CodeForces 97B Superset_those two points lie on same horizontal line;…-优快云博客

探讨如何根据特定条件构建平面点集的有效超集，确保任何两点间至少满足三个条件之一：共线、包含其他点或位于相同矩形内。通过算法实现，输出不超过2×10^5个整数坐标的点。

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A set of points on a plane is called good, if for any two points at least one of the three conditions is true:

those two points lie on same horizontal line;
those two points lie on same vertical line;
the rectangle, with corners in these two points, contains inside or on its borders at least one point of the set, other than these two. We mean here a rectangle with sides parallel to coordinates' axes, the so-called bounding box of the two points.

You are given a set consisting of n points on a plane. Find any good superset of the given set whose size would not exceed 2·10⁵ points.

Input

The first line contains an integer n (1 ≤ n ≤ 10⁴) — the number of points in the initial set. Next n lines describe the set's points. Each line contains two integers x_i and y_i ( - 10⁹ ≤ x_i, y_i ≤ 10⁹) — a corresponding point's coordinates. It is guaranteed that all the points are different.

Output

Print on the first line the number of points m (n ≤ m ≤ 2·10⁵) in a good superset, print on next m lines the points. The absolute value of the points' coordinates should not exceed 10⁹. Note that you should not minimize m, it is enough to find any good superset of the given set, whose size does not exceed 2·10⁵.

All points in the superset should have integer coordinates.

Sample test(s)

Input

2
1 1
2 2

Output

这题主要对第三个条件的理解，一开始我还以为是只有在两个点构成的矩形中没有任何点时才加入新的点，后来才知道，想多了，不管有没有点，直接加就行了。这样的话，，程序就相对的简单了。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <set>
#include <algorithm>

using namespace std;
#define maxn 10005
int n,cnt;
struct Node{
    int x;
    int y;
    bool operator < (const Node b)const {
        if(x == b.x){
            return y < b.y;
        }
        return x < b.x;
    }
};
Node arr[maxn];
set<Node> s;
set<Node>::iterator it;
int cmpX(Node a,Node b){
    return a.x < b.x;
}
int cmpY(Node a,Node b){
    return a.y < b.y;
}
void divide(int low,int high);
int main()
{
    scanf("%d",&n);
    cnt = 0;
     s.clear();
    for(int i = 0; i < n; i++){
        scanf("%d%d",&arr[i].x,&arr[i].y);
        s.insert(arr[i]);
    }

    sort(arr,arr+n,cmpX);

    divide(0,n-1);

    printf("%d\n",s.size());
      for(it = s.begin();it != s.end();it++){
          printf("%d %d\n",it->x,it->y);
      }
    return 0;
}
void divide(int low,int high){
    if(low  == high){
        return ;
    }
    int mid = (low + high) /2;
    for(int i = low; i <= high  ;i++){
        Node tmp;
        tmp.x = arr[mid].x;
        tmp.y = arr[i].y;
        s.insert(tmp);
    }

    divide(low,mid);
    divide(mid+1,high);
    return ;
}