POJ1087 DInic

n个插座，m个电器，k个适配器。

建图：

原点s到电器 容量1

电器到插座 容量1 

插座到终点 容量1 

插座到适配器 容量inf

要哭出来了。。。。。要建双向边！！！！不过方向容量0

#include<stdio.h>
#include<stdlib.h>
#include<algorithm>
#include<map>
#include<iostream>
#include<string>
using namespace std;
const int maxn = 505;
const int maxm = 20005;
const int inf = 99999999;
int cnt,head1[ maxn ];
int rece[ maxn ];
struct node{
    int from,to,val,next;
}edge[ maxm ];
map<string,int>mp;
int s,t;
void init(){
    cnt=0;
    mp.clear();
    memset( head1,-1,sizeof( head1 ));
}
void addedge( int a,int b,int c ){
    edge[ cnt ].from=a;
    edge[ cnt ].to=b;
    edge[ cnt ].val=c;
    edge[ cnt ].next=head1[ a ];
    head1[ a ]=cnt++;
    edge[ cnt ].from=b;
    edge[ cnt ].to=a;
    edge[ cnt ].val=0;
    edge[ cnt ].next=head1[ b ];
    head1[ b ]=cnt++;
}
int queue[ 1005 ];
int lev[ maxn ];
 int Dinic( int start,int end ){
     int max_flow=0;
     while(true){
        // printf("dinic100\n");
         int head,tail,id,i;
         head=tail=0;
         queue[tail++]=start;//队列从0开始，tail指向下一个入队元素的位置,head指向出队元素
         memset(lev,-1,sizeof(lev));
         lev[start]=0;
         while(head<tail){
             id=head1[queue[head++]];
             while(id!=-1){
                // printf("dinic200\n");
                 if(edge[id].val>0&&lev[edge[id].to]==-1){
                     lev[edge[id].to]=lev[edge[id].from]+1;
                     queue[tail++]=edge[id].to;
                     if(edge[id].to==end){
                         head=tail;
                         break;//end of while(head<tail)
                     }
                 }
                 id=edge[id].next;
             }
         }//分层，直到找到end为止，分层结束
 
         if(lev[end]==-1)
             break;//end of while(true)
 
         id=start;
         tail=0;
         while(true){//寻找增广路
        //     printf("dinic300\n");
             if(id==end){//找到了一条增广路
                 int flow=inf,flag=-1;
                 for(i=0;i<tail;i++){
                     if(edge[queue[i]].val<flow){
                         flag=i;
                         flow=edge[queue[i]].val;
                     }
                 //寻找最短的
                 }
                     for(i=0;i<tail;i++){
                        // printf("%d ",edge[queue[i]].from);
                         edge[queue[i]].val-=flow;
                         edge[queue[i]^1].val+=flow;
                     }
                    // printf("\nmax_flw:%d\n",max_flow);
                     if(flag!=-1){
                         max_flow+=flow;
                         tail=flag;
                         id=edge[queue[flag]].from;
                     }
                     else
                         return inf;
             }
             id=head1[id];
             while(id!=-1){
                // printf("dinic400\n");
                 if(edge[id].val>0&&(lev[edge[id].from]+1==lev[edge[id].to]))
                     break;
                 id=edge[id].next;
             }
             if(id!=-1){
                 queue[tail++]=id;
                 id=edge[id].to;
             }
             else{
                 if(tail==0)
                     break;
                 lev[edge[queue[tail-1]].to]=-1;
                 id=edge[queue[--tail]].from;
             } 
         }
     }
     return max_flow;
 }
 

int main(){
    int n,m,k;
    string name,pname;
    while( scanf("%d",&n)!=EOF ){
        init();
        s=0;
        int CNT=1;
        for( int i=0;i<n;i++ ){
            cin>>name;
            mp[ name ]=CNT;
            CNT++;
        }
        scanf("%d",&m);
        int t1,t2;
        int tot=m;
        while( m-- ){
            cin>>name>>pname;
            if( mp[ name ]==0 ){
                mp[ name ]=CNT;
                t1=CNT;
                CNT++;
            }
            else
                t1=mp[ name ];
            if( mp[ pname ]==0 ){
                mp[ pname ]=CNT;
                t2=CNT;
                CNT++;
            }
            else
                t2=mp[ pname ];
            addedge( s,t1,1 );
            addedge( t1,t2,1 );
        }
        scanf("%d",&k);
        while( k-- ){
            cin>>name>>pname;
            if( mp[ name ]==0 ){
                mp[ name ]=CNT;
                t1=CNT;
                CNT++;
            }
            else
                t1=mp[ name ];
            if( mp[ pname ]==0 ){
                mp[ pname ]=CNT;
                t2=CNT;
                CNT++;
            }
            else
                t2=mp[ pname ];
            addedge( t1,t2,inf );
        }
        for( int i=1;i<=n;i++ )
            addedge( i,maxn-1,1 );

        int ans=Dinic( 0,maxn-1 );
        printf("%d\n",tot-ans);
    }
    return 0;
}