Watchmen(排列，去重）

<span style="font-family: Arial, Helvetica, sans-serif; background-color: rgb(255, 255, 255);">Watchmen</span>

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

Description

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

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
|xi - xj| + |yi - yj|. 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
xi and
yi (|xi|, |yi| ≤ 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

3
1 1
7 5
1 5

Output

2

Input

6
0 0
0 1
0 2-1 1
0 1
1 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.

题目大意：有n个点，有两个人计算点与点之间的距离公式不同，一个用公式： |xi - xj| + |yi - yj|.。另一个用公式：

.，问有多少对点的距离，用这两个公式算出来时一样的。其实把两个公式合并化简后，可以得出
(xi - xj）× （yi - yj）=
0.  那么这个题目就是排序，，然后去掉重复的，，比如xi-xj = 0 && yi - y j  = 0 这样就可能重复判断

所以要去重。

代码：

#include <stdio.h>
using namespace std;
#include<iostream>
#include<algorithm>
struct N
{
int a,b;
}num[200100];
bool cmp1(N a , N b) //第一次排序
{
if ( a.a == b.a)
{
return a.b < b.b;
}
return a.a < b.a;
}
bool cmp2(N a , N b)  //第二次排序
{
if ( a.b == b.b)
{
return a.a < b.a;
}
return a.b < b.b;
}
int main()
{
int n;
while (~scanf("%d",&n ) )
{
for (int i = 0 ; i < n ; i++ )
{
scanf("%d%d",&num[i].a,&num[i].b);
}
sort(num,num+n,cmp1);
__int64 sum1 = 1;
__int64 sum2 = 1;
__int64 sum3=1;
__int64 res = 0;
for (int i = 1 ; i < n ; i++ )
{
if(num[i].a == num[i-1].a)
{
sum1 ++; //记录相同点
if ( num[i].b == num[i-1].b)
{
sum2++; //记录重合点
}
else
{
res -= sum2 * (sum2 - 1) / 2; //除去相同的对数
sum2 = 1;
}
}
else
{
res += sum1 * (sum1 -1 ) / 2;
res -= sum2 * (sum2 - 1) / 2;
sum1 = 1;
sum2 = 1;
}
}

res -= sum2 * (sum2 -1 ) / 2;

res += sum1 * (sum1 - 1) / 2; //查看结尾

sort(num,num+n,cmp2);
for (int i = 1 ; i < n ; i++ )
{
if(num[i].b == num[i-1].b)
{
sum3++;
}
else
{
res += sum3 * (sum3 -1 ) / 2;
sum3 = 1;
}
}

res += sum3 * (sum3 - 1) / 2;

printf("%I64d\n",res);
}
return 0;
}