【最大流，二分图匹配】【hdu2063】【过山车】

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();
    }
}