Nested Segments CodeForces - 976C （思维）

最新推荐文章于 2020-01-17 21:09:39 发布

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本文介绍了一种寻找一个区间是否完全包含另一个区间的算法。通过将区间按起始点升序排序，并在相同起始点时按终点降序排列的方法，实现高效查找。最后给出了完整的C++代码实现。

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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.

题意：给出n个区间[l,r]，问是否有一个区间包含另一个区间，如果有，输出被包含的区间的包含区间的下标，否则输出-1 -1

思路：按l的升序排序，如果l相等，那么r大的在前面，然后从左到有，依次比较第i和i-1个即可

#include "iostream"
#include "algorithm"
using  namespace std;
struct P
{
    int first,second,index;
};
const int Max=3e5+10;
P seg[Max];
bool cmp(P a,P b)
{
    if(a.first<=b.first){
        if(a.first==b.first) return a.second>b.second;
        else return 1;
    }
    else
        return 0;
}
int main()
{
    ios::sync_with_stdio(false);
    int n,flag=0;
    cin>>n;
    for(int i=0;i<n;i++){
        cin>>seg[i].first>>seg[i].second;
        seg[i].index=i+1;
    }
    sort(seg,seg+n,cmp);
    for(int i=1;i<n;i++){
        if(seg[i].second<=seg[i-1].second){
            flag=1;
            cout<<seg[i].index<<" "<<seg[i-1].index<<endl;
            break;
        }
    }
    if(!flag)
        cout<<"-1 -1"<<endl;
    return 0;
}