ) ;
	++ Time ;
	d[ss] = 0 ;
	cap[ss] = INF ;
	cur[ss] = -1 ;
	Q[tail ++] = ss ;
	while ( head != tail ) {
		int u = Q[head ++] ;
		vis[u] = Time - 1 ;
		for ( int i = H[u] ; ~i ; i = E[i].n ) {
			int v = E[i].v , c = E[i].c , w = E[i].w ;
			if ( c && d[v] > d[u] + w ) {
				d[v] = d[u] + w ;
				cap[v] = min ( cap[u] , c ) ;
				cur[v] = i ;
				if ( vis[v] != Time ) {
					vis[v] = Time ;
					Q[tail ++] = v ;
				}
			}
		}
	}
	if ( d[tt] == INF ) return 0 ;
	cost += d[tt] * cap[tt] ;
	flow += cap[tt] ;
	for ( int i = cur[tt] ; ~i ; i = cur[E[i ^ 1].v] ) {
		E[i].c -= cap[tt] ;
		E[i ^ 1].c += cap[tt] ;
	}
	return 1 ;
}

int mcmf () {
	flow = cost = 0 ;
	while ( spfa () ) ;
	return cost ;
}

void solve () {
	clear () ;
	s = n * m ;
	t = s + 1 ;
	ss = t + 1 ;
	tt = ss + 1 ;
	addedge ( ss , s , k , 0 ) ;
	addedge ( t , tt , k , 0 ) ;
	rep ( i , 0 , n ) rep ( j , 0 , m ) scanf ( "%d" , &a[i][j] ) ;
	rep ( i , 0 , n ) rep ( j , 0 , m ) {
		int ij = i * m + j ;
		if ( ( i + j ) & 1 ) {
			addedge ( s , ij , 1 , 0 ) ;
			if ( i < n - 1 ) addedge ( ij , ij + m , 1 , - a[i][j] * a[i + 1][j] ) ;
			if ( j < m - 1 ) addedge ( ij , ij + 1 , 1 , - a[i][j] * a[i][j + 1] ) ;
			if ( i ) addedge ( ij , ij - m , 1 , - a[i][j] * a[i - 1][j] ) ;
			if ( j ) addedge ( ij , ij - 1 , 1 , - a[i][j] * a[i][j - 1] ) ;
		} else addedge ( ij , t , 1 , 0 ) ;
	}
	printf ( "%d\n" , -mcmf () ) ;
}

int main () {
	clr ( vis , 0 ) ;
	Time = 0 ;
	while ( ~scanf ( "%d%d%d" , &n , &m , &k ) ) solve () ;
	return 0 ;
}