C. Watchmen

题目连接：

http://www.codeforces.com/contest/651/problem/C

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

3

1 1

7 5

1 5

Sample Output

2

Hint

题意

有n个点，然后让你算出有多少对点的欧几里得距离等于曼哈顿距离

题解：

平方之后，显然只要，只要(xi-xj)0或者(yi-yj)0就满足题意了

然后容斥一发，x相等+y相等-xy相等就好。

代码

#include<bits/stdc++.h>

using namespace std;

map<int,long long>x;

map<int,long long>y;

map<pair<int,int>,long long>xy;

int main()

{

    int n;

    scanf("%d",&n);

    for(int i=1;i<=n;i++)

    {

        int X,Y;

        scanf("%d%d",&X,&Y);

        x[X]++;

        y[Y]++;

        xy[make_pair(X,Y)]++;

    }

    map<int,long long>::iterator it;

    long long ans = 0;

    for(it=x.begin();it!=x.end();it++)

    {

        long long p = it->second;

        ans+=(p*(p-1)/2);

    }

    for(it=y.begin();it!=y.end();it++)

    {

        long long p = it->second;

        ans+=(p*(p-1)/2);

    }

    map<pair<int,int>,long long>::iterator it2;

    for(it2=xy.begin();it2!=xy.end();it2++)

    {

        long long p = it2->second;

        ans-=(p*(p-1)/2);

    }

    cout<<ans<<endl;

}