CodeForces - 650A(104/600)

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博士曼哈顿和朋友丹尼尔必须尽快警告处于危险中的监视者们。他们需要计算监视者之间的距离，但两人的计算方式不同。博士曼哈顿使用绝对差值，而丹尼尔使用欧几里得距离。题目要求找出两者计算结果相同的监视者对数。输入包含监视者数量和坐标，输出符合条件的对数。示例展示了计算过程和结果。

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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.

Example
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.

很容易能证明这种点必须是同一条x或者y

然后只要map大力去重就可以了

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<queue>
#include<functional>
#include<vector>
#include<cstring>
#include<string>
#include<map>
#include<cmath>
#include<stdio.h>
#include<stack>
#include<set>
using namespace std;
map<long long , long long >lsx, lsy;
long long  xx[200001], yy[200001],lisx[200001],lisy[200001],xg=0,yg=0;
long long  xxi[200001], yyi[200001];
map<long long , long long >mp[200001];
int main()
{
    long long  n,q,w;
    cin >> n;
    for (long long  a = 1; a <= n; a++)
    {
        scanf("%I64d%I64d", &q, &w);
        xx[a] = q, yy[a] = w;
        if (!lsx[q])
        {
            lisx[++lisx[0]] = q;
            lsx[q] = 1;
        }
        if (!lsy[w])
        {
            lisy[++lisy[0]] = w;
            lsy[w] = 1;
        }
    }
    lsx.clear(), lsy.clear();
    xg = lisx[0], yg = lisy[0];
    sort(lisx + 1, lisx + xg + 1);
    sort(lisy + 1, lisy + yg + 1);
    for (long long  a = 1; a <= xg; a++)lsx[lisx[a]] = a;
    for (long long  a = 1; a <= yg; a++)lsy[lisy[a]] = a;
    for (long long  a = 1; a <= n; a++)
    {
        xxi[lsx[xx[a]]]++;
        yyi[lsy[yy[a]]]++;
    }
    long long dan = 0;
    for (long long  a = 1; a <= xg; a++)dan += xxi[a]*(xxi[a] - 1)/2;
    for (long long  a = 1; a <= yg; a++)dan += yyi[a]*(yyi[a] - 1)/2;
    for (long long  a = 1; a <= n; a++)
    {
        long long  qq = lsx[xx[a]], ww = lsy[yy[a]];
        mp[qq][ww]++;
    }
    for (long long  a = 1; a <= xg; a++)
    {
        for (auto b : mp[a])
        {
            dan -= b.second*(b.second - 1) / 2;
        }
    }
    printf("%I64d", dan);
}