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。

 1 #include <stdio.h>
 2 #include <string.h>
 3 #include <iostream>
 4 #include <algorithm>
 5 #include <vector>
 6 #include <queue>
 7 #include <set>
 8 #include <map>
 9 #include <cstring>
10 #include <math.h>
11 #include <stdlib.h>
12 #include <time.h>
13 #include <stack>
14 #define clc(a,b) memset(a,b,sizeof(a))
15 #define LL long long
16 using namespace std;
17 const int maxn=200020;
18 const int inf=0x3f3f3f3f;
19 struct node{
20     LL x,y;
21 }a[maxn];
22 bool cmp1(node a,node b){
23      if(a.x==b.x) return a.y<b.y;
24      return a.x<b.x;
25 }
26
27 bool cmp2(node a,node b){
28     if(a.y==b.y) return a.x<b.x;
29     return a.y<b.y;
30 }
31
32 int n;
33 // int a[maxn],b[maxn];
34 int main(){
35     scanf("%d",&n);
36     LL sum=0;
37     for(int i=0;i<n;i++){
38         scanf("%I64d%I64d",&a[i].x,&a[i].y);
39     }
40     sort(a,a+n,cmp1);
41     LL sum1=0;
42     for(int i=0;i<n;i++){
43         int j=i;
44         sum1=0;
45         while(j<n&&a[i].x==a[j].x){
46             sum1++;
47             j++;
48         }
49         i=j-1;
50         sum+=sum1*(sum1-1)/2;
51     }
52     sort(a,a+n,cmp2);
53     sum1=0;
54     for(int i=0;i<n;i++){
55         int j=i;
56         sum1=0;
57         while(j<n&&a[i].y==a[j].y){
58             sum1++;
59             j++;
60         }
61         i=j-1;
62         sum+=sum1*(sum1-1)/2;
63     }
64     sum1=0;
65     for(int i=0;i<n;i++){
66         int j=i;
67         sum1=0;
68         while(j<n&&a[i].y==a[j].y&&a[j].x==a[i].x){
69             sum1++;
70             j++;
71         }
72         i=j-1;
73         sum-=sum1*(sum1-1)/2;
74     }
75     printf("%I64d\n",sum);
76     return 0;
77 }