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 = ;
 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 *****\n",u,v);
     edges.push_back((edge){u,v,cap,flow});
     edges.push_back((edge){v,u,,});
     int m;
     m = edges.size();
     g[u].push_back(m-);
     g[v].push_back(m-);
 }
 void init(int n)
 {
     int i;
     for(i = ;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,,sizeof(vis));
     queue<int>q;
     q.push(s);
     dis[s] = ;
     vis[s] = ;
     while(!q.empty())
     {
         int u,v;
         u = q.front();
         q.pop();
         for(int i = ;i < g[u].size();i++)
         {
             edge &e = edges[g[u][i]];
             v = e.v;
             if(!vis[v] && e.cap-e.flow > )
             {
                 vis[v] = ;
                 dis[v] = dis[u] +;
                 q.push(v);
             }
         }
     }
     return vis[t];
 }
 int dfs(int u,int a,int t)
 {
     if(u == t || a == )
     return a;
     int v;
     int flow = ,f;
     for(int& i = cur[u]; i < g[u].size();i++)
     {
         edge &e = edges[g[u][i]];
         v = e.v;
         if(dis[u]+ == dis[v] &&(f = dfs(v,min(a,e.cap-e.flow),t)))
         {
             e.flow += f;
             edges[g[u][i]^].flow -= f;
             flow += f;
             a -= f;
             if(a == )
             break;
         }
     }
     return flow;
 }
 int maxflow(int s,int t)
 {
     int flow = ;
     while(bfs(s,t,t))
     {
         memset(cur,,sizeof(cur));
         flow+=dfs(s,,t);
     }
     return flow;
 }
 struct node
 {
     int x1,y1,x2,y2;
 }px[],py[];
 int lap(int a,int b)
 {
     if(py[a].x1 == px[b].x1 && py[a].y1 == px[b].y1)
     return ;
     if(py[a].x2 == px[b].x1 && py[a].y2 == px[b].y1)
     return ;
     if(py[a].x1 == px[b].x2 && py[a].y1 == px[b].y2)
     return ;
     if(py[a].x2 == px[b].x2 && py[a].y2 == px[b].y2)
     return ;
     return ;
 }
 int main()
 {
     int n,m;
     while(scanf("%d%d",&n,&m)&&(m||n))
     {
         int i,j;
         int x,y;
         init(m+n+);
         for(i = ;i < n;i++)
         {
             scanf("%d %d",&x,&y);
             px[i].x1 = x;
             px[i].y2 =  px[i].y1 = y;
             px[i].x2 = x+;
         }
         for(j = ;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 = ;
         loop(,i,n)
         addedge(,i+,,);
         loop(,j,m)
         addedge(n+j+,m+n+,,);
         loop(,i,n)
         {
             loop(,j,m)
             {
                 if(lap(j,i))
                 addedge(i+,j+n+,,);
             }
         }
         cout<<m+n-maxflow(,m+n+)<<endl;
     }
 }