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=110;
const int INF=0x3f3f3f3f;
const int MOD=100000,STA=8000010;
const LL LNF=1LL<<60;
const double EPS=1e-8;
const double OO=1e15;
const int dx[4]={-1,0,1,0};
const int dy[4]={0,1,0,-1};
const int day[13]={0,31,28,31,30,31,30,31,31,30,31,30,31};
//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*100],id[N][N],y[N*100];
int tot;

struct Edge{
    int u,v;
}e[40010];
int first[N*100],next[40010];
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!=-1;i=next[i]){
        if(!vis[e[i].v]){
            vis[e[i].v]=1;
            if(!y[e[i].v] || dfs(y[e[i].v])){
                y[e[i].v]=u;
                return 1;
            }
        }
    }
    return 0;
}

int main()
{
 //   freopen("in.txt","r",stdin);
    int i,j,a,b,ans;
    while(~scanf("%d%d",&n,&m) && (n || m))
    {
        mem(id,0);
        tot=1;mt=0;mem(first,-1);
        for(i=0;i<n;i++){
            scanf("%d%d",&a,&b);
            if(!id[a][b])id[a][b]=tot++;
            if(!id[a+1][b])id[a+1][b]=tot++;
            adde(id[a][b],id[a+1][b]);
        }
        for(i=0;i<m;i++){
            scanf("%d%d",&a,&b);
            if(!id[a][b])id[a][b]=tot++;
            if(!id[a][b+1])id[a][b+1]=tot++;
            adde(id[a][b],id[a][b+1]);
        }

        ans=0;
        mem(y,0);
        for(i=1;i<tot;i++){
            mem(vis,0);
            if(dfs(i))ans++;
        }

        printf("%d
",ans>>1);
    }
    return 0;
}