P3033 [USACO11NOV]Cow Steeplechase G

发布时间 2023-03-28 23:01:34作者: towboat

给出 N平行于坐标轴的线段,要你选出尽量多的线段使得这些线段两两没有交点(顶点也算)。

横的与横的,竖的与竖的线段之间保证没有交点,输出最多能选出多少条线段。

 

#include <iostream>
#include <algorithm>
#include <cstring>
#include <queue>
using namespace std ;
const int N =1e4, M=3e6+2;
 const int inf =1e9;

 int b[N],x1[N],x2[N],y1[N],y2[N];
 
 int all=1,hd[N],go[M],w[M],nxt[M];
   
 int S,T,n,m;
 int dis[M],ans=0,now[M];
   
 void add_(int x,int y,int z){
     nxt[++all]=hd[x]; hd[x]=all; go[all]=y;
     w[all]=z;
     swap(x,y);
     nxt[++all]=hd[x]; hd[x]=all; go[all]=y;
     w[all]=0;
 }
  bool bfs(){
     for(int i=0;i<M;i++)dis[i]=inf;
     queue<int> q;
     q.push(S);
     now[S]=hd[S];
     dis[S]=0;
       
     while(q.empty()==0){
         int x=q.front();
         q.pop();
         for(int i=hd[x];i;i=nxt[i]){
             int y=go[i];
             if(w[i]>0&&dis[y]==inf){
                 dis[y]=dis[x]+1;
                 now[y]=hd[y];
                 q.push(y);
                 if(y==T) return 1;
             }
         }
     }
     return 0;
 }
 int dfs(int x,int sum){
     if(x==T) return sum;
     int k,res=0;
       
     for(int i=now[x];i&& sum ;i=nxt[i]){
         now[x]=i;
         int y=go[i];
         if(w[i]>0&&(dis[y]==dis[x]+1)){
             k=dfs(y,min(sum,w[i]));
             if(k==0) dis[y]=inf;
             w[i]-=k;
             w[i^1]+=k;
             res+=k;
             sum-=k;
         }
     }
     return res;
 }
 void dinic(){
 	int ans=0;
 	while(bfs()) ans+=dfs(S,inf);
 	cout<<n-ans<<endl;
 }
 signed main(){
 	cin>>n;
 	S=0,T=n+1;
 	int i,j;
 	for(i=1;i<=n;i++){
 		cin>>x1[i]>>y1[i]>>x2[i]>>y2[i];
 		if(x1[i]==x2[i]) b[i]=1;
 	    else b[i]=2;
 	}
 	
 	for(i=1;i<=n;i++) if(b[i]==1) add_(S,i,1);
 	for(i=1;i<=n;i++) if(b[i]==2) add_(i,T,1);
 	
 	for(i=1;i<=n;i++)
 	 for(j=1;j<=n;j++){
 	 	if(i!=j&&b[i]==1&&b[j]==2){
 	 		
 	    int Y1=y1[i],Y2=y2[i],Y3=y1[j]; if(Y1<Y2) swap(Y1,Y2);
 	    int X1=x1[j],X2=x2[j],X3=x1[i]; if(X1>X2) swap(X1,X2);
 	    
 	    if(Y1>=Y3&&Y2<=Y3&&X1<=X3&&X2>=X3)
 	  		add_(i,j,1);
 	  	}	
 	  }
 	  
 	dinic();
 }