codefroces 650A. Watchmen

最新推荐文章于 2025-06-07 10:07:37 发布

转载最新推荐文章于 2025-06-07 10:07:37 发布 · 56 阅读

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原文链接：http://www.cnblogs.com/shinianhuanniyijuhaojiubujian/p/7112489.html

本文介绍了一个关于Watchmen的距离计算问题，需要计算由Doctor Manhattan和Daniel分别使用不同距离公式得出相同结果的Watchmen对数。输入包含多个Watchmen的位置坐标，输出则是符合特定条件的Watchmen对的数量。

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 (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 $\text{[math]}$ .

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.

Examples

Input

Output

Input

Output

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 $\text{[math]}$ for Daniel. For pairs (1, 1), (1, 5) and (7, 5), (1, 5) Doctor Manhattan and Daniel will calculate the same distances.

加上相同的，减去重复的

#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
#include <cmath>
#include <vector>
#include <set>
#include <map>
#include <algorithm>
using namespace std;
typedef long long ll;
int main()
{
    map<ll,ll>ans,pos;
    map<pair<ll,ll>,ll>inf;
    ll x,y,n,cnt=0;
    cin>>n;
    while(n--)
    {
        cin>>x>>y;
        cnt+=ans[x]++;
        cnt+=pos[y]++;
        cnt-=inf[make_pair(x,y)]++;//减去重复的
    }
    cout<<cnt<<endl;
}

转载于:https://www.cnblogs.com/shinianhuanniyijuhaojiubujian/p/7112489.html