【SPOJ】13041 The Black Riders 二分最大流

传送门：【SPOJ】13041 The Black Riders

题目分析：直接将每个洞独立出来，第一类洞花矩阵中的时间，第二类洞花矩阵中的时间+C，以此建边，然后二分最长时间求最大流。只要不读错题就很容易写。。。

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 FOR( i , a , b ) for ( int i = ( a ) ; i <= ( b ) ; ++ i )
#define REV( 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 = 305 ;
const int MAXE = 50005 ;
const int INF = 0x3f3f3f3f ;

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

int H[MAXN] , cntE ;
int d[MAXN] , cnt[MAXN] , pre[MAXN] , cur[MAXN] ;
int Q[MAXN] , head , tail ;
int s , t , nv ;
int flow ;
int n , m ;
int K , C ;

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

void addedge ( int u , int v , int cost ) {
	E[cntE] = Edge ( v , cost , H[u] ) ;
	H[u] = cntE ++ ;
	E[cntE] = Edge ( u , cost , H[v] ) ;
	H[v] = cntE ++ ;
}

void rev_bfs () {
	head = tail = 0 ;
	CLR ( cnt , 0 ) ;
	CLR ( d , -1 ) ;
	Q[tail ++] = t ;
	d[t] = 0 ;
	cnt[0] = 1 ;
	while ( head != tail ) {
		int u = Q[head ++] ;
		for ( int i = H[u] ; ~i ; i = E[i].n ) {
			int v = E[i].v ;
			if ( ~d[v] ) {
				d[v] = d[u] + 1 ;
				cnt[d[v]] ++ ;
				Q[tail ++] = v ;
			}
		}
	}
}

int ISAP () {
	CPY ( cur , H ) ;
	rev_bfs () ;
	flow = 0 ;
	int u = pre[s] = s , i , pos , f , minv ;
	while ( d[s] < nv ) {
		if ( u == t ) {
			f = INF ;
			for ( i = s ; i != t ; i = E[cur[i]].v ) if ( f > E[cur[i]].c ) {
				f = E[cur[i]].c ;
				pos = i ;
			}
			for ( i = s ; i != t ; i = E[cur[i]].v ) {
				E[cur[i]].c -= f ;
				E[cur[i] ^ 1].c += f ;
			}
			flow += f ;
			u = pos ;
		}
		for ( i = cur[u] ; ~i ; i = E[i].n ) if ( E[i].c && d[E[i].v] + 1 == d[u] ) break ;
		if ( ~i ) {
			cur[u] = i ;
			pre[E[i].v] = u ;
			u = E[i].v ;
		} else {
			if ( 0 == -- cnt[d[u]] ) break ;
			for ( minv = nv , i = H[u] ; ~i ; i = E[i].n ) if ( E[i].c && minv > d[E[i].v] ) {
				minv = d[E[i].v] ;
				cur[u] = i ;
			}
			d[u] = minv + 1 ;
			cnt[d[u]] ++ ;
			u = pre[u] ;
		}
	}
	return flow ;
}

void solve () {
	int x , maxv = 0 ;
	clear () ;
	scanf ( "%d%d%d%d" , &n , &m , &K , &C ) ;
	s = 0 ;
	t = n + 2 * m + 1 ;
	nv = t + 1 ;
	FOR ( i , 1 , n ) {
		addedge ( s , i , 0 ) ;
		FOR ( j , 1 , m ) {
			scanf ( "%d" , &x ) ;
			addedge ( i , n + j , x ) ;
			addedge ( i , n + m + j , x + C ) ;
			if ( x > maxv ) maxv = x ;
		}
	}
	FOR ( i , 1 , m ) {
		addedge ( n + i , t , 0 ) ;
		addedge ( n + m + i , t , 0 ) ;
	}
	int l = 0 , r = maxv + C ;
	while ( l < r ) {
		int m = ( l + r ) >> 1 ;
		REP ( i , 0 , cntE ) {
			E[i].c = ( E[i].cost <= m ) ;
			E[++ i].c = 0 ;
		}
		if ( ISAP () >= K ) r = m ;
		else l = m + 1 ;
	}
	printf ( "%d\n" , l ) ;
}

int main () {
	int T ;
	scanf ( "%d" , &T ) ;
	while ( T -- ) solve () ;
	return 0 ;
}