codeforces Codeforces 650A Watchmen

题意：两点(x1,y1), (x2,y2)的曼哈顿距离=欧几里得距离

也就是：x1=x2或y1=y2，再删除重合点造成的重复计数即可。

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <cstring>
#include <math.h>
#include <stdlib.h>
#include <time.h>
#include <stack>
#define clc(a,b) memset(a,b,sizeof(a))
#define LL long long
using namespace std;
const int maxn=200020;
const int inf=0x3f3f3f3f;
struct node{
    LL x,y;
}a[maxn];
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){
    if(a.y==b.y) return a.x<b.x;
    return a.y<b.y;
}

int n;
// int a[maxn],b[maxn];
int main(){
    scanf("%d",&n);
    LL sum=0;
    for(int i=0;i<n;i++){
        scanf("%I64d%I64d",&a[i].x,&a[i].y);
    }
    sort(a,a+n,cmp1);
    LL sum1=0;
    for(int i=0;i<n;i++){
        int j=i;
        sum1=0;
        while(j<n&&a[i].x==a[j].x){
            sum1++;
            j++;
        }
        i=j-1;
        sum+=sum1*(sum1-1)/2;
    }
    sort(a,a+n,cmp2);
    sum1=0;
    for(int i=0;i<n;i++){
        int j=i;
        sum1=0;
        while(j<n&&a[i].y==a[j].y){
            sum1++;
            j++;
        }
        i=j-1;
        sum+=sum1*(sum1-1)/2;
    }
    sum1=0;
    for(int i=0;i<n;i++){
        int j=i;
        sum1=0;
        while(j<n&&a[i].y==a[j].y&&a[j].x==a[i].x){
            sum1++;
            j++;
        }
        i=j-1;
        sum-=sum1*(sum1-1)/2;
    }
    printf("%I64d
",sum);
    return 0;
}