【CodeForces】651C - Watchmen（排序，容斥原理）

题目链接：点击打开题目

C. 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

31 17 51 5

output

2

input

60 00 10 2-1 10 11 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.

这题有毒，cmp函数改了改就在第五组不RE了，RE了十几回也是够了。

代码如下：

#include <cstdio>#include <algorithm>using namespace std;struct node{__int64 x,y;}data[400022];bool cmp1(node x,node y)//谁能告诉我为什么这里变量写成a、b就RE!!! {return x.x<y.x || x.x==y.x && x.y<y.y;}bool cmp2(node x,node y){return x.y<y.y || x.y==y.y && x.x<y.x;}__int64 ans,n;int cnt;int main(){scanf ("%I64d\n",&n);for (int i = 1 ; i <= n ; i++)scanf ("%I64d%I64d\n",&data[i].x,&data[i].y);//先求x相同的 （同时统计出x、y均相同的个数） sort (data+1,data+n+1,cmp1);cnt = 1;for (int i = 2 ; i <= n ; i++){if (data[i].x == data[i-1].x){ans += cnt;cnt++;}elsecnt = 1;}//减去相同的cnt = 1;for (int i = 2 ; i <= n ; i++){if (data[i].x == data[i-1].x && data[i].y == data[i-1].y){ans -= cnt;cnt++;}elsecnt = 1;}//再求y相同的sort (data+1,data+1+n,cmp2);cnt = 1;for (int i = 2 ; i <= n ; i++){if (data[i].y == data[i-1].y){ans += cnt;cnt++;}elsecnt = 1;}//最后输出printf ("%I64d\n",ans);return 0;}