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\n",ans>>1);

119     }

120     return 0;

121 }