题目链接

题目

题目描述

The cows are journeying north to Thunder Bay in Canada to gain cultural enrichment and enjoy a vacation on the sunny shores of Lake Superior. Bessie, ever the competent travel agent, has named the Bullmoose Hotel on famed Cumberland Street as their vacation residence. This immense hotel has N (1 ≤ N ≤ 50,000) rooms all located on the same side of an extremely long hallway (all the better to see the lake, of course).
The cows and other visitors arrive in groups of size Di (1 ≤ Di ≤ N) and approach the front desk to check in. Each group i requests a set of Di contiguous rooms from Canmuu, the moose staffing the counter. He assigns them some set of consecutive room numbers r..r+Di-1 if they are available or, if no contiguous set of rooms is available, politely suggests alternate lodging. Canmuu always chooses the value of r to be the smallest possible.
Visitors also depart the hotel from groups of contiguous rooms. Checkout i has the parameters Xi and Di which specify the vacating of rooms Xi ..Xi +Di-1 (1 ≤ Xi ≤ N-Di+1). Some (or all) of those rooms might be empty before the checkout.
Your job is to assist Canmuu by processing M (1 ≤ M < 50,000) checkin/checkout requests. The hotel is initially unoccupied.

输入描述

• Line 1: Two space-separated integers: N and M
• Lines 2..M+1: Line i+1 contains request expressed as one of two possible formats: (a) Two space separated integers representing a check-in request: 1 and Di (b) Three space-separated integers representing a check-out: 2, Xi, and Di

输出描述

• Lines 1.....: For each check-in request, output a single line with a single integer r, the first room in the contiguous sequence of rooms to be occupied. If the request cannot be satisfied, output 0.

示例1

输入

10 6
1 3
1 3
1 3
1 3
2 5 5
1 6

输出

1
4
7
0
5

题解

知识点：线段树，二分。

考虑连续区间的权值最大值，需要维护三个信息，区间连续空房个数 \(emp\) 、区间从左端点开始的连续空房个数 \(lemp\) 、区间从右端点开始的连续空房的个数 \(remp\) 。

合并时， \(mx\) 为左右子区间 \(mx\) 和左子区间 \(remp\) 加右子区间 \(lemp\) 之和取最大值。 \(lemp,remp\) 需要考虑跨越左右区间的特殊情况，因此需要维护区间长度 \(len\) 。例如，若左子区间的 \(lemp\) 等于左子区间的 \(len\) ，那么区间 \(lemp\) 就是左子区间 \(len\) 加右子区间 \(lemp\) 否则继承左子区间的 \(lemp\) 即可，区间 \(remp\) 同理。

因此，区间信息需要维护 \(len,emp,lemp,remp\) 。

接下来是查找第一个连续空位大于等于 \(val\) 的位置，利用线段树上二分即可解决。首先判断是否存在，之后分三类情况：

1. 若左子区间 \(emp\) 大于等于 \(val\) ，则查询左子区间。
2. 否则若跨越左右子区间的空位，即左子区间的 \(remp\) 加右子区间的 \(lemp\) 之和，大于等于 \(val\) ，则直接返回位置。
3. 否则查询右子区间。

区间修改维护一个信息，修改种类 \(upd\) ， 有三个值 \(0/1/-1\) 表示未修改、全部入住、全部离开。 修改很朴素，看代码就行。

时间复杂度 \(O((n+m) \log n)\)

空间复杂度 \(O(n)\)

代码

#include <bits/stdc++.h>
using namespace std;
using ll = long long;

struct T {
    int len;
    int lemp, remp;
    int emp;
    static T e() { return{ 0,0,0,0 }; }
    friend T operator+(const T &a, const T &b) {
        return{
            a.len + b.len,
            a.lemp == a.len ? a.len + b.lemp : a.lemp,
            b.remp == b.len ? a.remp + b.len : b.remp,
            max({a.emp,b.emp,a.remp + b.lemp})
        };
    }
};

struct F {
    int upd;
    static F e() { return { 0 }; }
    T operator()(const T &x) {
        if (upd == 1)
            return{
                x.len,
                0,0,
                0
        };
        if (upd == -1)
            return{
                x.len,
                x.len,x.len,
                x.len
        };
        return x;
    }
    F operator()(const F &g) {
        return { upd ? upd : g.upd };
    }
};

class SegmentTreeLazy {
    int n;
    vector<T> node;
    vector<F> lazy;

    void push_down(int rt) {
        node[rt << 1] = lazy[rt](node[rt << 1]);
        lazy[rt << 1] = lazy[rt](lazy[rt << 1]);
        node[rt << 1 | 1] = lazy[rt](node[rt << 1 | 1]);
        lazy[rt << 1 | 1] = lazy[rt](lazy[rt << 1 | 1]);
        lazy[rt] = F::e();
    }

    void update(int rt, int l, int r, int x, int y, F f) {
        if (r < x || y < l) return;
        if (x <= l && r <= y) return node[rt] = f(node[rt]), lazy[rt] = f(lazy[rt]), void();
        push_down(rt);
        int mid = l + r >> 1;
        update(rt << 1, l, mid, x, y, f);
        update(rt << 1 | 1, mid + 1, r, x, y, f);
        node[rt] = node[rt << 1] + node[rt << 1 | 1];
    }

    int query(int rt, int l, int r, int val) {
        if (l == r) return l;
        push_down(rt);
        int mid = l + r >> 1;
        if (node[rt << 1].emp >= val) return query(rt << 1, l, mid, val);
        if (node[rt << 1].remp + node[rt << 1 | 1].lemp >= val) return mid - node[rt << 1].remp + 1;
        return query(rt << 1 | 1, mid + 1, r, val);
    }

public:
    SegmentTreeLazy(int _n = 0) { init(_n); }

    void init(int _n) {
        n = _n;
        node.assign(n << 2, T::e());
        lazy.assign(n << 2, F::e());

        function<void(int, int, int)> build = [&](int rt, int l, int r) {
            if (l == r) return node[rt] = { 1,1,1,1 }, void();
            int mid = l + r >> 1;
            build(rt << 1, l, mid);
            build(rt << 1 | 1, mid + 1, r);
            node[rt] = node[rt << 1] + node[rt << 1 | 1];
        };
        build(1, 1, n);
    }

    void update(int x, int y, F f) { update(1, 1, n, x, y, f); }

    int query(int val) {
        if (node[1].emp < val) return 0;
        return query(1, 1, n, val);
    }
};

int main() {
    std::ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
    int n, m;
    cin >> n >> m;
    SegmentTreeLazy sgt(n);
    while (m--) {
        int op;
        cin >> op;
        if (op == 1) {
            int d;
            cin >> d;
            int pos = sgt.query(d);
            if (pos) sgt.update(pos, pos + d - 1, { 1 });
            cout << pos << '\n';
        }
        else {
            int x, d;
            cin >> x >> d;
            sgt.update(x, x + d - 1, { -1 });
        }
    }
    return 0;
}