【HDU】3917 Road constructions 最大权闭合子图

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传送门：【HDU】3917 Road constructions

题目分析：最大权闭包问题。

题意不好理解。。。有一点要特别注意，就是如果选择一个公司u，且他承包的道路中有a->b，且如果存在b->c是公司v承包的，那么那个公司也要一起选走。。（理解为u依赖v，可以同时存在一个公司依赖多个公司，那么这时如果选择了一个公司，那么他所依赖的公司都要选）

具体建图就是：对每个公司，val = 税收-补贴。如果val>0那么和源点建边，容量等于val，如果val<0那么和汇点建边，容量等于-val。对于所有类似上述关系u依赖v的，建边<u,v>，容量为INF。跑一遍最小割，正权边权值和-最小割容量即答案。

关于最小割中最大权闭合子图的问题可以看看Amber的论文~

获益匪浅

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 )const int MAXN = 5005 ;const int MAXQ = 5005 ;const int MAXE = 1000005 ;const int INF = 0x3f3f3f3f ;struct Edge {int v , n , c ;Edge () {}Edge ( int v , int c , int n ) : v ( v ) , c ( c ) , n ( n ) {}} ;struct Net {Edge E[MAXE] ;int H[MAXN] , cntE ;int d[MAXN] , num[MAXN] , cur[MAXN] , pre[MAXN] ;int Q[MAXQ] , head , tail ;int s , t , nv ;int n , m ;int flow ;int val[MAXN] ;Edge A[MAXE] ;int Ha[MAXN] , cntA ;void init () {cntE = 0 ;CLR ( H , -1 ) ;cntA = 0 ;CLR ( Ha , -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 ;}flow += f ;u = pos ;}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] ) {cur[u] = i ;mmin = d[E[i].v] ;}d[u] = mmin + 1 ;num[d[u]] ++ ;u = pre[u] ;}}return flow ;}void add ( int u , int v , int idx ) {A[cntA] = Edge ( v , idx , Ha[u] ) ;Ha[u] = cntA ++ ;}void solve () {int k ;int u , v , a , b ;int sum = 0 ;init () ;CLR ( val , 0 ) ;s = 0 ;t = m + 1 ;nv = t + 1 ;FOR ( i , 1 , m )scanf ( "%d" , &val[i] ) ;scanf ( "%d" , &k ) ;REP ( i , 0 , k ) {scanf ( "%d%d%d%d" , &u , &v , &a , &b ) ;val[a] -= b ;add ( u , v , a ) ;}FOR ( i , 1 , m ) {if ( val[i] == 0 )continue ;if ( val[i] > 0 )addedge ( s , i ,  val[i] ) , sum += val[i] ;elseaddedge ( i , t , -val[i] ) ;}FOR ( i , 1 , n )for ( int j = Ha[i] ; ~j ; j = A[j].n ) {int v = A[j].v ;for ( int k = Ha[v] ; ~k ; k = A[k].n )addedge ( A[j].c , A[k].c , INF ) ;}printf ( "%d\n" , sum - ISAP () ) ;}} x ;int main () {while ( ~scanf ( "%d%d" , &x.n , &x.m ) && ( x.n || x.m ) )x.solve () ;return 0 ;}

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