对应POJ题目：点击打开链接
Double Queue
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 11741 | Accepted: 5335 |
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
2 1 20 14 1 30 3 2 1 10 99 3 2 2 0
Sample Output
0 20 30 10 0
题意：
有3个操作：（1 x y）表示以y为优先级在队列里插入一对数（x y），其中每一对数的x和y均不同；（2）表示输出优先级最高的一对数中的x，（3）表示输出优先级最低的一对数中的x；（0）表示退出。
思路：
各种二叉排序树，还有STL里的map，set也可AC，贴AVL模板
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#define MAX(x, y) ((x)>(y)?(x):(y))
typedef struct NODE
{
int val, data;
int bf; //平衡因子（左子树高度与右子树高度之差）
int h; //以当前结点为根结点的数的高度
NODE *l, *r;
}Node;
class BalanceTree
{
public:
void Init()
{
rt = NULL;
}
int Height(Node *T)
{
if(NULL == T) return 0;
return T->h;
}
int BF(Node *l, Node *r)
{
if(NULL == l && NULL == r) return 0;
else if(NULL == l) return -r->h;
else if(NULL == r) return l->h;
else return l->h - r->h;
}
Node *LL_rotate(Node *A)
{
Node *B;
B = A->l;
A->l = B->r;
B->r = A;
A->h = MAX(Height(A->l), Height(A->r)) + 1;
B->h = MAX(Height(B->l), Height(B->r)) + 1;
A->bf = BF(A->l, A->r);
B->bf = BF(B->l, B->r);
return B;
}
Node *RR_rotate(Node *A)
{
Node *B;
B = A->r;
A->r = B->l;
B->l = A;
A->h = MAX(Height(A->l), Height(A->r)) + 1;
B->h = MAX(Height(B->l), Height(B->r)) + 1;
A->bf = BF(A->l, A->r);
B->bf = BF(B->l, B->r);
return B;
}
Node *LR_rotate(Node *A)
{
Node *C;
A->l = RR_rotate(A->l);
C = LL_rotate(A);
return C;
}
Node *RL_rotate(Node *A)
{
Node *C;
A->r = LL_rotate(A->r);
C = RR_rotate(A);
return C;
}
void Insert(int v, int e)
{
Insert_(rt, v, e);
}
void Insert_(Node *&T, int v, int e)
{
if(NULL == T){
T = (Node *)malloc(sizeof(Node));
T->val = v;
T->data = e;
T->bf = 0;
T->h = 1;
T->l = T->r = NULL;
return;
}
if(e < T->data) Insert_(T->l, v, e);
else Insert_(T->r, v, e);
T->h = MAX(Height(T->l), Height(T->r)) + 1;
T->bf = BF(T->l, T->r);
if(T->bf > 1 || T->bf < -1){ //T结点失衡
if(T->bf > 0 && T->l->bf > 0) T = LL_rotate(T); //如果T->bf > 0 则肯定有左儿子
else if(T->bf < 0 && T->r->bf < 0) T = RR_rotate(T); //如果T->bf < 0 则肯定有右儿子
else if(T->bf > 0 && T->l->bf < 0) T = LR_rotate(T);
else if(T->bf < 0 && T->r->bf > 0) T = RL_rotate(T);
}
}
void Delete(int flag) //flag为1表示找最大值
{
if(NULL == rt){
printf("0\n");
return;
}
Node *tmp = rt;
if(!flag) while(tmp->l) tmp = tmp->l;
else while(tmp->r) tmp = tmp->r;
printf("%d\n", tmp->val);
Delete_(rt, tmp->data);
}
void Delete_(Node *&T, int key)
{
if(NULL == T) return;
if(key < T->data){ //从左边删除
Delete_(T->l, key);
T->bf = BF(T->l, T->r); //计算删除后T结点的平衡因子
if(T->bf < -1){ //T结点左边高度变小（删了左边的元素）导致失衡
if(1 == T->r->bf) T = RL_rotate(T); //RL型
else T = RR_rotate(T); //平衡因子为0或-1
}
}
else if(key > T->data){ //从右边删除
Delete_(T->r, key);
T->bf = BF(T->l, T->r); //计算删除后T结点的平衡因子
if(T->bf > 1){ //T结点右边高度变小（删了右边的元素）导致失衡
if(-1 == T->l->bf) T = LR_rotate(T); //LR型
else T = LL_rotate(T); //平衡因子为0或1
}
}
else{
if(T->l && T->r){ //左右都不为空
Node *tmp = T->l; //用被删除节点的左子树的最大值代替被删除节点
while(tmp->r) tmp = tmp->r;
T->data = tmp->data;
T->val = tmp->val;
Delete_(T->l, tmp->data);
T->bf = BF(T->l, T->r); //计算删除后T结点的平衡因子
if(T->bf < -1){ //T结点左边高度变小（删了左边的元素）导致失衡
if(1 == T->r->bf) T = RL_rotate(T); //RL型
else T = RR_rotate(T); //平衡因子为0或-1
}
}
else{
Node *tmp = T;
if(T->l) T = T->l; //有左儿子
else if(T->r) T = T->r; //有右儿子
else{ //T无儿子
free(T);
T = NULL;
}
if(T) free(tmp);
}
}
}
void Show()
{
InOrder(rt);
printf("\n");
}
void InOrder(Node *T)
{
if(NULL == T) return;
InOrder(T->l);
printf("%d(%d) ", T->val, T->data);
InOrder(T->r);
}
void Free()
{
FreeTree(rt);
}
void FreeTree(Node *T)
{
if(NULL == T) return;
FreeTree(T->l);
FreeTree(T->r);
free(T);
}
private:
Node *rt;
};
BalanceTree bt;
int main()
{
#if 0
freopen("in.txt","r",stdin);
#endif
int op, v, e;
bt.Init();
while(scanf("%d", &op), op)
{
if(1 == op){
scanf("%d%d", &v, &e);
bt.Insert(v, e);
}
else if(2 == op)
bt.Delete(1);
else if(3 == op)
bt.Delete(0);
}
bt.Free();
return 0;
}
