使用unordered_set，对pair类型写一个哈希函数，查找操作的时间复杂度为平均 O(1)，可AC

#include <iostream>
#include <unordered_set>

using namespace std;

#define x first
#define y second
typedef pair<int, int> PII;
const int N = 55, M = 1010;
int n, l, s;
int b[N][N];
PII tree[M];

// 提供哈希函数
struct pair_hash {
    template <class T1, class T2>
    std::size_t operator() (const std::pair<T1, T2>& p) const {
        auto hash1 = std::hash<T1>{}(p.first);
        auto hash2 = std::hash<T2>{}(p.second);
        return hash1 ^ (hash2 << 1); // 组合两个哈希值
    }
};

unordered_set<PII, pair_hash> st;

int main()
{
    cin >> n >> l >> s;
    for (int i = 0; i < n; i ++) cin >> tree[i].x >> tree[i].y, st.insert(tree[i]);
    for (int i = s; i >= 0; i --)
        for (int j = 0; j <= s; j ++)
            cin >> b[i][j];

    int res = 0;
    for (int k = 0; k < n; k ++) {
        int tx = tree[k].x, ty = tree[k].y;
        if (tx + s <= l && ty + s <= l) {
            bool flag = true;
            for (int i = tx; i <= tx + s; i ++) {
                for (int j = ty; j <= ty + s; j ++) {
                    int bx = i - tx, by = j - ty;
                    int temp = 0;
                    if (st.find({i, j}) != st.end()) temp = 1;
                    if (temp != b[bx][by]) {
                        flag = false;
                        break;
                    }
                }
                if (!flag) break;
            }
            if (flag) res ++;
        }
    }

    cout << res << endl;

    return 0;
}

用set，查找操作的时间复杂度为 O(log n)，大数据TLE

#include <iostream>
#include <set>

using namespace std;

#define x first
#define y second
typedef pair<int, int> PII;
const int N = 55, M = 1010;
int n, l, s;
int b[N][N];
PII tree[M];
set<pair<int, int>> st;

int main()
{
    cin >> n >> l >> s;
    for (int i = 0; i < n; i ++) cin >> tree[i].x >> tree[i].y, st.insert(tree[i]);
    for (int i = s; i >= 0; i --)
        for (int j = 0; j <= s; j ++)
            cin >> b[i][j];

    int res = 0;
    for (int k = 0; k < n; k ++) {
        int tx = tree[k].x, ty = tree[k].y;
        if (tx + s <= l && ty + s <= l) {
            bool flag = true;
            for (int i = tx; i <= tx + s; i ++) {
                for (int j = ty; j <= ty + s; j ++) {
                    int bx = i - tx, by = j - ty;
                    int temp = 0;
                    if (st.count({i, j})) temp = 1;
                    if (temp != b[bx][by]) {
                        flag = false;
                        break;
                    }
                }
                if (!flag) break;
            }
            if (flag) res ++;
        }
    }

    cout << res << endl;

    return 0;
}