Codeforces 651C - Watchmen

最新推荐文章于 2016-05-27 14:39:06 发布

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本文介绍了一个算法问题，即计算平面上n个守望者位置中，由Doctor Manhattan和Daniel Dreiberg定义的距离相等的守望者对的数量。Doctor Manhattan使用曼哈顿距离，而Daniel使用欧几里得距离。

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 $\text{[math]}$ .

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 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

/*
Codeforces 651C - Watchmen

给你很多组a,b  然后通过两种计算方法判断答案是否相同
1.  |xi-xj|+|yi-yj|
2. sqrt((xi-xj)*(xi-xj) + (yi-yj)*(yi-yj))

然后发现只有当xi = xj 或 yi = yj时答案相同
但同时要排除(xi,yi)=(xj,yj)的情况

hhh-2016-03-07 17:04:24
*/

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <stack>
#include <map>
using namespace std;
typedef long long ll;
typedef long double ld;
#define lson (i<<1)
#define rson ((i<<1)|1)
const int maxn = 10010;

ll a;
ll b;
map<ll,ll> mp1;
map<ll,ll> mp2;
map< pair<ll,ll>,ll > mp;
int main()
{
    int T,n,m;
    while(scanf("%d",&n) != EOF)
    {
        ll ans = 0;
        ll num = 0;
        mp1.clear();
        mp2.clear();
        mp.clear();
        for(int i = 1; i <= n; i++)
        {
            scanf("%I64d%I64d",&a,&b);

            num += mp[ make_pair(a,b)];
            mp[ make_pair(a,b)] ++;

            if(mp1.find(a) == mp1.end())
                mp1[a] = 1;
            else
            {
                ans += mp1[a];
                mp1[a]++;
            }

            if(mp2.find(b) == mp2.end())
                mp2[b] = 1;
            else
            {
                ans += mp2[b];
                mp2[b]++;
            }
        }
        printf("%I64d\n",(ll)ans-num);
    }
    return 0;
}