解题思路

思维题，讲道理，我一开始打算去二维康康能不能用什么办法处理出这个地图来……想了一会发现行不通，最小的复杂度都要遍历整个递推。

那么再看，我们要最后的地毯，倒叙看呢？！震惊，直接找到那个点倒叙第一个被覆盖的编号就行了！！！

#include <bits/stdc++.h>
#pragma GCC optimize(2)
#pragma GCC optimize(3)
using namespace std;
#define js ios::sync_with_stdio(false);cin.tie(0); cout.tie(0)
typedef long long ll;
const ll MOD = 1e9 + 7;
inline ll read() { ll s = 0, w = 1; char ch = getchar(); while (ch < 48 || ch > 57) { if (ch == '-') w = -1; ch = getchar(); }    while (ch >= 48 && ch <= 57) s = (s << 1) + (s << 3) + (ch ^ 48), ch = getchar();    return s * w; }
inline void write(ll x) { if (!x) { putchar('0'); return; } char F[200]; ll tmp = x > 0 ? x : -x; if (x < 0)putchar('-');    int cnt = 0;    while (tmp > 0) { F[cnt++] = tmp % 10 + '0';        tmp /= 10; }    while (cnt > 0)putchar(F[--cnt]); }
inline ll gcd(ll x, ll y) { return y ? gcd(y, x % y) : x; }
ll qpow(ll a, ll b) { ll ans = 1;    while (b) { if (b & 1)    ans *= a;        b >>= 1;        a *= a; }    return ans; }    ll qpow(ll a, ll b, ll mod) { ll ans = 1; while (b) { if (b & 1)(ans *= a) %= mod; b >>= 1; (a *= a) %= mod; }return ans % mod; }
inline int lowbit(int x) { return x & (-x); }

const int N = 1e4 + 7;
int x[N], y[N], lenx[N], leny[N];

int main() {
    int n = read();
    for (int i = 1; i <= n; ++i)
        x[i] = read(), y[i] = read(), lenx[i] = read(), leny[i] = read();
    int xx = read(), yy = read();
    for (int i = n; i; --i) {
        if (xx >= x[i] and xx <= x[i] + lenx[i] and yy >= y[i] and yy <= y[i] + leny[i])
            return write(i), putchar(10), 0;
    }
    return 0;
}