嘟嘟嘟cf
嘟嘟嘟luogu


剛開始我看成了對於一個點\(i\)，存在一個點\(j\)滿足三個條件之一，而不是任意的\(j\)。結果自然\(gg\)了，第二個點就\(WA\)了。
也不知怎麽來的思路：平面分治。
先把所有點按\(x\)排序，然後規定一個中間點\(a_{mid}\)。兩邊的點向中間點作投影，這樣對於任意的在左半部分的點\(i\)和任意的在右半部分的點\(j\)，必定滿足條件。
然後我們在分治到\([L, mid - 1]\)和\([mid + 1, R]\)中，解決同側點的問題。

#include<cstdio>
#include<iostream>
#include<cmath>
#include<algorithm>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<vector>
#include<stack>
#include<queue>
#include<map>
using namespace std;
#define enter puts("") 
#define space putchar(‘ ‘)
#define Mem(a, x) memset(a, x, sizeof(a))
#define rg register
typedef long long ll;
typedef double db;
const int INF = 0x3f3f3f3f;
const db eps = 1e-8;
const int maxn = 2e5 + 5;
inline ll read()
{
  ll ans = 0;
  char ch = getchar(), last = ‘ ‘;
  while(!isdigit(ch)) last = ch, ch = getchar();
  while(isdigit(ch)) ans = (ans << 1) + (ans << 3) + ch - ‘0‘, ch = getchar();
  if(last == ‘-‘) ans = -ans;
  return ans;
}
inline void write(ll x)
{
  if(x < 0) x = -x, putchar(‘-‘);
  if(x >= 10) write(x / 10);
  putchar(x % 10 + ‘0‘);
}

int n, cnt;
struct Node
{
  int x, y;
  bool operator < (const Node& oth)const
  {
    return x < oth.x || (x == oth.x && y < oth.y);
  }
  bool operator == (const Node& oth)const
  {
    return x == oth.x && y == oth.y;
  }
}a[maxn];

void solve(int L, int R)
{
  if(L > R) return;
  int mid = (L + R) >> 1;
  for(int i = L; i <= R; ++i)
    a[++n] = (Node){a[mid].x, a[i].y};
  solve(L, mid - 1); solve(mid + 1, R);
}

int main()
{
  n = read();
  for(int i = 1; i <= n; ++i) a[i].x = read(), a[i].y = read();
  sort(a + 1, a + n + 1);
  solve(1, n);
  sort(a + 1, a + n + 1);
  int _n = unique(a + 1, a + n + 1) - a - 1;
  write(_n), enter;
  for(int i = 1; i <= _n; ++i) write(a[i].x), space, write(a[i].y), enter;
  return 0;
}
 

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