poj 3168 Barn Expansion 几何yy

题链：http://poj.org/problem?

id=3168

Barn Expansion

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 2087		Accepted: 544

Description

Farmer John has N (1 <= N <= 25,000) rectangular barns on his farm, all with sides parallel to the X and Y axes and integer corner coordinates in the range 0..1,000,000. These barns do not overlap although they may share corners and/or sides with other barns. 



Since he has extra cows to milk this year, FJ would like to expand some of his barns. A barn has room to expand if it does not share a corner or a wall with any other barn. That is, FJ can expand a barn if all four of its walls can be pushed outward by at least
some amount without bumping into another barn. If two barns meet at a corner, neither barn can expand. 



Please determine how many barns have room to expand.

Input

Line 1: A single integer, N 



Lines 2..N+1: Four space-separated integers A, B, C, and D, describing one barn. The lower-left corner of the barn is at (A,B) and the upper right corner is at (C,D).

Output

Line 1: A single integer that is the number of barns that can be expanded.

Sample Input

5
0 2 2 7
3 5 5 8
4 2 6 4
6 1 8 6
0 0 8 1

Sample Output

2

Hint

Explanation of the sample: 



There are 5 barns. The first barn has its lower-left corner at (0,2) and its upper-right corner at (2,7), and so on. 



Only two barns can be expanded --- the first two listed in the input. All other barns are each in contact with at least one other barn.

题意：给若干个矩形，直接仅仅有接触，没有重叠。计算出有多少矩形是不和其它矩形有接触。

做法：

把两条横向边和纵向边分解开来。各自存入数组 hh，和ss。

然后排序。以横向为例。先按高度排序，高度同样的 按左边的坐标从小到大排序。

然后for一遍，注意下推断重合时，之前的那个矩形pre也要标记成有接触。

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <limits.h>
#include <malloc.h>
#include <ctype.h>
#include <math.h>
#include <string>
#include <iostream>
#include <algorithm>
using namespace std;
#include <stack>
#include <queue>
#include <vector>
#include <deque>
#include <set>
#include <map>

struct point
{
	int s,x,id;//下 左 负
	int z,y;
	point()
	{}
	point(int _x,int _s,int _z,int _y,int _id)
	{
		s=_s,x=_x,z=_z,y=_y,id=_id;
	}
};
point hh[1000010]; //放横的
point ss[1001000];
int has[26000];
int cmph(point a,point b)
{
	if(a.s!=b.s)
	return a.s<b.s;
	return a.z<b.z;
}
int cmps(point a,point b)
{
	if(a.z!=b.z)
	return a.z<b.z;
	return a.x<b.x;
}
int main()
{
	int n;
	while(scanf("%d",&n)!=EOF)
	{
		memset(has,0,sizeof has);
		int h=0;
		int s=0;
		 for(int i=0;i<n;i++)
		 {
			 int l,x,r,sh;
			 scanf("%d%d",&l,&x);
			 scanf("%d%d",&r,&sh);
			 hh[h++]=point(x,x,l,r,i);
			 hh[h++]=point(sh,sh,l,r,i);

			 ss[s++]=point(x,sh,l,l,i);
			 ss[s++]=point(x,sh,r,r,i);
		 }
		 sort(hh,hh+h,cmph);
		 sort(ss,ss+s,cmps);
		 int z,y;
		 int pre;
		 for(int i=0;i<h;i++)
		 {
			  if(i==0)
			  {
				  z=hh[i].z;
				  y=hh[i].y;
				  pre=hh[i].id;
			  }
			  else if(hh[i-1].s==hh[i].s)
			  {
				  if(hh[i].z<=y) //在之前的范围内
				  {
					  has[pre]=1;
					  has[hh[i].id]=1;
				  }
				  else  //不在之前范围内
				  {
					  z=hh[i].z;
					  y=hh[i].y;
					  pre=hh[i].id;
				  }
				  if(hh[i].y>y)//扩展右边
					  y=hh[i].y;
			  }
			  else//不在一个高度时
			  {
				  z=hh[i].z;
				  y=hh[i].y;
				  pre=hh[i].id;
			  }
		 }
		 int xi,sh;
		 for(int i=0;i<=s;i++)
		 {
			// printf("x%d s%d l%d  id%d\n",ss[i].x,ss[i].s,ss[i].z);
			 if(i==0)
			 {
				 xi=ss[i].x;
				 sh=ss[i].s;
				 pre=ss[i].id;
			 }
			 else if(ss[i-1].y==ss[i].y)
			 {
				 if(ss[i].x<=sh)
				 {
					has[ss[i].id]=1;
					has[pre]=1;
				 }
				 else
				 {
					 xi=ss[i].x;
					 sh=ss[i].s;
					 pre=ss[i].id;
				 }
				 if(ss[i].s>sh)
					 sh=ss[i].s;
			 }
			 else
			 {
				 xi=ss[i].x;
				 sh=ss[i].s;
				 pre=ss[i].id;
			 }
		 }
		 int ans=0;
		 for(int i=0;i<n;i++)
		 {
			 if(has[i])
			 {
				// printf("id%d ",i);
				 ans++;
			 }
		 }
		 printf("%d\n",n-ans);
	}
	return 0;
}
/*
8 4
4 1 0 0 0 0 1 0
0 0 0 1 0 1 0 0
0 2 1 1 3 0 4 0
0 0 0 4 1 1 1 0
*/