CodeForces - 976C(结构体排序思维)

CodeForces - 976C

You are given a sequence a₁, a₂, ..., a_n of one-dimensional segments numbered 1 through n. Your task is to find two distinct indices i and j such that segment a_i lies within segment a_j.

Segment [l₁, r₁] lies within segment [l₂, r₂] iff l₁ ≥ l₂ and r₁ ≤ r₂.

Print indices i and j. If there are multiple answers, print any of them. If no answer exists, print -1 -1.

Input

The first line contains one integer n (1 ≤ n ≤ 3·10⁵) — the number of segments.

Each of the next n lines contains two integers l_i and r_i (1 ≤ l_i ≤ r_i ≤ 10⁹) — the i-th segment.

Output

Print two distinct indices i and j such that segment a_i lies within segment a_j. If there are multiple answers, print any of them. If no answer exists, print -1 -1.

Examples

Input

Output

2 1

Input

Output

-1 -1

Note

In the first example the following pairs are considered correct:

(2, 1), (3, 1), (4, 1), (5, 1) — not even touching borders;
(3, 2), (4, 2), (3, 5), (4, 5) — touch one border;
(5, 2), (2, 5) — match exactly.

emm这题就是结构体排序先按照他们所给定的l值从小到大排序然后l值相等的就按照r排序......这样子刚开始设定的是lrid都是1的，然后....就是遍历一下2开始的结构体看看后面那个是否在我们设定的u的里面或者说看看我们设定的u是够在这个遍历的结构体里面如果有就直接输出吧

要是没有的话 l 有两种情况是在我们假定的u的范围里面还有一种是在u外面

1. 如果是在u的外面那么这个u也就没有用了直接更新新的

2.如果l在里面的话因为这个l一定比那个u的l大那么就看他的r 如果r比 u的r小已经被输出了所以就是在外面 ....... 同样这样子也不可能再有东西在这个里面了其中要注意的就是当n==1的时候是输出-1-1这边我输出1 1 wa了两发

代码:

#include<bits/stdc++.h>
using namespace std;
int n;
const int maxn = 3*1e5+100;
struct node{
	int l,r; 
	int id;   
}s[maxn]; 
bool cmp(node a,node b){
	if(a.l!=b.l) return a.l<b.l;
	else return a.r<b.r;
}
int main()
{
	int n;
	scanf("%d",&n);
	for(int i = 1;i<=n;i++)
	{
		scanf("%d%d",&s[i].l,&s[i].r);
		s[i].id = i;
	}	
	sort(s+1,s+1+n,cmp);//保证了后面的l一定是大于前面的l0 
	if(n==1){
		printf("-1 -1\n");//这边我还以为是直接输出1 1 结果不是~~~ 
		return 0;
	}
	int l = s[1].l;
	int r = s[1].r;
	int u = s[1].id;
	for(int i = 2;i<=n;i++)
	{
		if(s[i].l>=l&&s[i].r<=r)
		{
			printf("%d %d\n",s[i].id,u);
			return 0;
		}//这个肯定要放在前面
		else if(s[i].l==l&&s[i].r>=r){
			printf("%d %d\n",u,s[i].id);
			return 0;
		}else if(s[i].l>=r)// l在边界外面   
		{
			l = s[i].l;
			r= s[i].r;
			u = s[i].id;
		}else if(s[i].r>r)
		{
			l = s[i].l;
			r= s[i].r;
			u = s[i].id;
		} 
	}
	printf("-1 -1\n");
	return 0;
}

CodeForces - 976C(结构体排序思维)

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