HDU 3367 Hotel

Refer to the practice of this blog: Please stamp it and explain it as follows:

The consecutive longest interval that meets the condition in the query interval is usually a question of interval merging.

Four domains are added for the node. Lx: number of consecutive empty rooms on the left of the slave interval. RX: number of consecutive empty rooms on the right of the range. Ax: Total number of consecutive empty rooms in this interval

Col: records the status of the residents in the specified range. 1 indicates full occupancy, 0 indicates full occupancy, and-1 indicates that a room can be occupied.

When querying whether there is a continuous number of empty rooms num, first query the left, when tree [v]. lx> = num is used to recursively return the left interval, and then query the center when tree [V * 2]. lx + tree [V * 2 + 1]. RX> = num directly returns the leftmost endpoint. The last query is on the right when tree [v]. RX> = num recursion right interval. Col records the status of the interval and uses the lazy idea, that is, when tree [v]. col = 1 or tree [v]. col = 0 (when the room is full or cannot be fully occupied), first pass the status to the left and right subtree, and set the col of this interval to-1.

Update the range [st, ST + num-1], that is, when the range is set to null or set to full, you must first pass the status of the range (all empty or full) to the left and right subtree, then recursion, and finally update the Left and Right sub-trees. Because the sub-tree state changes, the father state will certainly change.

After reading this afternoon, I finally understood it and looked at my blog. Forgive me for being a zombie. =

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 hasN(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 sizeDi(1 ≤Di≤ N) and approach the front desk to check in. Each groupIRequests a setDiContiguous rooms from canmuu, the moose staffing the counter. He assigns them some set of consecutive room numbersR..R+Di-1 if they are available or, if no contiguous set of rooms is available, politely suggests alternate lodging. canmuu always chooses the valueRTo be the smallest possible. Visitors also depart the hotel from groups of contiguous rooms. CheckoutIHas the parametersXIAndDiWhich specify the vacating of roomsXI..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 processingM(1 ≤M<50,000) checkin/Checkout requests. The hotel is initially unoccupied. Input * Line 1: two space-separated integers:NAndM * Lines 2 ..M+ 1: LineI+ 1 contains request expressed as one of two possible formats: (a) two space separated integers representing a check-in request: 1 andDi(B) Three space-separated integers representing a check-out: 2,XI, AndDi Output * Lines 1...: For each check-in request, output a single line with a single integerR, 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 Source Usaco 2008 February gold, a tear. It involves lazy ideas, and markup =. =, Comment again after dinner #include <cstdio>#include <cstring>#include <cctype>#include <algorithm>using namespace std;#define lson l , m , rt << 1#define rson m + 1 , r , rt << 1 | 1const int maxn = 50005;int lsum[maxn<<2] , rsum[maxn<<2] , msum[maxn<<2];int cover[maxn<<2];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 >> 1); msum[rt<<1|1] = lsum[rt<<1|1] = rsum[rt<<1|1] = cover[rt] ? 0 : (m >> 1); cover[rt] = -1; }}void PushUp(int rt,int m) { lsum[rt] = lsum[rt<<1]; rsum[rt] = rsum[rt<<1|1]; if (lsum[rt] == m - (m >> 1)) lsum[rt] += lsum[rt<<1|1]; if (rsum[rt] == (m >> 1)) rsum[rt] += rsum[rt<<1]; msum[rt] = max(lsum[rt<<1|1] + rsum[rt<<1] , max(msum[rt<<1] , msum[rt<<1|1]));}void build(int l,int r,int rt) { msum[rt] = lsum[rt] = rsum[rt] = r - l + 1; cover[rt] = -1; if (l == r) return ; int m = (l + r) >> 1; build(lson); build(rson);}void update(int L,int R,int c,int l,int r,int rt) { if (L <= l && r <= R) { msum[rt] = lsum[rt] = rsum[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 (m < R) update(L , R , c , rson); PushUp(rt , r - l + 1);}int query(int w,int l,int r,int rt) { if (l == r) return l; PushDown(rt , r - l + 1); int m = (l + r) >> 1; if (msum[rt<<1] >= w) return query(w , lson); else if (rsum[rt<<1] + lsum[rt<<1|1] >= w) return m - rsum[rt<<1] + 1; return query(w , rson);}int main() { int n,m; int u,v,w; scanf("%d%d",&n,&m); build(1 , n , 1); while (m --) { scanf("%d",&u); if (u== 1) { scanf("%d",&w); if (msum[1] < w) puts("0"); else { int pos= query(w, 1 , n , 1); printf("%d\n",pos); update(pos , pos + w - 1 , 1 , 1 , n , 1); } } else { scanf("%d%d",&v,&w); update(v, v + w - 1 , 0 , 1 , n , 1); } } return 0;}