【POJ】3189 Steady Cow Assignment 二分最大流

传送门：【POJ】3189 Steady Cow Assignment

题目分析：同学说自己写的超时。。。让我看看，然后我敲了一个就过了。。。Orz。。。这逗比一定是敲搓了。。

具体思路是二分区间长度，然后枚举这个区间长度覆盖的区间范围，找到一个就可以退出了，满足要求的条件是：流量等于牛的数量。

赤裸裸的二分最大流。。。

没啥trick啊。。。

代码如下：

#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 - 1 ; i >= b ; -- i )
#define FOV( 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 = 1100 ;
const int MAXQ = 1100 ;
const int MAXE = 50000 ;
const int INF = 0x3f3f3f3f ;

struct Edge {
	int v , c , n ;
	Edge () {}
	Edge ( int v , int c , int n ) : v ( v ) , c ( c ) , n ( n ) {}
} ;

struct NetWork {
	Edge E[MAXE] ;
	int H[MAXN] , cntE ;
	int d[MAXN] , num[MAXN] , pre[MAXN] , cur[MAXN] ;
	int Q[MAXN] , head , tail ;
	int s , t , nv ;
	int flow ;
	int n , m ;
	
	int G[MAXN][20] ;
	int cap[MAXN] ;
	
	void init () {
		cntE = 0 ;
		CLR ( H , -1 ) ;
	}
	
	void addedge ( int u , int v , int c ) {
		E[cntE] = Edge ( v , c , H[u] ) ;
		H[u] = cntE ++ ;
		E[cntE] = Edge ( u , 0 , H[v] ) ;
		H[v] = cntE ++ ;
	}
	
	void rev_bfs () {
		CLR ( d , -1 ) ;
		CLR ( num , 0 ) ;
		head = tail = 0 ;
		Q[tail ++] = t ;
		d[t] = 0 ;
		num[d[t]] = 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] )
					continue ;
				d[v] = d[u] + 1 ;
				num[d[v]] ++ ;
				Q[tail ++] = v ;
			}
		}
	}
	
	int ISAP () {
		CPY ( cur , H ) ;
		rev_bfs () ;
		flow = 0 ;
		int u = pre[s] = s ;
		while ( d[s] < nv ) {
			if ( u == t ) {
				int f = INF , pos ;
				for ( int i = s ; i != t ; i = E[cur[i]].v )
					if ( f > E[cur[i]].c ) {
						f = E[cur[i]].c ;
						pos = i ;
					}
				for ( int i = s ; i != t ; i = E[cur[i]].v ) {
					E[cur[i]].c -= f ;
					E[cur[i] ^ 1].c += f ;
				}
				u = pos ;
				flow += f ;
			}
			for ( int &i = cur[u] ; ~i ; i = E[i].n )
				if ( E[i].c && d[u] == d[E[i].v] + 1 )
					break ;
			if ( ~cur[u] ) {
				pre[E[cur[u]].v] = u ;
				u = E[cur[u]].v ;
			}
			else {
				if ( 0 == ( -- num[d[u]] ) )
					break ;
				int mmin = nv ;
				for ( int i = H[u] ; ~i ; i = E[i].n )
					if ( E[i].c && mmin > d[E[i].v] ) {
						mmin = d[E[i].v] ;
						cur[u] = i ;
					}
				d[u] = mmin + 1 ;
				num[d[u]] ++ ;
				u = pre[u] ;
			}
		}
		return flow ;
	}
	
	int ok ( int mid ) {
		FOR ( l , 1 , m - mid + 1 ) {
			init () ;
			FOR ( i , 1 , n ) {
				addedge ( s , i , 1 ) ;
				FOR ( j , l , l + mid - 1 )
					addedge ( i , n + G[i][j] , 1 ) ;
			}
			FOR ( i , 1 , m )
				addedge ( n + i , t , cap[i] ) ;
			ISAP () ;
			if ( flow == n )
				return 1 ;
		}
		return 0 ;
	}
	
	void solve () {
		s = 0 ;
		t = n + m + 1 ;
		nv = t + 1 ;
		FOR ( i , 1 , n )
			FOR ( j , 1 , m )
				scanf ( "%d" , &G[i][j] ) ;
		FOR ( i , 1 , m )
			scanf ( "%d" , &cap[i] ) ;
		int l = 1 , r = m + 1 ;
		while ( l < r ) {
			int mid = ( l + r ) >> 1 ;
			if ( ok ( mid ) )
				r = mid ;
			else
				l = mid + 1 ;
		}
		printf ( "%d\n" , r ) ;
	}
} x ;

int main () {
	while ( ~scanf ( "%d%d" , &x.n , &x.m ) )
		x.solve () ;
	return 0 ;
}