本文主要是介绍【ZOJ】2071 Technology Trader 最大权闭合子图，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】2071 Technology Trader
题目分析：最大权闭合子图问题。源点向订单建边，容量为利益，汇点向组件建边，容量为成本，原图所有边变成容量无穷大的边，最后跑一遍最小割，订单利益和减去最小割容量就是最大净利润。
输出方案就从源点跑一遍dfs，能从源点到达的所有点都标记上。然后看从源点出发的边的弧尾是否被标记，被标记表示被使用，然后再看从汇点出发的点，如果被标记也表示被使用。然后就没有然后了。。。。
代码如下：
#include <map>
#include <string>
#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 ;
typedef int type_f ;const int MAXN = 355 ;
const int MAXQ = 355 ;
const int MAXE = 100000 ;
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 ) , n ( n ) {}
} ;struct Node {char s[50] ;int val ;void input () {scanf ( "%s%d" , s , &val ) ;}
} ;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_f flow ;int n , m ;Node A[MAXN] , B[MAXN] ;map < string , int > mp ;bool 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 ( 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_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 ;}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 dfs ( int u ) {vis[u] = 1 ;for ( int i = H[u] ; ~i ; i = E[i].n )if ( !vis[E[i].v] && E[i].c )dfs ( E[i].v ) ;}void solve () {int p ;char buf[50] ;int sum = 0 ;mp.clear () ;init () ;scanf ( "%d" , &n ) ;REP ( i , 0 , n ) {A[i].input () ;mp[A[i].s] = i ;}scanf ( "%d" , &m ) ;REP ( i , 0 , m ) {B[i].input () ;sum += B[i].val ;scanf ( "%d" , &p ) ;while ( p -- ) {scanf ( "%s" , buf ) ;addedge ( n + i , mp[buf] , INF ) ;}}s = n + m , t = s + 1 , nv = t + 1 ;REP ( i , 0 , n )addedge ( i , t , A[i].val ) ;REP ( i , 0 , m )addedge ( s , i + n , B[i].val ) ;printf ( "%d\n" , sum - ISAP () ) ;CLR ( vis , 0 ) ;dfs ( s ) ;int cnt1 = 0 , cnt2 = 0 ;//ordersfor ( int i = H[s] ; ~i ; i = E[i].n )if ( vis[E[i].v] )++ cnt2 ;printf ( "%d\n" , cnt2 ) ;for ( int i = H[s] ; ~i ; i = E[i].n )if ( vis[E[i].v] )printf ( "%s\n" , B[E[i].v - n].s ) ;//componentsfor ( int i = H[t] ; ~i ; i = E[i].n )if ( vis[E[i].v] )++ cnt1 ;printf ( "%d\n" , cnt1 ) ;for ( int i = H[t] ; ~i ; i = E[i].n )if ( vis[E[i].v] )printf ( "%s\n" , A[E[i].v].s ) ;}
} e ;int main () {int T ;scanf ( "%d" , &T ) ;while ( T -- ) {e.solve () ;if ( T )printf ( "\n" ) ;}return 0 ;
}这篇关于【ZOJ】2071 Technology Trader 最大权闭合子图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！
