Educational Codeforces Round 43 (Rated for Div. 2) C. Nested Segments

本文链接：https://blog.youkuaiyun.com/albertluf/article/details/80196251

本文介绍了一种在一维空间中寻找嵌套区间的高效算法。通过将区间按左端点排序并考虑右端点的最大值，可以在O(n log n)的时间复杂度内找到至少完全包含在另一个区间内的两个不同区间。

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

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 li and ri (1 ≤ li ≤ ri ≤ 109) — the i-th segment.

Output

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

Examples
inputCopy
5
1 10
2 9
3 9
2 3
2 9
outputCopy
2 1
inputCopy
3
1 5
2 6
6 20
outputCopy
-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.

题解：
按照L从小到大排序，则后面的L一定大于等于前面的，
只需满足当前的R小于等于之前出现的R的最大值即可。
注意：若L相同则按R从大到小排序
代码：
#include<bits/stdc++.h>
using namespace std;
const int maxn=300005;
struct node
{
    int l,r,id;
}c[maxn];
bool cmp(const node &a,const node &b)
{
    if(a.l!=b.l)return a.l<b.l;
    return a.r>b.r;
}
int main()
{
    int n;scanf("%d",&n);
    for(int i=1;i<=n;i++)
    {
        scanf("%d%d",&c[i].l,&c[i].r);
        c[i].id=i;
    }
    sort(c+1,c+n+1,cmp);
    int ma=-1,id;bool bb=1;
    for(int i=1;i<=n;i++)
    {
        if(c[i].r<=ma)
        {
            printf("%d %d\n",c[i].id,id);
            bb=0;break;
        }
        if(ma<c[i].r)
        {
            ma=c[i].r;id=c[i].id;
        }
    }
    if(bb)printf("-1 -1\n");
    return 0;
}