【POJ3481】【splay】Double Queue

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转载最新推荐文章于 2026-03-30 10:50:11 发布 241 阅读 AI 写同款· GEO 优化 ›

Description

The new founded Balkan Investment Group Bank (BIG-Bank) opened a new office in Bucharest, equipped with a modern computing environment provided by IBM Romania, and using modern information technologies. As usual, each client of the bank is identified by a positive integer K and, upon arriving to the bank for some services, he or she receives a positive integer priority P. One of the inventions of the young managers of the bank shocked the software engineer of the serving system. They proposed to break the tradition by sometimes calling the serving desk with the lowest priority instead of that with the highest priority. Thus, the system will receive the following types of request:

0	The system needs to stop serving
1 K P	Add client K to the waiting list with priority P
2	Serve the client with the highest priority and drop him or her from the waiting list
3	Serve the client with the lowest priority and drop him or her from the waiting list

Your task is to help the software engineer of the bank by writing a program to implement the requested serving policy.

Input

Each line of the input contains one of the possible requests; only the last line contains the stop-request (code 0). You may assume that when there is a request to include a new client in the list (code 1), there is no other request in the list of the same client or with the same priority. An identifier K is always less than 10⁶, and a priority P is less than 10⁷. The client may arrive for being served multiple times, and each time may obtain a different priority.

Output

For each request with code 2 or 3, the program has to print, in a separate line of the standard output, the identifier of the served client. If the request arrives when the waiting list is empty, then the program prints zero (0) to the output.

Sample Input

Sample Output

Source

Southeastern Europe 2007

【分析】

本来写一个splay模板，想先过了这题然后去刷3580和维修数列的...

结果交上去果断T了，我大惊，莫非双旋比单旋还慢！

最后D了半天...发现是内存池的原因，知道真相的我眼泪掉下来.....话说我写内存池为什么就没成功过一次....

