【POJ】3084 Panic Room 最小割

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传送门：【POJ】3084 Panic Room

题目分析：POJ今天的C++逗我玩呢。。G++AC，C++RE。。。

对于终点，和汇点建边，容量INF，对所有有入侵者的房间和源点建边，容量INF。

对所有房间A可以绝对进入房间B的关系，建边(A,B,INF),(B,A,1)。

这样如果最终答案为INF，说明无论怎么上锁，入侵者都能达到终点，否则最小割就是答案。

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 long long LL ;typedef int type_c ;const int MAXN = 25 ;const int MAXQ = 25 ;const int MAXE = 1000000 ;const int INF = 0x3f3f3f3f ;struct Edge {int v , n ;type_c c ;Edge () {}Edge ( int v , type_c c , int n ) : v ( v ) , c ( c ) , n ( n ) {}} ;struct Net {Edge E[MAXE] ;int H[MAXN] , cntE ;int d[MAXN] , cur[MAXN] , pre[MAXN] , num[MAXN] ;int Q[MAXQ] , head , tail ;int s , t , nv ;type_c flow ;int n , ed ;int G[25][25] ;void init () {cntE = 0 ;CLR ( H , -1 ) ;}void addedge ( int u , int v , type_c c , type_c rc = 0 ) {E[cntE] = Edge ( v ,  c , H[u] ) ;H[u] = cntE ++ ;E[cntE] = Edge ( u , rc , 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] == -1 ) {Q[tail ++] = v ;d[v] = d[u] + 1 ;num[d[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 ;for ( mmin = nv , 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 solve () {char str[5] ;int x , v ;CLR ( G , 0 ) ;init () ;scanf ( "%d%d" , &n , &ed ) ;s = n ;t = s + 1 ;nv = t + 1 ;addedge ( ed , t , INF ) ;REP ( i , 0 , n ) {scanf ( "%s%d" , str , &x ) ;if ( str[0] == 'I' )addedge ( s , i , INF ) ;while ( x -- ) {scanf ( "%d" , &v ) ;++ G[i][v] ;}}REP ( i , 0 , n )REP ( j , 0 , n )if ( G[i][j] )addedge ( i , j , INF , G[i][j] ) ;ISAP () ;if ( flow == INF )printf ( "PANIC ROOM BREACH\n" ) ;elseprintf ( "%d\n" , flow ) ;}} e ;int main () {int T ;scanf ( "%d" , &T ) ;while ( T -- )e.solve () ;return 0 ;}

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