本文主要是介绍【ZOJ】2332 Gems 最大流——判断满流，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】2332 Gems
题目分析：首先我们设立源点s，汇点t，s向所有宝石建边，容量为题目中给出的，然后所有可行的转换，向两个宝石之间建无向边，容量为INF，接下来所有的宝石向自己相应的类型建边，容量INF，所有的宝石向自己相应的颜色建边，容量INF。最后，所有的类型以及颜色向汇点建边，容量为题目中给出的。最后跑一遍最大流，如果满流，说明所有的宝石都成功的限制条件下分给了男主角以及女主角，输出Yes，否则输出No。
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 = 128 ;
const int MAXQ = 128 ;
const int MAXE = 1024 ;
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 , q ;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 solve () {int r1 , c1 , r2 , c2 , x ;scanf ( "%d%d" , &n , &m ) ;int nm = n * m , sum = 0 ;s = nm + n + m , t = s + 1 , nv = t + 1 ;init () ;REP ( i , 0 , n ) {int tmp = 0 ;REP ( j , 0 , m ) {scanf ( "%d" , &x ) ;addedge ( s , i * m + j , x ) ;addedge ( i * m + j , nm + i , INF ) ;addedge ( i * m + j , nm + n + j , INF ) ;sum += x ;}}scanf ( "%d" , &q ) ;while ( q -- ) {scanf ( "%d%d%d%d" , &r1 , &c1 , &r2 , &c2 ) ;addedge ( r1 * m + c1 , r2 * m + c2 , INF , INF ) ;}REP ( i , 0 , n ) {scanf ( "%d" , &x ) ;addedge ( nm + i , t , x ) ;}REP ( i , 0 , m ) {scanf ( "%d" , &x ) ;addedge ( nm + n + i , t , x ) ;}ISAP () ;if ( sum == flow )printf ( "Yes\n" ) ;elseprintf ( "No\n" ) ;}
} e ;int main () {int T ;scanf ( "%d" , &T ) ;while ( T -- )e.solve () ;return 0 ;
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