学步园

传送门：【POJ】1815 Friendship

题目分析：枚举都能过。。。Orz。。。一开始建图就是普通的将一个点 i 拆成两个点i，i'，建边( i , i' , 1 )，对关系( u , v )建边( u' , v , INF ) , ( v' , u , INF )，然后跑一遍最大流作为初始最小割容量。

接下来按照字典序升序枚举所有点，枚举时，假设删除该点，如果跑出来的最大流比初始最小割的容量小，则说明这个是割点，然后减小初始最小割容量。如果不是割点，还原该点到图中，一直枚举到结束。

这题无语了。。。枚举竟然还能过。。。。。

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 REV( i , a , b ) for ( int i = a - 1 ; i >= b ; -- i )
#define FOR( i , a , b ) for ( int i = a ; 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 )

typedef int type_c ;

const int MAXN = 405 ;
const int MAXQ = 405 ;
const int MAXE = 20005 ;
const int INF = 0x3f3f3f3f ;

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

struct Net {
	Edge E[MAXE] ;
	int H[MAXN] , cntE , cntEE ;
	int d[MAXN] , num[MAXN] , pre[MAXN] , cur[MAXN] ;
	int Q[MAXQ] , head , tail ;
	int s ,t , nv ;
	type_c flow ;
	
	int n , m ;
	bool vis[MAXN] ;
	int G[MAXN][MAXN] ;
	int ans[MAXN] ;
	
	void init () {
		cntE = 0 ;
		CLR ( H , -1 ) ;
	}
	
	void addedge ( int u , int v , type_c 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 ;
			}
		}
	}
	
	type_c ISAP () {
		CPY ( cur , H ) ;
		rev_bfs () ;
		flow = 0 ;
		int u = pre[s] = s , i , pos , mmin ;
		while ( d[s] < nv ) {
			if ( u == t ) {
				type_c 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 ;
				}
				u = pos ;
				flow += f ;
			}
			for ( i = cur[u] ; ~i ; i = E[i].n )
				if ( E[i].c && d[u] == d[E[i].v] + 1 )
					break ;
			if ( ~i ) {
				cur[u] = i ;
				pre[E[i].v] = u ;
				u = E[i].v ;
			}
			else {
				if ( 0 == -- num[d[u]] )
					break ;
				mmin = nv ;
				for ( 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 ;
	}
	
	void dfs ( int u ) {
		vis[u] = 1 ;
		for ( int i = H[u] ; ~i ; i = E[i].n )
			if ( E[i].c && !vis[E[i].v] )
				dfs ( E[i].v ) ;
	}
	
	void solve () {
		FOR ( i , 1 , n )
			FOR ( j , 1 , n )
				scanf ( "%d" , &G[i][j] ) ;
		if ( G[s][t] ) {
			printf ( "NO ANSWER!\n" ) ;
			return ;
		}
		init () ;
		s += n ;
		nv = n << 1 | 1 ;
		FOR ( i , 1 , n )
			addedge ( i , i + n , 1 ) ;
		FOR ( i , 1 , n )
			FOR ( j , 1 , n )
				if ( i != j && G[i][j] )
					addedge ( i + n , j , INF ) ;
		int cnt = 0 ;
		int maxflow = ISAP () ;
		FOR ( i , 1 , n ) {
			-- i ;//下标-1易于计算
			REP ( j , 0 , cntE )
				E[j].c = E[j].rc ;
			E[i << 1].c = E[i << 1].rc = 0 ;
			if ( ISAP () < maxflow ) {
				ans[cnt ++] = i + 1 ;
				if ( 0 == -- maxflow )
					break ;
			}
			else
				E[i << 1].c = E[i << 1].rc = 1 ;
			++ i ;//还原
		}
		printf ( "%d\n" , cnt ) ;
		REP ( i , 0 , cnt )
			printf ( "%d%c" , ans[i] , i < cnt - 1 ? ' ' : '\n' ) ;
	}
} e ;

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