灯塔(LightHouse)

Description

As shown in the following figure, If another lighthouse is in gray area, they can beacon each other.

For example, in following figure, (B, R) is a pair of lighthouse which can beacon each other, while (B, G), (R, G) are NOT.

Input

1st line: N

2nd ~ (N + 1)th line: each line is X Y, means a lighthouse is on the point (X, Y).

Output

How many pairs of lighthourses can beacon each other

( For every lighthouses, X coordinates won't be the same , Y coordinates won't be the same )

Example

Input

3
2 2
4 3
5 1

Output

1

Restrictions

For 90% test cases: 1 <= n <= 3 * 105

For 95% test cases: 1 <= n <= 106

For all test cases: 1 <= n <= 4 * 106

For every lighthouses, X coordinates won't be the same , Y coordinates won't be the same.

1 <= x, y <= 10^8

Time: 2 sec

Memory: 256 MB

Hints

The range of int is usually [-231, 231 - 1], it may be too small.

第一眼看到这题时第一反应是用树状数组，但是发现数据太大，开不了10的6次方的二维数组，看了别人的博客才知道是用归并排序。

先理解一下题意，要求有多少对两两互相照亮的灯塔，怎么样才能是两两互相照亮的呢，两点的斜率为正，也就是按x递增排序，y也递增。

这样咱们就可以把n个点对x进行排序，找到y有多少对是顺序对就行，这就可以用到归并排序，在对左右两个集合合并时，i，j分别为左右两个集合的

指针，如果le[i]<ri[j]，正序对加上n2-j+1对，就这样去考虑。还有就是那里不能用<algorithm>，可以用stdlib.h里的qsort（）。

 #include <iostream>
 #include <cstring>
 #include <cstdio>
 #include <cstdlib>

 using namespace std;

 struct w
 {
        int x;
        int y;
 }k[];
 int yi[];
 long long ans;
 int le[],ri[];

 int cmp( const void *a ,const void *b)
 {
     return (*(w *)a).x > (*(w *)b).x ;
 }

 void merge(int l,int mi,int r)
 {
      int i,j,p,n1=mi-l+,n2=r-mi;
      const int MAX=;

      for (int i=;i<=n1;i++)
      le[i]=k[l+i-].y;
      for (int i=;i<=n2;i++)
      ri[i]=k[mi+i].y;

       le[n1+]=MAX;ri[n2+]=MAX;
       i=;j=;
       for (int p=l;p<=r;p++)
       {
           if (le[i]>ri[j])
           k[p].y=ri[j++];
           else
           {
               k[p].y=le[i++];
               ans+=n2-j+;
           //    cout <<ans<<endl;
           }
       }
 }
 void mergesort(int l,int r)
 {
      if (l==r)
      return;
      int mid=l+(r-l)/;
      mergesort(l,mid);
      mergesort(mid+,r);
      merge(l,mid,r);
 }

 int main()
 {
     int n;
     cin>>n;
     for (int i=;i<n;i++)
     cin>>k[i].x>>k[i].y;
     qsort(k,n,sizeof(k[]),cmp);
     ans=;
     mergesort(,n-);
     cout <<ans<<endl;
     return ;
 }