Codeforces 863e

Turn Off The TV

题意：给出n个区间，求一个被其他区间覆盖的区间的编号

思路：模拟，先按左端点从小到大排序，再按右端点从大到小排序，枚举每一个区间，维护一个R，表示i之前的区间覆盖的最右端，因为i之前每一个区间的左端点都比i的左端点小，因此不需要考虑i的左端点了

AC代码：

#include "iostream"
#include "iomanip"
#include "string.h"
#include "stack"
#include "queue"
#include "string"
#include "vector"
#include "set"
#include "map"
#include "algorithm"
#include "stdio.h"
#include "math.h"
#pragma comment(linker, "/STACK:102400000,102400000")
#define bug(x) cout<<x<<" "<<"UUUUU"<<endl;
#define mem(a,x) memset(a,x,sizeof(a))
#define step(x) fixed<< setprecision(x)<<
#define mp(x,y) make_pair(x,y)
#define pb(x) push_back(x)
#define ll long long
#define endl ("
")
#define ft first
#define sd second
#define lrt (rt<<1)
#define rrt (rt<<1|1)
using namespace std;
const ll mod=1e9+7;
const ll INF = 1e18+1LL;
const int inf = 1e9+1e8;
const double PI=acos(-1.0);
const int N=2e5+100;

struct Node{
    int  l, r, id;
    bool friend operator< (Node a, Node b){
        if(a.l==b.l) return a.r>b.r;
        return a.l<b.l;
    }
}a[N];
int n,R=-100;
int main(){
    ios::sync_with_stdio(false),cin.tie(0),cout.tie(0);
    cin>>n;
    for(int i=1; i<=n; ++i){
        cin>>a[i].l>>a[i].r;
        a[i].id=i;
    }
    sort(a+1,a+1+n);
    for(int i=1; i<=n; ++i){
        if(R+1>=a[i+1].l && a[i+1].r>=a[i].r){
            cout<<a[i].id<<endl;
            return 0;
        }
        if(R>=a[i].r){
            cout<<a[i].id<<endl;
            return 0;
        }
        R=max(R,a[i].r);
    }
    cout<<-1<<endl;
    return 0;
}