C. Rectangles

题目链接：http://codeforces.com/contest/1028/problem/C

题目大意：就是给出n个矩形，求任意一个至少n-1个矩形都覆盖的点；

C. Rectangles

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

You are given nn rectangles on a plane with coordinates of their bottom left and upper right points. Some (n−1)(n−1) of the given nn rectangles have some common point. A point belongs to a rectangle if this point is strictly inside the rectangle or belongs to its boundary.

Find any point with integer coordinates that belongs to at least (n−1)(n−1) given rectangles.

Input

The first line contains a single integer nn (2≤n≤1326742≤n≤132674) — the number of given rectangles.

Each the next nn lines contains four integers x1x1, y1y1, x2x2 and y2y2 (−109≤x1<x2≤109−109≤x1<x2≤109, −109≤y1<y2≤109−109≤y1<y2≤109) — the coordinates of the bottom left and upper right corners of a rectangle.

Output

Print two integers xx and yy — the coordinates of any point that belongs to at least (n−1)(n−1) given rectangles.

Examples

input

Copy

3
0 0 1 1
1 1 2 2
3 0 4 1

output

Copy

1 1

input

Copy

3
0 0 1 1
0 1 1 2
1 0 2 1

output

Copy

1 1

input

Copy

4
0 0 5 5
0 0 4 4
1 1 4 4
1 1 4 4

output

Copy

1 1

input

Copy

5
0 0 10 8
1 2 6 7
2 3 5 6
3 4 4 5
8 1 9 2

output

Copy

3 4

Note

The picture below shows the rectangles in the first and second samples. The possible answers are highlighted.

The picture below shows the rectangles in the third and fourth samples.

 

这个代码是从大佬那里看来的，时间1996ms，很险；

原理和之前的求

#include<iostream>
#include<set>
using namespace std;
multiset<int > mstx1,mstx2,msty1,msty2;
struct NODE{
	int x1;
	int y1;
	int x2;
	int y2;
}node[140000];
int main(){
	int n;
	cin>>n;
	int a;
	for(int i=0;i<n;i++){
		cin>>node[i].x1>>node[i].y1>>node[i].x2>>node[i].y2;
		mstx1.insert(node[i].x1);
		msty1.insert(node[i].y1);
		mstx2.insert(node[i].x2);
		msty2.insert(node[i].y2); 
	}
	for(int i=0;i<n;i++){
		mstx1.erase(mstx1.find(node[i].x1));
		msty1.erase(msty1.find(node[i].y1));
		mstx2.erase(mstx2.find(node[i].x2));
		msty2.erase(msty2.find(node[i].y2));
		if(*mstx2.begin()>=*mstx1.rbegin() && *msty2.begin()>=*msty1.rbegin()){
			cout<<*mstx2.begin()<<' '<<*msty2.begin()<<endl;
			return 0;
		}
		mstx1.insert(node[i].x1);
		mstx2.insert(node[i].x2);
		msty1.insert(node[i].y1);
		msty2.insert(node[i].y2);
	}
	return 0;
}