hdu 4619 Warm up 2 网络流 最小割

题意：告诉你一些骨牌，然后骨牌的位置与横竖，这样求最多保留多少无覆盖的方格。

这样的话有人用二分匹配，因为两个必定去掉一个，我用的是最小割，因为保证横着和竖着不连通即可。

#include <stdio.h>
#include <string.h>
#include <vector>
#include <iostream>
#include <queue>
#define loop(s,i,n) for(i = s;i < n;i++)
using namespace std;
const int maxn = 2050;
struct edge
{
    int u,v,cap,flow;
};
vector<edge>edges;
vector<int>g[maxn],G[maxn];
void addedge(int u,int v,int cap,int flow)
{
    //printf("********u %d ****v %d *****
",u,v);
    edges.push_back((edge){u,v,cap,flow});
    edges.push_back((edge){v,u,0,0});
    int m;
    m = edges.size();
    g[u].push_back(m-2);
    g[v].push_back(m-1);
}
void init(int n)
{
    int i;
    for(i = 0;i <= n;i++)
    {
        g[i].clear();
    }
    edges.clear();
}
int vis[maxn],dis[maxn],cur[maxn];
int bfs(int s,int t,int n)
{
    memset(vis,0,sizeof(vis));
    queue<int>q;
    q.push(s);
    dis[s] = 0;
    vis[s] = 1;
    while(!q.empty())
    {
        int u,v;
        u = q.front();
        q.pop();
        for(int i = 0;i < g[u].size();i++)
        {
            edge &e = edges[g[u][i]];
            v = e.v;
            if(!vis[v] && e.cap-e.flow > 0)
            {
                vis[v] = 1;
                dis[v] = dis[u] +1;
                q.push(v);
            }
        }
    }
    return vis[t];
}
int dfs(int u,int a,int t)
{
    if(u == t || a == 0)
    return a;
    int v;
    int flow = 0,f;
    for(int& i = cur[u]; i < g[u].size();i++)
    {
        edge &e = edges[g[u][i]];
        v = e.v;
        if(dis[u]+1 == dis[v] &&(f = dfs(v,min(a,e.cap-e.flow),t)))
        {
            e.flow += f;
            edges[g[u][i]^1].flow -= f;
            flow += f;
            a -= f;
            if(a == 0)
            break;
        }
    }
    return flow;
}
int maxflow(int s,int t)
{
    int flow = 0;
    while(bfs(s,t,t))
    {
        memset(cur,0,sizeof(cur));
        flow+=dfs(s,1000050,t);
    }
    return flow;
}
struct node
{
    int x1,y1,x2,y2;
}px[2050],py[1050];
int lap(int a,int b)
{
    if(py[a].x1 == px[b].x1 && py[a].y1 == px[b].y1)
    return 1;
    if(py[a].x2 == px[b].x1 && py[a].y2 == px[b].y1)
    return 1;
    if(py[a].x1 == px[b].x2 && py[a].y1 == px[b].y2)
    return 1;
    if(py[a].x2 == px[b].x2 && py[a].y2 == px[b].y2)
    return 1;
    return 0;
}
int main()
{
    int n,m;
    while(scanf("%d%d",&n,&m)&&(m||n))
    {
        int i,j;
        int x,y;
        init(m+n+5);
        for(i = 0;i < n;i++)
        {
            scanf("%d %d",&x,&y);
            px[i].x1 = x;
            px[i].y2 =  px[i].y1 = y;
            px[i].x2 = x+1;
        }
        for(j = 0;j < m;j++)
        {
            scanf("%d %d",&x,&y);
            py[j].x1 = py[j].x2 = x;
            py[j].y1 = py[j].y2 = y;
            py[j].y2++;
        }
        int cnt;
        cnt = 0;
        loop(0,i,n)
        addedge(0,i+1,1,0);
        loop(0,j,m)
        addedge(n+j+1,m+n+1,1,0);
        loop(0,i,n)
        {
            loop(0,j,m)
            {
                if(lap(j,i))
                addedge(i+1,j+n+1,1005,0);
            }
        }
        cout<<m+n-maxflow(0,m+n+1)<<endl;
    }
}