Watchmen(Codeforces 650A)

                                             A. Watchmen

time limit per test 3 seconds memory limit per test 256 megabytes input standard input output standard output

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.

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

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.

题意是问有多少对的欧几里得距离等于曼哈顿距离

进行两次排序

#include <cstdio>
#include <cmath>
#include <cstring>
#include <iostream>
#include <algorithm>
#define MAX_N 200005
using namespace std;

struct node{
	__int64 x, y;
} dis[MAX_N];

bool cmp1(node a, node b) {
	if (a.x == b.x) {
		return a.y > b.y;
	}
	return a.x > b.x;
}
bool cmp2(node a, node b) {
	return a.y > b.y;
}

int main() {
	__int64 n;
	while (scanf("%I64d", &n) != EOF) {
		__int64 ans = 0;
		for (__int64 i = 0; i < n; i++) {
			scanf("%I64d%I64d", &dis[i].x, &dis[i].y);
		}
		__int64 cnt1 = 1, cnt2 = 1;
		sort(dis, dis + n, cmp1);
		for (__int64 i = 1; i < n;i++) {
			if (dis[i - 1].x == dis[i].x) {
				cnt1++;
				if (dis[i - 1].y == dis[i].y) 	cnt2++;
				else {
					ans -= (cnt2 - 1) * cnt2 / 2;
					cnt2 = 1;
				}
			}
			else {
				ans += cnt1 * (cnt1 - 1) / 2;
				cnt1 = 1;
				ans -= (cnt2 - 1) * cnt2 / 2;
				cnt2 = 1; 
			}	
		}
		ans += cnt1 * (cnt1 - 1) / 2 - cnt2 * (cnt2 - 1) / 2;
		sort(dis, dis + n, cmp2);
		cnt1 = 1;
		for (int i = 1; i < n; i++) {
			if (dis[i - 1].y == dis[i].y) {
				if (dis[i - 1].y == dis[i].y) cnt1++;
			}
			else {
				ans += cnt1 * (cnt1 - 1) / 2;
				cnt1 = 1;
			}
		}
		ans += cnt1 * (cnt1 - 1) / 2;
		printf("%I64d\n", ans);
 	}
	return 0;
}

 

转载于:https://www.cnblogs.com/cniwoq/p/6770950.html