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  • Lara Croft and the New Game CodeForces

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

    Input
    5
    1 10
    2 9
    3 9
    2 3
    2 9
    Output
    2 1
    Input
    3
    1 5
    2 6
    6 20
    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.

    关于cmp的写法我博客有专门文章讲

    #include<stdio.h>
    #include<string.h>
    #include<iostream>
    #include<cstring>
    #include<cmath>
    #include<algorithm>
    using namespace std;
    #define LL long long int
    const int N = 1e5 + 7;
    const int MOD = 1e9+7;
    struct point
    {
        LL l,r,index;
        bool operator < (const point &tmp) const
        {
            return l == tmp.l ? r > tmp.r : l < tmp.l;
        }
    } num[300004];
    
    int main()
    {
        LL n;
        cin>>n;
        for(LL i=0; i<n; i++)
        {
            scanf("%lld %lld",&num[i].l,&num[i].r);
            //cin>>num[i].l>>num[i].r;
            num[i].index = i + 1;
        }
        //cout<<"1"<<endl;
        sort(num,num+n);
        LL i1=-1,i2=-1;
        for(LL i=1; i<n; i++)
        {
            if(num[i-1].r >= num[i].r )
            {
                i1 = num[i-1].index;
                i2 = num[i].index;
                break;
            }
        }
    
        cout<<i2<<" "<<i1<<endl;
    
    }
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  • 原文地址:https://www.cnblogs.com/upstart/p/8982263.html
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