  1 #include <iostream>
  2 #include <cstdio>
  3 #include <algorithm>
  4 #include <cstring>
  5 #include <vector>
  6 #include <utility>
  7 #include <iomanip>
  8 #include <string>
  9 #include <cmath>
 10 #include <queue>
 11 #include <assert.h>
 12 #include <map>
 13 #include <ctime>
 14 #include <cstdlib>
 15 
 16 const int MAXN = 100001 + 10;
 17 const int INF = 0x7fffffff;
 18 const int SIZE = 450;
 19 const int maxnode =  10000000 + 10;
 20 using namespace std;
 21 typedef int ll;
 22 struct SPLAY{
 23        struct Node{
 24               int val, size;
 25               int num;//num代表编号 
 26               Node *parent, *ch[2];
 27               
 28               Node(){
 29                      val = size = 0;
 30                      num = 0;
 31                      parent = ch[0] = ch[1] = NULL;       
 32               }
 33               int cmp(){
 34                   if (parent == NULL) return -1;
 35                   if (parent->ch[0]->num == num) return 0;
 36                   if (parent->ch[1]->num == num) return 1;
 37               }
 38               void update(){
 39                    size = 1;
 40                    if (ch[0] != NULL) size += ch[0]->size;
 41                    if (ch[1] != NULL) size += ch[1]->size;
 42               }
 43        }*root, *nil, _nil, mem[1222222];//我写内存池就没成功过
 44        
 45        /*struct MEMPOOR{//内存池 
 46               queue< Node >Q;
 47               Node mem[maxnode];
 48               Node *nil;
 49               
 50               void init(Node *&t){
 51                    for (int i = 0; i < maxnode; i++) Q.push( mem[i] );
 52                    nil = t;
 53               }
 54               Node *get(){
 55                    Node *p = &(Q.front());
 56                    Q.pop();
 57                    p->parent = p->ch[0] = p->ch[1] = nil;
 58                    p->num = p->val = 0;
 59                    p->size = 1;
 60                    return p;
 61               }
 62               //回收内存 
 63               void push(Node *&p){
 64                    Q.push(*(p));
 65                    p->parent->ch[p->cmp()] = nil;
 66               }
 67        }poor;*/
 68        int tot, cnt;//计数 
 69        /*void debug(){//测试内存池 
 70             poor.init();
 71             Node *a = poor.get();
 72             a->val = 10;
 73             Node *b = poor.get();
 74             b->val = 5;
 75             a->ch[0] = b;
 76             printf("%d\n", poor.Q.size());
 77             printf("%d\n", a->val);
 78             printf("%d\n", b->cmp());
 79        }*/
 80        Node *NEW(){
 81             Node *p = &mem[cnt++];
 82             p->parent = p->ch[0] = p->ch[1] = nil;
 83             p->num = p->val = 0;
 84             p->size = 1;
 85             return p;
 86        }
 87        void init(){
 88             //循环哨兵的定义 
 89             nil = &_nil;
 90             _nil.parent = _nil.ch[0] = _nil.ch[1] = nil;
 91             
 92             tot = 0;
 93             cnt = 0;
 94             root = nil;
 95             insert(root, -INF, -INF);
 96             insert(root, INF, INF);
 97        }
 98        void rotate(Node *t, int d){
 99             Node *p = t->parent;//t的右旋对于p来说也是右旋
100             t = p->ch[d ^ 1];
101             p->ch[d ^ 1] = t->ch[d];
102             t->ch[d]->parent = p;
103             t->ch[d] = p;
104             t->parent = p->parent;
105             //注意，这里要更新的原因在于t并不是引用 
106             if (t->parent != nil){
107                if (t->parent->ch[0] == p) t->parent->ch[0] = t;
108                else if (t->parent->ch[1] == p)  t->parent->ch[1] = t;
109             }
110             p->parent = t;
111             if (t->parent == nil) root = t;
112             //不用换回去了... 
113             p->update();
114             t->update();
115        }
116        //将x旋转为y的子树 
117        void splay(Node *x, Node *y){
118             while (x->parent != y){
119                   if (x->parent->parent == y) rotate(x, (x->cmp() ^ 1));
120                   else{//不然连转两下 
121                        rotate(x->parent, x->parent->cmp() ^ 1);
122                        rotate(x, x->cmp() ^ 1);
123                   }
124                   x->update();
125             }
126        }
127        //编号和权值 
128        void insert(Node *&t, int num, int val){
129             if (t == nil){
130                t = NEW();
131                t->val = val;
132                t->num = num;
133                return;
134             }
135             Node *x = t;
136             while (1){
137                   int dir = (val > x->val);
138                   if (x->ch[dir] == nil){
139                      x->ch[dir] = NEW();
140                      x->ch[dir]->val = val;
141                      x->ch[dir]->num = num;
142                      x->ch[dir]->parent = x;
143                      splay(x->ch[dir], nil);
144                      return;
145                   }else x = x->ch[dir];
146             }
147             return;
148        }
149        void debug(){
150             /*init();
151             root = nil;
152             insert(root, 1, 1);
153             //printf("%d", root->val);
154             insert(root, 2, 2);
155             insert(root, 0, 0);
156             insert(root, 4, 4);
157             print(root);
158             //printf("%d\n", root->size); */
159        }
160        //找到val值为k的数的名次 
161        int find(Node *t, int k){
162            if (t == nil) return -1;//-1代表找不到 
163            int tmp = t->ch[0]->size;
164            if (k == t->val) return tmp + 1;
165            else if (k < t->val) return find(t->ch[0], k);
166            else return find(t->ch[1], k) + tmp + 1;
167        }
168        //找到第k小的权值并将其splay到根 
169        void get(Node *t, Node *y, int k){
170            Node *x = t;
171            while (1){
172                  int tmp = x->ch[0]->size;
173                  if ((tmp + 1) == k) break;
174                  if (k <= tmp) x = x->ch[0];
175                  else {x = x->ch[1]; k -= tmp + 1;}
176            }
177            splay(x, y);
178        }
179        //删除val值为k的节点 
180        void Delete(int k){
181             int tmp = find(root, k);
182             get(root, nil, tmp - 1);
183             get(root, root, tmp + 1);
184             //卡出来 
185             //poor.push(root->ch[1]->ch[0]);
186             root->ch[1]->ch[0] = nil;
187             root->ch[1]->update();
188             root->update();
189             return;
190        }
191        void print(Node *t){
192             if (t == nil) return;
193             print(t->ch[0]);
194             printf("%d ", t->val);
195             print(t->ch[1]);
196        }
197        
198 }A;
199 int n, m;
200 
201 
202 void work(){
203      //A.root = nil;
204      A.init();//A.tot记录了A中的元素个数 
205      int t;
206      while (scanf("%d", &t) && t){
207             if (t == 2){
208                if (A.tot == 0) {printf("0\n");continue;}
209                A.get(A.root, A.nil, A.tot + 1);
210                printf("%d\n", A.root->num);
211                A.Delete(A.root->val);
212                A.tot--;
213             }else if (t == 3){
214                if (A.tot == 0) {printf("0\n");continue;}
215                A.get(A.root, A.nil, 2);
216                printf("%d\n", A.root->num);
217                A.Delete(A.root->val);
218                A.tot--;
219             }else{
220                   int val, num;
221                   scanf("%d%d", &num, &val);
222                   A.insert(A.root, num, val);
223                   A.tot++;
224             }
225      }
226 }
227 
228 
229 int main(){
230     #ifdef LOCAL
231     freopen("data.txt", "r", stdin);
232     freopen("out.txt", "w", stdout);
233     #endif
234     work();
235     //A.debug();
236     return 0; 
237 }