本文主要是介绍【ZOJ】2532 Internship 最小割——关键割边，希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！
传送门：【ZOJ】2532 Internship
题目分析：题目意思很明显，问你能否增加一条边的容量使得流量增加，就是让求关键割边。关键割边怎么求？首先按照题意建图跑一遍最小割。之后在残余网络上进行dfs，将从源点s出发能到的点标记为1，将从汇点t出发能到的点重标记为2。一条边为关键割边当且仅当它为正向边且弧头标记为1，弧尾标记为2。
PS：注意dfs的细节，s出发沿正向残余网络，t出发沿反向残余网络。
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 ; 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 ;
typedef int type_f ;const int MAXN = 105 ;
const int MAXQ = 105 ;
const int MAXE = 2333 ;
const int INF = 0x3f3f3f3f ;struct Edge {int u , v , n ;type_c c ;Edge () {}Edge ( int u , int v , type_c c , int n ) : u ( u ) , v ( v ) , c ( c ) , n ( n ) {}
} ;struct Net {Edge E[MAXE] ;int H[MAXN] , cntE ;int d[MAXN] , num[MAXN] , pre[MAXN] , cur[MAXN] ;int Q[MAXQ] , head , tail ;int s , t , nv ;type_f flow ;int n , m , l ;int vis[MAXN] ;void init () {cntE = 0 ;CLR ( H , -1 ) ;}void addedge ( int u , int v , type_c c , type_c rc = 0 ) {E[cntE] = Edge ( u , v , c , H[u] ) ;H[u] = cntE ++ ;E[cntE] = Edge ( v , 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 ) {d[v] = d[u] + 1 ;Q[tail ++] = v ;num[d[v]] ++ ;}}}}type_f ISAP () {CPY ( cur , H ) ;rev_bfs () ;flow = 0 ;int u = pre[s] = s , i , pos , mmin ;while ( d[s] < nv ) {if ( u == t ) {type_f 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[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 dfs ( int u , int x ) {vis[u] = x + 1 ;for ( int i = H[u] ; ~i ; i = E[i].n ) {int v = E[i].v ;if ( vis[v] != x + 1 && E[i ^ x].c )dfs ( E[i].v , x ) ;}}void solve () {int u , v , c ;init () ;t = 0 , s = n + m + 1 , nv = n + m + 2 ;REP ( i , 0 , l ) {scanf ( "%d%d%d" , &u , &v, &c ) ;addedge ( u , v , c ) ;}FOR ( i , 1 , n )addedge ( s , i , INF ) ;ISAP () ;CLR ( vis , 0 ) ;dfs ( s , 0 ) ;dfs ( t , 1 ) ;int flg = 0 ;REP ( i , 0 , cntE ) {if ( vis[E[i].u] == 1 && vis[E[i].v] == 2 ) {if ( flg )printf ( " " ) ;flg = 1 ;printf ( "%d" , ( i >> 1 ) + 1 ) ;}++ i ;}printf ( "\n" ) ;}
} e ;int main () {while ( ~scanf ( "%d%d%d" , &e.n , &e.m , &e.l ) && e.n )e.solve () ;return 0 ;
}这篇关于【ZOJ】2532 Internship 最小割——关键割边的文章就介绍到这儿，希望我们推荐的文章对编程师们有所帮助！
