題目地址
題意：有長度為n米的停車場，停車的要求是要與前面一輛車至少隔a米，和後一輛車至少隔b米（只要符合要求就可以停入，不管之後會不會打破這個要求。PS：我就是想了好久沒有想通），有m個操作，有兩種操作型別：

1. 1 x 把長度為x米的車停入停車場（一定要符合停車要求）
2. 2 x 將第x個操作中停入的車開出

思路：把停車場擴建為n+a+b，把每輛車都變為x+a+b來查詢，這樣就可以直接停車了，但是車子還是隻有x的長度，所以在update的時候是用了pos+a~pos+a+y-1（為什麼要減一，因為他的邊可以重複利用）去更新，但是因為第一輛車多佔用了a個，所以輸出位置的時候要減去a就是pos了
PS：要這題模板的

#include <iostream>
#include <cstring>
#include <string>
#include <queue>
#include <vector>
#include <map>
#include <set>
#include <stack>
#include <cmath>
#include <cstdio>
#include <algorithm>
#define N 50010
#define LL long long
#define inf 0x3f3f3f3f
 

#define gol ans<<1
#define gor ans<<1|1
#define lson l,mid,ans<<1
#define rson mid+1,r,ans<<1|1
using namespace std;
const LL mod = 1e9 + 7;
const double eps = 1e-9;
struct node {
    int l, r, lazy;
    int llen, rlen, maxlen;
}tree[N<<2];
struct nope {
    int l, r;
}now;
map
 
<int, nope> mapp;
struct Segment__Tree {
    void pushDown(int ans, int num) {
        if (tree[ans].lazy != -1) {
            tree[gol].lazy = tree[ans].lazy;
            tree[gor].lazy = tree[ans].lazy;
            tree[gol].llen = tree[gol].rlen = tree[gol].maxlen = (tree[ans].lazy ? 0 : num - (num >> 1));
            tree[gor].llen = tree[gor].rlen = tree[gor].maxlen = (tree[ans].lazy ? 0 : (num >> 1));
            tree[ans].lazy = -1;
        }
    }
    void pushUp(int ans, int num) {
        tree[ans].llen = tree[gol].llen;
        tree[ans].rlen = tree[gor].rlen;
        tree[ans].maxlen = max(tree[gol].maxlen, tree[gor].maxlen);
        if (tree[ans].llen == num - (num >> 1)) {
            tree[ans].llen += tree[gor].llen;
        }
        if (tree[ans].rlen == num >> 1) {
            tree[ans].rlen += tree[gol].rlen;
        }
        tree[ans].maxlen = max(tree[ans].maxlen, tree[gol].rlen + tree[gor].llen);
    }
    void build(int l, int r, int ans) {
        tree[ans].l = l;
        tree[ans].r = r;
        tree[ans].lazy = -1;
        tree[ans].llen = tree[ans].rlen = tree[ans].maxlen = r - l + 1;
        if (l == r) {
            return;
        }
        int mid = (l + r) >> 1;
        build(lson);
        build(rson);
    }
    void updata(int l, int r, int ans, int num) {//更新區間
        if (l <= tree[ans].l&&r >= tree[ans].r) {
            tree[ans].lazy = num;
            tree[ans].llen = tree[ans].rlen = tree[ans].maxlen = (num ? 0 : tree[ans].r - tree[ans].l + 1);
            return;
        }
        pushDown(ans, tree[ans].r - tree[ans].l + 1);
        int mid = (tree[ans].l + tree[ans].r) >> 1;
        if (l <= mid) {
            updata(l, r, gol, num);
        }
        if (mid < r) {
            updata(l, r, gor, num);
        }
        pushUp(ans, tree[ans].r - tree[ans].l + 1);
    }
    int query(int ans, int num) {//查詢長度為num的空閒區間
        if (tree[ans].l == tree[ans].r) {
            return tree[ans].l;
        }
        pushDown(ans, tree[ans].r - tree[ans].l + 1);
        int mid = (tree[ans].l + tree[ans].r) >> 1;
        if (tree[gol].maxlen >= num) {
            return query(gol, num);
        }
        if (tree[gol].rlen + tree[gor].llen >= num) {
            return mid - tree[gol].rlen + 1;
        }
        return query(gor, num);
    }
};
int main() {
    cin.sync_with_stdio(false);
    int n, m;
    int a, b;
    int x, y;
    Segment__Tree trees;
    while (cin >> n >> a >> b) {
        cin >> m;
        mapp.clear();
        n = n + a + b;
        trees.build(1, n, 1);
        for (int i = 1; i <= m; i++) {
            cin >> x >> y;
            if (x == 1) {
                if (tree[1].maxlen < y + a + b) {
                    cout << -1 << endl;
                }
                else {
                    int pos = trees.query(1, y + a + b);
                    now.l = pos + a;
                    now.r = now.l + y - 1;
                    mapp[i]=now;
                    trees.updata(now.l, now.r, 1, 1);
                    cout << pos - 1 << endl;
                }
            }
            else {
                trees.updata(mapp[y].l, mapp[y].r, 1, 0);
            }
        }
    }
    return 0;
}