CF Watchmen 【思维+数学】

E - Watchmen

Time Limit:3000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u

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Description

Watchmen are in a danger and Doctor Manhattan together with his friend Daniel Dreiberg should warn them as soon as possible. There aren watchmen on a plane, the i-th watchman is located at point (x_i, y_i).

They need to arrange a plan, but there are some difficulties on their way. As you know, Doctor Manhattan considers the distance between watchmen i and j to be |x_i - x_j| + |y_i - y_j|. Daniel, as an ordinary person, calculates the distance using the formula .

The success of the operation relies on the number of pairs (i, j) (1 ≤ i < j ≤ n), such that the distance between watchman i and watchmen j calculated by Doctor Manhattan is equal to the distance between them calculated by Daniel. You were asked to compute the number of such pairs.

Input

The first line of the input contains the single integer n (1 ≤ n ≤ 200 000) — the number of watchmen.

Each of the following n lines contains two integers x_i and y_i (|x_i|, |y_i| ≤ 10⁹).

Some positions may coincide.

Output

Print the number of pairs of watchmen such that the distance between them calculated by Doctor Manhattan is equal to the distance calculated by Daniel.

Sample Input

Input

31 17 51 5

Output

2

Input

60 00 10 2-1 10 11 1

Output

11

Hint

In the first sample, the distance between watchman 1 and watchman 2 is equal to |1 - 7| + |1 - 5| = 10 for Doctor Manhattan and 

 for Daniel. For pairs (1, 1), (1, 5) and (7, 5), (1, 5) Doctor Manhattan and Daniel will calculate the same distances.

题确实挺好的，明白了题意：求x1=x2||y1=y2的对数，直接循环出错

看博客思路，用结构题分别对x和y排序，算对数，再减x,y都相等时重复的

代码：

#include <cstdio>#include <algorithm>using namespace std;struct Node{int x;int y;}a[200005];int cmp1(Node a,Node b){if(a.x==b.x)return a.y<b.y;return a.x<b.x;}int cmp2(Node a,Node b){if(a.y==b.y)return a.x<b.x;return a.y<b.y;}int main(){int n,i,j;long long ans,cnt,sum;while(~scanf("%d",&n)){for(i=0;i<n;i++)scanf("%d%d",&a[i].x,&a[i].y);ans=1;cnt=1;sum=0;sort(a,a+n,cmp1);//第一次求对数for(i=1;i<n;i++){if(a[i].x==a[i-1].x){ans++;if(a[i].y==a[i-1].y)//重复的对数{cnt++;}else{sum-=cnt*(cnt-1)/2;cnt=1;}}else{sum+=ans*(ans-1)/2;sum-=cnt*(cnt-1)/2;cnt=1;ans=1;}}if(ans!=1)sum+=ans*(ans-1)/2;if(cnt!=1)sum-=cnt*(cnt-1)/2;sort(a,a+n,cmp2);//第二次求对数ans=1;for(i=1;i<n;i++){if(a[i].y==a[i-1].y){ans++;}else{sum+=ans*(ans-1)/2;ans=1;}}if(ans!=1){sum+=ans*(ans-1)/2;}printf("%lld\n",sum);}return 0;}