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.

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：

任意找出两个区间，使得一个区间含于另一个区间。不存在输出-1 -1；

思路：

一、可以对区间按照左边界从小到大，右边界从大到小排序。遍历数组，如果存在前一个区间含于本区间，则输出；否则不存在。可以简单验证一下。

扫描二维码关注公众号，回复： 527090 查看本文章

代码

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
#define inf 1<<29
#define MAXN 300005
#define ll long long
int n,m;
struct node
{
    int l,r,id;
}p[MAXN];
bool cmp(node a,node b)
{
    return a.l==b.l? a.r>b.r:a.l<b.l;
}

int main()
{
    scanf("%d",&n);
    for(int i=0;i<n;i++)
    {
        scanf("%d%d",&p[i].l,&p[i].r);
        p[i].id=i;
    }
    sort(p,p+n,cmp);
    for(int i=0;i<n-1;i++)
    {
        if(p[i+1].l>=p[i].l && p[i+1].r<=p[i].r)
        {
            printf("%d %d\n",p[i+1].id+1,p[i].id+1);
            exit(0);
        }
    }
    printf("-1 -1\n");
    return 0;
}

思路二：利用splay，每次插入一个区间，插入时如果碰到一个区间含于或包含它，则跳出程序。否则按照左区间的大小进行插入。全部成功插入则输出-1 -1.

#include <cstdio>
#include<cstring>
#include <algorithm>
#define maxn 300005
using namespace std;
const int inf=~0u>>2;
#define lc(x) ch[(x)][0]
#define min(x,y) (x)>(y)?(y):(x)
int fa[maxn],ch[maxn][2],root,ind=1;
int a[maxn][2];
struct data
{
    int l,r,id;
}k[maxn];
inline void rotate(int p)
{
    int q=fa[p],y=fa[q],x=ch[q][1]==p;
    ch[q][x]=ch[p][x^1];fa[ch[q][x]]=q;
    ch[p][x^1]=q;fa[q]=p;
    fa[p]=y;
    if(y)
    {
        if(ch[y][0]==q)ch[y][0]=p;
        else if(ch[y][1]==q)ch[y][1]=p;
    }
}
inline void splay(int x)
{
    for(int y;y=fa[x];rotate(x))
        if(fa[y])
            rotate((x==lc(y))==(y==lc(fa[y]))?y:x);
    root=x;
}
inline void insert(int x,int vl,int vr,int id)
{
    int y;
    while(true)
    {
        if(k[x].l<=vl && vr<=k[x].r)
        {
            printf("%d %d\n",id,k[x].id);
            exit(0);
        }
        if(vl<=k[x].l && k[x].r<=vr)
        {
            printf("%d %d\n",k[x].id,id);
            exit(0);
        }
        y=ch[x][k[x].l<vl];
        if(y==0)
        {
            y=++ind;
            k[y].l=vl;
            k[y].r=vr;
            k[y].id=id;
            ch[y][0]=ch[y][1]=0;
            fa[y]=x;
            ch[x][k[x].l<vl]=y;
            break;
        }
        x=y;
    }
    splay(y);
}

int main()
{
    int n,ans=0;
    scanf("%d",&n);
    for(int i=1;i<=n;i++)
    {
        scanf("%d%d",&a[i][0],&a[i][1]);
    }
    root=1;
    k[1].l=a[1][0];
    k[1].r=a[1][1];
    k[1].id=1;
    ch[root][0]=ch[root][1]=0;
    fa[root]=0;
    for(int i=2;i<=n;i++)
    {
        insert(root,a[i][0],a[i][1],i);
    }
    printf("-1 -1\n");
    return 0;
}

Educational Codeforces Round 43 (Rated for Div. 2) C. Nested Segments【排序/splay】

题意：

思路：

猜你喜欢