题目链接：Codeforces 46D Parking Lot

题目大意：一个街道，长为N，每辆车停进来的时候必须和前面间隔B米，和后面间隔F米，现在用两种操作，1是停进来一个长为x的车，2是第x辆车开走。

解题思路：区间合并，建一颗长度为N + B + F的线段树，然后每次停车进去的时候都查询x + B + F的区间，然后修改的时候只修改x的长度。

#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

const int maxn = 100505;

#define lson(x) ((x)<<1)
#define rson(x) (((x)<<1)|1)
int lc[maxn << 2], rc[maxn << 2], set[maxn << 2];
int L[maxn << 2], R[maxn << 2], S[maxn << 2]; 

inline int length(int u) {
    return rc[u] - lc[u] + 1;
}

inline void maintain (int u, int v) {
    S[u] = L[u] = R[u] = (v ? length(u) : 0);
}

inline void pushup (int u) {
    S[u] = max(max(S[lson(u)], S[rson(u)]), R[lson(u)] + L[rson(u)]);
    L[u] = L[lson(u)] + (L[lson(u)] == length(lson(u)) ? L[rson(u)] : 0);
    R[u] = R[rson(u)] + (R[rson(u)] == length(rson(u)) ? R[lson(u)] : 0);
}

inline void pushdown(int u) {
    if (set[u] != -1) {
        maintain(lson(u), set[u]);
        maintain(rson(u), set[u]);
        set[u] = -1;
    }
}

void build (int u, int l, int r) {
    lc[u] = l;
    rc[u] = r;
    set[u] = -1;

    if (l == r) {
        maintain(u, 1);
        return;
    }

    int mid = (l + r) / 2;
    build(lson(u), l, mid);
    build(rson(u), mid + 1, r);
    pushup(u);
}

void modify (int u, int l, int r, int v) {
    if (l <= lc[u] && rc[u] <= r) {
        maintain(u, v);
        return;
    }

    pushdown(u);
    int mid = (lc[u] + rc[u]) / 2;
    if (l <= mid)
        modify(lson(u), l, r, v);
    if (r > mid)
        modify(rson(u), l, r, v);
    pushup(u);
}

int query(int u, int k) {
    if (S[u] < k)
        return -1;

    if (lc[u] == rc[u])
        return lc[u];

    pushdown(u);
    int mid = (lc[u] + rc[u]) / 2, ret;
    if (S[lson(u)] >= k)
        ret = query(lson(u), k);
    else if (R[lson(u)] + L[rson(u)] >= k)
        ret = mid - R[lson(u)] + 1;
    else
        ret = query(rson(u), k);
    pushup(u);
    return ret;
}

int N, M, B, F;
int op[105][2];

int main () {
    scanf("%d%d%d%d", &N, &B, &F, &M);
    build(1, 0, B + F + N - 1);

    int x, y;
    for (int i = 1; i <= M; i++) {
        scanf("%d%d", &x, &y);
        if (x == 1) {
            op[i][0] = query(1, y + B + F);
            if (op[i][0] != -1) {
                //op[i][0] = 
                op[i][1] = op[i][0] + y - 1;
                modify(1, op[i][0] + B, op[i][1] + B, 0);
            }
            printf("%d\n", op[i][0]);
        } else
            modify(1, op[y][0] + B, op[y][1] + B, 1);
    }
    return 0;
}

Codeforces 46D Parking Lot(线段树)