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