题意：裸的求二分图匹配

建立一个源点 连向一边所有的点 容量为1;

另外一边点都连向汇点  容量为1;

二分图的边容量也为1

源点汇点求一遍最大流即可

#include <cstdio>

#include <cstdlib>

#include <cmath>

#include <cstring>

#include <ctime>

#include <algorithm>

#include <iostream>

#include <sstream>

#include <string>

#define oo 0x13131313

using namespace std;

const int MAXN=2000+5;

const int MAXM=10000+5;

const int INF=0x3f3f3f3f;

struct Edge

{

    int to,next,cap,flow;

    void get(int a,int b,int c,int d)

    {

        to=a;next=b;cap=c;flow=d;

    }

}edge[MAXM];

int tol;

int head[MAXN];

int gap[MAXN],dep[MAXN],pre[MAXN],cur[MAXN];

void init()

{

    tol=0;

    memset(head,-1,sizeof(head));

}

//单向图三个参数，无向图四个参数

void addedge(int u,int v,int w,int rw=0)

{

    edge[tol].get(v,head[u],w,0);head[u]=tol++;

    edge[tol].get(u,head[v],rw,0);head[v]=tol++;

}

int sap(int start,int end,int N)

{

    memset(gap,0,sizeof(gap));

    memset(dep,0,sizeof(dep));

    memcpy(cur,head,sizeof(head));

    int u=start;

    pre[u]=-1;

    gap[0]=N;

    int ans=0;

    while(dep[start]<N)

    {

        if(u==end)

        {

            int Min=INF;

            for(int i=pre[u];i!=-1;i=pre[edge[i^1].to])

                if(Min>edge[i].cap-edge[i].flow)

                   Min=edge[i].cap-edge[i].flow;

            for(int i=pre[u];i!=-1;i=pre[edge[i^1].to])

            {

                edge[i].flow+=Min;

                edge[i^1].flow-=Min;

            }

            u = start;

            ans+=Min;

            continue;

        }

        bool flag=false;

        int v;

        for(int i=cur[u];i !=-1;i=edge[i].next)

        {

            v=edge[i].to;

            if(edge[i].cap-edge[i].flow&&dep[v]+1==dep[u])

            {

                flag=true;

                cur[u]=pre[v]=i;

                break;

            }

        }

        if(flag)

        {

            u=v;

            continue;

        }

        int Min=N;

        for(int i=head[u];i!=-1;i=edge[i].next)

            if(edge[i].cap-edge[i].flow&&dep[edge[i].to]<Min)

        {

            Min=dep[edge[i].to];

            cur[u]=i;

        }

        gap[dep[u]]--;

        if(!gap[dep[u]]) return ans;

        dep[u]=Min+1;

        gap[dep[u]]++;

        if(u!=start) u=edge[pre[u]^1].to;

    }

    return ans;

}

int KK,MM,NN;

void input()

{

    int a,b;

    for(int i=1;i<=KK;i++)

    {

        scanf("%d%d",&a,&b);

        addedge(a,b+MM,1);

    }

}

void solve()

{

    int ANS;

    for(int i=1;i<=MM;i++)

    {

        addedge(MM+NN+1,i,1);

    }

    for(int i=MM+1;i<=MM+NN;i++)

    {

        addedge(i,MM+NN+2,1);

    }

    ANS=sap(MM+NN+1,MM+NN+2,MM+NN+2);

    printf("%d\n",ANS);

}

void File()

{

    freopen("a.in","r",stdin);

    freopen("a.out","w",stdout);

}

int main()

{

  //  File();

    while(cin>>KK>>MM>>NN&&KK)

    {

        init();

        input();

        solve();

    }

}