​​传送门​​​ 思路：一开始想的直接暴力，发现根本行不通，会runtime error,看了题解之后发现可以直接判断范围。
题意：可以开一个二维数组倒序来看，逆序扫一边发现点正好在所在范围，直接输出即可。

/**
* From:
* Qingdao Agricultural University
* Created by XiangwangAcmer
* Date : 2019-09-29-19.40.30
* Talk is cheap.Show me your code.
*/
#include<iostream>
#include<cstring>
#include<algorithm>
#include<cstdio>
#include<cstdlib>
#include<queue>
#include<cmath>
#include<cctype>
#include<stack>
#include<map>
#include<string>
#include<cstdlib>
#define ll long long
using namespace std;
const ll maxn = 1e4 + 5;
const ll minn = 1e9 + 5;
const ll mod = 1000000007;
const int inf = 0x3f3f3f3f;
vector<int>v[maxn];
const long long LIMIT = 4294967295LL;
bool row[maxn], col[maxn];
bool flag = 0;
int a[maxn][4];
queue<int>q;
int main() {
    ios::sync_with_stdio(false);
    int cnt = 1, n;
    cin >> n;
    int x,y;
    for(int i = 1; i <= n; i++)
        cin >> a[i][0] >> a[i][1] >> a[i][2] >> a[i][3];
    cin>>x>>y;
    int flag=0;
    for(int i=n;i>=1;i--)
    {
        if(a[i][0]<=x&&(a[i][0]+a[i][2])>=x&&a[i][1]<=y&&(a[i][1]+a[i][3])>=y)
        {
            flag=i;
            break;
        }
    }
    if(flag)
        cout<<flag<<endl;
    else cout<<-1<<endl;
    return 0;
}