旅馆房间管理(Hotel G) 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 <iostream>
#include <vector>
#include <algorithm>
using namespace std;

int n, m;
vector<int> pre, suf, mx, tag;

void pull(int u, int l, int r) {
    int mid = (l + r) >> 1;
    int lenL = mid - l + 1;
    int lenR = r - mid;
    pre[u] = pre[u << 1] + (pre[u << 1] == lenL ? pre[u << 1 | 1] : 0);
    suf[u] = suf[u << 1 | 1] + (suf[u << 1 | 1] == lenR ? suf[u << 1] : 0);
    mx[u] = max({mx[u << 1], mx[u << 1 | 1], suf[u << 1] + pre[u << 1 | 1]});
}

void apply(int u, int l, int r, int t) {
    // t = 0: empty, t = 1: occupied
    int val = (t == 0) ? (r - l + 1) : 0;
    pre[u] = suf[u] = mx[u] = val;
    tag[u] = t;
}

void push(int u, int l, int r) {
    if (tag[u] != -1) {
        int mid = (l + r) >> 1;
        apply(u << 1, l, mid, tag[u]);
        apply(u << 1 | 1, mid + 1, r, tag[u]);
        tag[u] = -1;
    }
}

void build(int u, int l, int r) {
    tag[u] = -1;
    pre[u] = suf[u] = mx[u] = r - l + 1;
    if (l == r) return;
    int mid = (l + r) >> 1;
    build(u << 1, l, mid);
    build(u << 1 | 1, mid + 1, r);
}

void update(int u, int l, int r, int ql, int qr, int t) {
    if (ql <= l && r <= qr) {
        apply(u, l, r, t);
        return;
    }
    push(u, l, r);
    int mid = (l + r) >> 1;
    if (ql <= mid) update(u << 1, l, mid, ql, qr, t);
    if (qr > mid) update(u << 1 | 1, mid + 1, r, ql, qr, t);
    pull(u, l, r);
}

int queryFirst(int u, int l, int r, int x) {
    if (l == r) return l;
    push(u, l, r);
    int mid = (l + r) >> 1;
    if (mx[u << 1] >= x) return queryFirst(u << 1, l, mid, x);
    if (suf[u << 1] + pre[u << 1 | 1] >= x) return mid - suf[u << 1] + 1;
    return queryFirst(u << 1 | 1, mid + 1, r, x);
}

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    if (!(cin >> n >> m)) return 0;
    pre.assign(4 * n + 5, 0);
    suf.assign(4 * n + 5, 0);
    mx.assign(4 * n + 5, 0);
    tag.assign(4 * n + 5, -1);
    build(1, 1, n);
    for (int i = 0; i < m; ++i) {
        int op;
        cin >> op;
        if (op == 1) {
            int x;
            cin >> x;
            if (mx[1] < x) {
                cout << 0 << "\n";
            } else {
                int p = queryFirst(1, 1, n, x);
                cout << p << "\n";
                update(1, 1, n, p, p + x - 1, 1);
            }
        } else {
            int x, y;
            cin >> x >> y;
            update(1, 1, n, x, min(n, x + y - 1), 0);
        }
    }
    return 0;
}