HDU-4619 Warm up 2 二分匹配

　　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 ;
 }