Codeforences #345 C
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.
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.
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.
3 1 1 7 5 1 5
2
6 0 0 0 1 0 2 -1 1 0 1 1 1
11
In the first sample, the distance between watchman 1 and watchman 2 is equal to |1 - 7| + |1 - 5| = 10 for Doctor Manhattan and 
【题意】给了n组数据，每组有两个值，问在n组数据中，有多少对满足两个距离相等，其中由距离公式平方化简，容易推出当两个x或者两个y相等就满足条件吗，最后再去掉重复元素。
【AC代码】
#include <stdio.h>
#include <iostream>
#include <algorithm>
#include <map>
using namespace std;
#define ll long long
map<int,int>m1,m2;
map<pair<int,int>,int>mp;
int main()
{
int n,a,b;
ll ans = 0;
scanf("%d",&n);
for(int i=1; i<=n; i++)
{
scanf("%d%d",&a,&b);
ans -= mp[make_pair(a,b)];
ans += m1[a];
ans += m2[b];
m1[a]++;
m2[b]++;
mp[make_pair(a,b)]++;
}
printf("%I64d\n",ans);
return 0;
}
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