1850G-The-Morning-Star · znzryb
```markdown
---
title: 1850G-The-Morning-Star
categories:
- codeforces题解
tags:
- null
date: 2025-03-05 11:48:05
---
我们都知道，想要找到同x或者同y的点的数量易如反掌。
那么，同对角线的那？
其实也很简单，找x+y || x-y的值是相同的点的数量就行
```cpp
// Problem: G. The Morning Star
// Contest: Codeforces - Codeforces Round 886 (Div. 4)
// URL: https://codeforces.com/problemset/problem/1850/G
// Memory Limit: 256 MB
// Time Limit: 2000 ms
// by znzryb
//
// Powered by CP Editor (https://cpeditor.org)
#include <bits/stdc++.h>
#define FOR(i, a, b) for (long long i = (a); i <= (b); ++i)
#define ROF(i, a, b) for (long long i = (a); i >= (b); --i)
#define all(x) x.begin(),x.end()
#define CLR(i,a) memset(i,a,sizeof(i))
#define fi first
#define se second
#define pb push_back
#define SZ(a) ((int) a.size())
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<ll,ll> pll;
typedef array<ll,3> arr;
typedef double DB;
typedef pair<DB,DB> pdd;
constexpr ll MAXN=static_cast<ll>(2e5)+10,INF=static_cast<ll>(1e18)+3;
ll N,M,T;
pll xy[MAXN];
bool cmpy(const pll &a,const pll &b){
if(a.se!=b.se) return a.se<b.se;
return a.fi<b.fi;
}
bool cmpxPy(const pll &a,const pll &b){
if(a.fi+a.se!=b.fi+b.se) return a.fi+a.se<b.fi+b.se;
return a.fi<b.fi;
}
bool cmpxMy(const pll &a,const pll &b){
if(a.fi-a.se!=b.fi-b.se) return a.fi-a.se<b.fi-b.se;
return a.fi<b.fi;
}
inline void solve(){
cin>>N;
for(int i=1;i<=N;++i){
cin>>xy[i].fi>>xy[i].se;
}
sort(xy+1,xy+1+N);
ll ans=0;
for(int i=1;i<=N;++i){
pll uppConst={xy[i].fi,INF},lowConst={xy[i].fi,-INF};
ans+=upper_bound(xy+1,xy+1+N,uppConst)-lower_bound(xy+1,xy+1+N,lowConst)-1;
}
// #ifdef LOCAL
// cerr << ans << '\n';
// #endif
sort(xy+1,xy+1+N,cmpy);
for(int i=1;i<=N;++i){
pll uppConst={INF,xy[i].se},lowConst={-INF,xy[i].se};
ans+=upper_bound(xy+1,xy+1+N,uppConst,cmpy)-lower_bound(xy+1,xy+1+N,lowConst,cmpy)-1;
}
// #ifdef LOCAL
// cerr << ans << '\n';
// #endif
// 同y和同x的匹配完了，同对角线的怎么统计？
// 哈哈，其实和上面的一样的思路，同一对角线，x-y||x+y的值是一样的
sort(xy+1,xy+1+N,cmpxPy);
for(int i=1;i<=N;++i){
ll aPb=xy[i].fi+xy[i].se;
pll uppConst={INF,aPb-INF},lowConst={-INF,aPb+INF};
ans+=upper_bound(xy+1,xy+1+N,uppConst,cmpxPy)-lower_bound(xy+1,xy+1+N,lowConst,cmpxPy)-1;
}
// #ifdef LOCAL
// cerr << ans << '\n';
// #endif
sort(xy+1,xy+1+N,cmpxMy);
for(int i=1;i<=N;++i){
ll aMb=xy[i].fi-xy[i].se;
pll uppConst={INF,-aMb+INF},lowConst={-INF,-aMb-INF};
ans+=upper_bound(xy+1,xy+1+N,uppConst,cmpxMy)-lower_bound(xy+1,xy+1+N,lowConst,cmpxMy)-1;
}
cout<<ans<<"\n";
}
int main()
{
ios::sync_with_stdio(false);
cin.tie(0);cout.tie(0);
cin>>T;
while(T--){
solve();
}
return 0;
}
/*
AC
https://codeforces.com/problemset/submission/1850/309011870
*/
```
```