Codeforces Round 97B 点分治

B. Superset

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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 xi and yi ( - 109 ≤ xi, yi ≤ 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.

Examples

input

Copy

2
1 1
2 2

output

Copy

3
1 1
2 2
1 2

题意：给定n个点，请添加一些点，使任意两点满足：

①在同一条水平线或竖直线上

②或构成一个矩形框住其他点。

输出添加最少点后，满足条件的点集。

思路：将点集按x从小到大排序，取中间的点m的x坐标做一条直线l，取其他的点在其上的投影，将这些点加入点集，此时点m与其他所有点之间都已满足条件且在l两端的任意两点也互相满足条件，之后将区间二分递归操作，即可得到最后的点集。

代码：

 #include"bits/stdc++.h"
 #include"cstdio"
 #include"map"
 #include"set"
 #include"cmath"
 #include"queue"
 #include"vector"
 #include"string"
 #include"cstring"
 #include"ctime"
 #include"iostream"
 #include"cstdlib"
 #include"algorithm"
 #define db double
 #define ll long long
 #define vec vector<ll>
 #define mt  vector<vec>
 #define ci(x) scanf("%d",&x)
 #define cd(x) scanf("%lf",&x)
 #define cl(x) scanf("%lld",&x)
 #define pi(x) printf("%d\n",x)
 #define pd(x) printf("%f\n",x)
 #define pl(x) printf("%lld\n",x)
 //#define rep(i, x, y) for(int i=x;i<=y;i++)
 #define rep(i, n) for(int i=0;i<n;i++)
 const int N   = 1e6 + ;
 const int mod = 1e9 + ;
 const int MOD = mod - ;
 const int inf = 0x3f3f3f3f;
 const db  PI  = acos(-1.0);
 const db  eps = 1e-;
 using namespace std;
 typedef pair<int,int> P;
 P a[];
 set <P> s;
 void dfs(int l,int r)
 {
     int mid=(l+r)>>;
     int x=a[mid].first;
     for(int i=l;i<=r;i++){
        s.emplace(P(x,a[i].second));//去重+不改变顺序
     }
     if(l<mid){
         dfs(l,mid-);
     }
     if(r>mid){
         dfs(mid+,r);
     }

 }
 int main()
 {
     int n;
     ci(n);
     for(int i=;i<n;i++){
         ci(a[i].first),ci(a[i].second);
     }
     sort(a,a+n);
     for(int i=;i<n;i++) s.insert(a[i]);
     dfs(,n-);
     pi(s.size());
     for(auto &it: s){//遍历
         printf("%d %d\n",it.first,it.second);
     }
 }