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。。。
  1 //STATUS:C++_AC_31MS_480KB
  2 #include <functional>
  3 #include <algorithm>
  4 #include <iostream>
  5 //#include <ext/rope>
  6 #include <fstream>
  7 #include <sstream>
  8 #include <iomanip>
  9 #include <numeric>
 10 #include <cstring>
 11 #include <cassert>
 12 #include <cstdio>
 13 #include <string>
 14 #include <vector>
 15 #include <bitset>
 16 #include <queue>
 17 #include <stack>
 18 #include <cmath>
 19 #include <ctime>
 20 #include <list>
 21 #include <set>
 22 #include <map>
 23 using namespace std;
 24 //using namespace __gnu_cxx;
 25 //define
 26 #define pii pair<int,int>
 27 #define mem(a,b) memset(a,b,sizeof(a))
 28 #define lson l,mid,rt<<1
 29 #define rson mid+1,r,rt<<1|1
 30 #define PI acos(-1.0)
 31 //typedef
 32 typedef __int64 LL;
 33 typedef unsigned __int64 ULL;
 34 //const
 35 const int N=110;
 36 const int INF=0x3f3f3f3f;
 37 const int MOD=100000,STA=8000010;
 38 const LL LNF=1LL<<60;
 39 const double EPS=1e-8;
 40 const double OO=1e15;
 41 const int dx[4]={-1,0,1,0};
 42 const int dy[4]={0,1,0,-1};
 43 const int day[13]={0,31,28,31,30,31,30,31,31,30,31,30,31};
 44 //Daily Use ...
 45 inline int sign(double x){return (x>EPS)-(x<-EPS);}
 46 template<class T> T gcd(T a,T b){return b?gcd(b,a%b):a;}
 47 template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;}
 48 template<class T> inline T lcm(T a,T b,T d){return a/d*b;}
 49 template<class T> inline T Min(T a,T b){return a<b?a:b;}
 50 template<class T> inline T Max(T a,T b){return a>b?a:b;}
 51 template<class T> inline T Min(T a,T b,T c){return min(min(a, b),c);}
 52 template<class T> inline T Max(T a,T b,T c){return max(max(a, b),c);}
 53 template<class T> inline T Min(T a,T b,T c,T d){return min(min(a, b),min(c,d));}
 54 template<class T> inline T Max(T a,T b,T c,T d){return max(max(a, b),max(c,d));}
 55 //End
 56 
 57 int n,m;
 58 int vis[N*100],id[N][N],y[N*100];
 59 int tot;
 60 
 61 struct Edge{
 62     int u,v;
 63 }e[40010];
 64 int first[N*100],next[40010];
 65 int mt;
 66 
 67 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!
 68 {
 69     e[mt].u=a;e[mt].v=b;
 70     next[mt]=first[a];first[a]=mt++;
 71     e[mt].u=b;e[mt].v=a;
 72     next[mt]=first[b];first[b]=mt++;
 73 }
 74 
 75 int dfs(int u)
 76 {
 77     int i;
 78     for(i=first[u];i!=-1;i=next[i]){
 79         if(!vis[e[i].v]){
 80             vis[e[i].v]=1;
 81             if(!y[e[i].v] || dfs(y[e[i].v])){
 82                 y[e[i].v]=u;
 83                 return 1;
 84             }
 85         }
 86     }
 87     return 0;
 88 }
 89 
 90 int main()
 91 {
 92  //   freopen("in.txt","r",stdin);
 93     int i,j,a,b,ans;
 94     while(~scanf("%d%d",&n,&m) && (n || m))
 95     {
 96         mem(id,0);
 97         tot=1;mt=0;mem(first,-1);
 98         for(i=0;i<n;i++){
 99             scanf("%d%d",&a,&b);
100             if(!id[a][b])id[a][b]=tot++;
101             if(!id[a+1][b])id[a+1][b]=tot++;
102             adde(id[a][b],id[a+1][b]);
103         }
104         for(i=0;i<m;i++){
105             scanf("%d%d",&a,&b);
106             if(!id[a][b])id[a][b]=tot++;
107             if(!id[a][b+1])id[a][b+1]=tot++;
108             adde(id[a][b],id[a][b+1]);
109         }
110 
111         ans=0;
112         mem(y,0);
113         for(i=1;i<tot;i++){
114             mem(vis,0);
115             if(dfs(i))ans++;
116         }
117 
118         printf("%d
",ans>>1);
119     }
120     return 0;
121 }