Codeforces Round #345 (Div. 2) C. Watchmen

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

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.

给你n个点的坐标,叫你求((xj-xi)+(yj-yi))*((xj-xi)+(yj-yi))=(xj-xi)*(xj-xi)+(yj-yi)*(yj-yi)

思路:如果该式子成立有三种情况

1.xi=xj        

2.yi=yj 

3.xi=xj&&yi=yj

x相等的对数加y相等的对数减去xy都相等的对数就好了。

此题需要注意的一点是求x,y都相等的对数的时候不能利用x*(1e9+1)+y进行hash,因为x可能为负数,所以(0,-1)和(-1,1e9)的hash值是相同的。应该乘以(1e10+1)!!!!!!

或者对原来的x,y进行统一排序,然后判断x,y同时相等的数目

#include<bits/stdc++.h>
using namespace std;
const int maxn=201000;
typedef __int64 ll;
struct node{
    int x,y;
}a[maxn];
const ll M=1e9;
map<ll,int>mp1,mp2;
bool cmp(node x,node y){
    if(x.x!=y.x)
        return x.x<y.x;
    return x.y<y.y;
}

int main(){
    int n;
    while(scanf("%d",&n)!=EOF){
        for(int i=1;i<=n;i++)
            scanf("%d%d",&a[i].x,&a[i].y);
        mp1.clear(),mp2.clear();
        __int64 ans=0;
        for(int i=1;i<=n;i++){
            ans=ans+mp1[a[i].x],mp1[a[i].x]++;
            ans=ans+mp2[a[i].y],mp2[a[i].y]++;
        }
        sort(a+1,a+n+1,cmp);
        int cnt=0;
        for(int i=2;i<=n;i++){
            if(a[i].x==a[i-1].x&&a[i].y==a[i-1].y){
                cnt++;
                ans-=cnt;
            }
            else
                cnt=0;
        }
        printf("%I64d\n",ans);
    }
    return 0;
}

#include<bits/stdc++.h>
using namespace std;
const int INF=0x3f3f3f3f;
const int maxn=201000;
typedef __int64 ll;
typedef unsigned __int64 ull;
__int64 x[maxn],y[maxn];
const ull M=1e10;
map<ll,int>mp1,mp2;
map<ull,int>mp3;

int main(){
    int n;
    while(scanf("%d",&n)!=EOF){
        for(int i=1;i<=n;i++)
            scanf("%I64d%I64d",&x[i],&y[i]);
        mp1.clear(),mp2.clear(),mp3.clear();
        __int64 ans=0;
        for(int i=1;i<=n;i++){
            ans=ans+mp1[x[i]],mp1[x[i]]++;
            ans=ans+mp2[y[i]],mp2[y[i]]++;
            ans=ans-mp3[(M+1)*x[i]+y[i]],mp3[(M+1)*x[i]+y[i]]++;
        }
        printf("%I64d\n",ans);
    }
    return 0;
}