题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=4619
一看就知道是二分匹配题目，对每个点拆点建立二分图，最后答案除2。因为这里是稀疏图，用邻接表处理。。。
//STATUS:C++_AC_31MS_480KB
#include <functional>
#include <algorithm>
#include <iostream>
//#include <ext/rope>
#include <fstream>
#include <sstream>
#include <iomanip>
#include <numeric>
#include <cstring>
#include <cassert>
#include <cstdio>
#include <string>
#include <vector>
#include <bitset>
#include <queue>
#include <stack>
#include <cmath>
#include <ctime>
#include <list>
#include <set>
#include <map>
using namespace std;
//using namespace __gnu_cxx;
//define
#define pii pair<int,int>
#define mem(a,b) memset(a,b,sizeof(a))
#define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1
#define PI acos(-1.0)
//typedef
typedef __int64 LL;
typedef unsigned __int64 ULL;
//const
const int N=;
const int INF=0x3f3f3f3f;
const int MOD=,STA=;
const LL LNF=1LL<<;
const double EPS=1e-;
const double OO=1e15;
const int dx[]={-,,,};
const int dy[]={,,,-};
const int day[]={,,,,,,,,,,,,};
//Daily Use ...
inline int sign(double x){return (x>EPS)-(x<-EPS);}
template<class T> T gcd(T a,T b){return b?gcd(b,a%b):a;}
template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;}
template<class T> inline T lcm(T a,T b,T d){return a/d*b;}
template<class T> inline T Min(T a,T b){return a<b?a:b;}
template<class T> inline T Max(T a,T b){return a>b?a:b;}
template<class T> inline T Min(T a,T b,T c){return min(min(a, b),c);}
template<class T> inline T Max(T a,T b,T c){return max(max(a, b),c);}
template<class T> inline T Min(T a,T b,T c,T d){return min(min(a, b),min(c,d));}
template<class T> inline T Max(T a,T b,T c,T d){return max(max(a, b),max(c,d));}
//End int n,m;
int vis[N*],id[N][N],y[N*];
int tot; struct Edge{
int u,v;
}e[];
int first[N*],next[];
int mt; void adde(int a,int b) //对于一条边，需建立双向边，一个容量为cap，反向边容量为0!
{
e[mt].u=a;e[mt].v=b;
next[mt]=first[a];first[a]=mt++;
e[mt].u=b;e[mt].v=a;
next[mt]=first[b];first[b]=mt++;
} int dfs(int u)
{
int i;
for(i=first[u];i!=-;i=next[i]){
if(!vis[e[i].v]){
vis[e[i].v]=;
if(!y[e[i].v] || dfs(y[e[i].v])){
y[e[i].v]=u;
return ;
}
}
}
return ;
} int main()
{
// freopen("in.txt","r",stdin);
int i,j,a,b,ans;
while(~scanf("%d%d",&n,&m) && (n || m))
{
mem(id,);
tot=;mt=;mem(first,-);
for(i=;i<n;i++){
scanf("%d%d",&a,&b);
if(!id[a][b])id[a][b]=tot++;
if(!id[a+][b])id[a+][b]=tot++;
adde(id[a][b],id[a+][b]);
}
for(i=;i<m;i++){
scanf("%d%d",&a,&b);
if(!id[a][b])id[a][b]=tot++;
if(!id[a][b+])id[a][b+]=tot++;
adde(id[a][b],id[a][b+]);
} ans=;
mem(y,);
for(i=;i<tot;i++){
mem(vis,);
if(dfs(i))ans++;
} printf("%d\n",ans>>);
}
return ;
}