POJ 3667 - Hotel(线段树+区间合并)

Description

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 D_i (1 ≤ D_i ≤ N) and approach the front desk to check in. Each group i requests a set of D_icontiguous rooms from Canmuu, the moose staffing the counter. He assigns them some set of consecutive room numbers r..r+D_i-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 X_i and D_i which specify the vacating of rooms X_i..X_i +D_i-1 (1 ≤ X_i ≤ N-D_i+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.

Input

* 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 D_i (b) Three space-separated integers representing a check-out: 2, X_i, and D_i

Output

* 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.

Sample Input

10 61 31 31 31 32 5 51 6

Sample Output

14705

题意：n个人要住房，如果旅馆有足够空闲房间，那么只住进最左边的那些客房。

最基础的区间合并了。弱鸡表示还是得好好学习。

#include <stdio.h>#include <string.h>#include <cmath>#include <algorithm>using namespace std;#define lson l,m,rt<<1#define rson m+1,r,rt<<1|1const int maxn = 1e5 + 10;int lsum[maxn << 2], rsum[maxn << 2], msum[maxn << 2], cover[maxn << 2];void PushUp(int rt, int m){lsum[rt] = lsum[rt << 1];rsum[rt] = rsum[rt << 1 | 1];if (lsum[rt << 1] == m - m / 2) lsum[rt] += lsum[rt << 1 | 1];if (rsum[rt << 1 | 1] == m / 2) rsum[rt] += rsum[rt << 1];msum[rt] = max(rsum[rt << 1] + lsum[rt << 1 | 1], max(msum[rt << 1], msum[rt << 1 | 1]));}void PushDown(int rt, int m){if (cover[rt] != -1){cover[rt << 1] = cover[rt << 1 | 1] = cover[rt];msum[rt << 1] = lsum[rt << 1] = rsum[rt << 1] = cover[rt] ? 0 : m - m / 2;msum[rt << 1 | 1] = lsum[rt << 1 | 1] = rsum[rt << 1 | 1] = cover[rt] ? 0 : m / 2;cover[rt] = -1;}}void build(int l, int r, int rt){lsum[rt] = rsum[rt] = msum[rt] = r - l + 1;cover[rt] = -1;//0表示里面没住人，1表示有，-1表示初始状态或者不确定。if (l == r) return;int m = (l + r) >> 1;build(lson);build(rson);}int query(int c, int l, int r, int rt){if (l == r) return l;PushDown(rt, r - l + 1);int m = (l + r) >> 1;if (c <= msum[rt << 1])return query(c, lson);else if (c <= rsum[rt << 1] + lsum[rt << 1 | 1])return m - rsum[rt << 1] + 1;elsereturn query(c, rson);}void update(int L, int R, int c, int l, int r, int rt){if (L <= l&&r <= R){lsum[rt] = rsum[rt] = msum[rt] = c ? 0 : r - l + 1;cover[rt] = c;return;}PushDown(rt, r - l + 1);int m = (l + r) >> 1;if (L <= m) update(L, R, c, lson);if (R > m) update(L, R, c, rson);PushUp(rt, r - l + 1);}int main(){int n, T;while (scanf("%d%d", &n, &T) != EOF){int op;build(1, n, 1);while (T--){scanf("%d", &op);if (op == 1){int a;scanf("%d", &a);if (a > msum[1])puts("0");else{int p = query(a, 1, n, 1);printf("%d\n", p);update(p, p + a - 1, 1, 1, n, 1);}}else{int a, b;scanf("%d%d", &a, &b);update(a, a + b - 1, 0, 1, n, 1);}}}return 0;}