codeforces 651C. Watchmen(排列，去重)

题目链接：http://codeforces.com/contest/651/problem/C

C. Watchmen

time limit per test
3 seconds

memory limit per test
256 megabytes

input
standard input

output
standard output

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.

Examples

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

Note

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 <bits/stdc++.h>
using namespace std;
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 temp = 1;       //几个是相同的
__int64 tt = 1;         //重复的个数
__int64 res = 0;
for (int i = 1 ; i < n ; i++ )
{
if(num[i].a == num[i-1].a)      //xi == xi-1
{
temp ++;
if ( num[i].b == num[i-1].b)
{
tt++;
}
else
{
res -= tt * (tt - 1) / 2;   //去重
tt = 1;
}
}
else
{
res += temp * (temp -1 ) / 2;       //排列组合，从temp个两两组合的个数
res -= tt * (tt - 1) / 2;
tt = 1;
temp = 1;
}
}
if ( tt != 1)       //判断结尾是不是有些没有去重
{
res -= tt * (tt -1 ) / 2;
}
tt = 1;
if ( temp != 1 )        //判断结尾有些是不是没有计算
{
res += temp * (temp - 1) / 2;
}
//printf("%I64d\n",res);
temp = 1;
sort(num,num+n,cmp2);       //第二次排序
for (int i = 1 ; i < n ; i++ )
{
if(num[i].b == num[i-1].b)
{
temp ++;
}
else
{
res += temp * (temp -1 ) / 2;
temp = 1;
}
}
if ( temp != 1 )
{
res += temp * (temp - 1) / 2;
}
printf("%I64d\n",res);
}
return 0;
}