poj 3667 Hotel - 线段树

时间：2017-03-23 22:04:41 阅读：353 评论：0 收藏：0 [点我收藏+]

标签：inline lan tput namespace lag cep res while ble

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_i contiguous 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 rto 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

Sample Output

　　题意是说写一个数据结构，支持一下操作

　　1.Check-in 找一段连续的大小为k的空房间，使开始的位置尽量小

　　2.Check-out 使一段连续的房间变为空房间。

　　这里就直接讲正解了吧。用线段树，维护以下三个量

　　1.从最左边开始的一段连续的空房间的长度

　　2.从最右边开始的一段连续的空房间的长度

　　3.当前区间最长的一段连续的空房间长度

技术分享

　　现在来思考一下维护，左边和右边连续比较简单，特殊处理当整个区间都是空闲的房间的情况。

技术分享

　　至于最长的连续的空闲房间呢，先用两段的len值更新它，然后再看一下中间合并后多出来的那一段是否可以更新len值。

　　Check-out操作现在可以完成了，和普通的线段树区间修改是一样的。

　　Check-in操作运用一点"人生的经验"（贪心）,先看左边的len是否满足，满足的话就递归左边，再看中间是否满足，满足就返回结果（这个起点是可以算出来的），最后看右边是否满足。查询结束后，当结果不为0时才去把一段连续的空房间变为已住满的房间（我真的没有想出来边查询边打标记）。

Code

  1 /**
  2  * poj
  3  * Problem#3667
  4  * Accepted
  5  * Time:1735ms
  6  * Memory:5388k
  7  */
  8 #include<iostream>
  9 #include<fstream> 
 10 #include<sstream>
 11 #include<algorithm>
 12 #include<cstdio>
 13 #include<cstring>
 14 #include<cstdlib>
 15 #include<cctype>
 16 #include<cmath>
 17 #include<ctime>
 18 #include<map>
 19 #include<stack>
 20 #include<set>
 21 #include<queue>
 22 #include<vector>
 23 #ifndef WIN32
 24 #define AUTO "%lld"
 25 #else
 26 #define AUTO "%I64d"
 27 #endif
 28 using namespace std;
 29 typedef bool boolean;
 30 #define inf 0xfffffff
 31 #define smin(a, b) (a) = min((a), (b))
 32 #define smax(a, b) (a) = max((a), (b))
 33 #define max3(a, b, c)    max((a), max((b), (c)))
 34 template<typename T>
 35 inline boolean readInteger(T& u) {
 36     char x;
 37     int aFlag = 1;
 38     while(!isdigit((x = getchar())) && x != ‘-‘ && x != -1);
 39     if(x == -1)    {
 40         ungetc(x, stdin);
 41         return false;
 42     }
 43     if(x == ‘-‘) {
 44         aFlag = -1;
 45         x = getchar();
 46     }
 47     for(u = x - ‘0‘; isdigit((x = getchar())); u = u * 10 + x - ‘0‘);
 48     u *= aFlag;
 49     ungetc(x, stdin);
 50     return true;
 51 }
 52 
 53 typedef class SegTreeNode {
 54     public:
 55         int len;
 56         int llen;
 57         int rlen;
 58         int lazy;
 59         SegTreeNode* left, *right;
 60         SegTreeNode(const int len = 1, const int llen = 1, const int rlen = 1):len(len), llen(llen), rlen(rlen), lazy(0) {
 61             left = right = NULL;
 62         }
 63         
 64         inline void pushUp(int l, int r, int mid) {
 65             len = max3(left->len, right->len, left->rlen + right->llen);
 66             llen = left->llen, rlen = right->rlen;
 67             if(left->llen == (mid - l + 1))    llen = left->llen + right->llen;
 68             if(right->rlen == (r - mid))    rlen = left->rlen + right->rlen;    
 69         }
 70         
 71         inline void pushDown(int l, int r, int mid) {
 72             this->left->len = this->left->llen = this->left->rlen = (lazy > 0) ? (mid - l + 1) : (0);
 73             this->right->len= this->right->llen= this->right->rlen= (lazy > 0) ? (r - mid) : (0);
 74             this->left->lazy = this->right->lazy = lazy;
 75             lazy = 0;
 76         }
 77 }SegTreeNode;
 78 
 79 typedef class SegTree {
 80     public:
 81         SegTreeNode *root;
 82         
 83         SegTree():root(NULL) {         }
 84         SegTree(int s) {
 85             build(root, 1, s);
 86         }
 87         
 88         void build(SegTreeNode*& node, int l, int r) {
 89             node = new SegTreeNode();
 90             if(l == r)        return;
 91             int mid = (l + r) >> 1;
 92             build(node->left, l, mid);
 93             build(node->right, mid + 1, r);
 94             node->pushUp(l, r, mid);
 95         }
 96         
 97         void update(SegTreeNode*& node, int l, int r, int cl, int cr, int val) {
 98             int mid = (l + r) >> 1;
 99             if(node->lazy && l != r)    node->pushDown(l, r, mid);
100             if(cl == l && r == cr) {
101                 node->lazy = (val) ? (1) : (-1);
102                 node->len = node->llen = node->rlen = val * (r - l + 1);
103                 return;
104             }
105             if(cr <= mid)    update(node->left, l, mid, cl, cr, val);
106             else if(cl > mid)    update(node->right, mid + 1, r, cl, cr, val);
107             else {
108                 update(node->left, l, mid, cl, mid, val);
109                 update(node->right, mid + 1, r, mid + 1, cr, val);
110             }
111             node->pushUp(l, r, mid);
112         }
113         
114         int query(SegTreeNode*& node, int l, int r, int len) {
115             int mid = (l + r) >> 1;
116             if(node->lazy && l != r)    node->pushDown(l, r, mid);
117             if(r - l + 1 == len)    return l;
118             if(l == r)    return 0;
119             if(node->left->len >= len)    return query(node->left, l, mid, len);
120             else if(node->left->rlen + node->right->llen >= len && node->left->rlen > 0)
121                 return mid - node->left->rlen + 1;
122             else if(node->right->len >= len)    return query(node->right, mid + 1, r, len);
123             return 0;
124         }
125         
126 }SegTree;
127 
128 int n, m;
129 SegTree st;
130 
131 inline void solve() {
132     readInteger(n);
133     readInteger(m);
134     st = SegTree(n);
135     for(int i = 1, a, b, c; i <= m; i++) {
136         readInteger(a);
137         readInteger(b);
138         if(a == 1) {
139             c = st.query(st.root, 1, n, b);
140             if(c != 0)    st.update(st.root, 1, n, c, b + c - 1, 0);
141             printf("%d\n", c);
142         } else {
143             readInteger(c);
144             st.update(st.root, 1, n, b, b + c - 1, 1);
145         }
146     }
147 }
148 
149 int main() {
150     solve();
151     return 0;
152 }

poj 3667 Hotel - 线段树

标签：inline lan tput namespace lag cep res while ble

原文地址：http://www.cnblogs.com/yyf0309/p/6607093.html

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