园丁的烦恼(The Gardeners Trouble) 题解

思路分析

按题目要求使用对应数据结构与算法模拟/递推求解，注意数据范围边界与时间复杂度。

参考代码

#include <algorithm>
#include <iostream>
#include <vector>
using namespace std;

struct Op {
    int x, y, type, id;
    bool operator<(const Op& o) const {
        if (x != o.x) return x < o.x;
        return type < o.type; // point (0) before query (+/-1)
    }
};

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    int n, m;
    if (!(cin >> n >> m)) return 0;
    vector<Op> ops;
    ops.reserve(n + 4 * m);
    vector<int> ys;
    ys.reserve(n + 2 * m);
    for (int i = 0; i < n; ++i) {
        int x, y;
        cin >> x >> y;
        ops.push_back({x, y, 0, 0});
        ys.push_back(y);
    }
    for (int i = 0; i < m; ++i) {
        int a, b, c, d;
        cin >> a >> b >> c >> d;
        ops.push_back({c, d, 1, i});
        ops.push_back({a - 1, d, -1, i});
        ops.push_back({c, b - 1, -1, i});
        ops.push_back({a - 1, b - 1, 1, i});
        ys.push_back(d);
        ys.push_back(b - 1);
    }
    sort(ys.begin(), ys.end());
    ys.erase(unique(ys.begin(), ys.end()), ys.end());
    sort(ops.begin(), ops.end());

    int K = (int)ys.size();
    vector<int> bit(K + 1, 0);
    auto add = [&](int idx, int v) {
        for (; idx <= K; idx += idx & -idx) bit[idx] += v;
    };
    auto sum = [&](int idx) {
        int s = 0;
        for (; idx > 0; idx -= idx & -idx) s += bit[idx];
        return s;
    };

    vector<int> ans(m, 0);
    for (const auto& op : ops) {
        int y_idx = (int)(lower_bound(ys.begin(), ys.end(), op.y) - ys.begin()) + 1;
        if (op.type == 0) {
            add(y_idx, 1);
        } else {
            ans[op.id] += op.type * sum(y_idx);
        }
    }
    for (int i = 0; i < m; ++i) cout << ans[i] << "\n";
    return 0;
}