SPOJ ADAQUEUE - Ada and Queue

题目大意：有四种操作，back输出队尾，去列尾；front输出队首，去队首；reverse颠倒队列；push_back N尾端插入N；toFront N首端插入N。如果操作失败，也就是队空时，希望输出队首队尾，就输出"No job for Ada?"

解题思路：用双向队列，就是简单的调用函数了,颠倒时，也由一个变量标记，使正常接下来的操作尾变头，头变尾。或者模拟，队首队尾分别标记，一个方向变量初始为1表示未颠倒过队列，颠倒时方向变量变号，首尾标记交换。

双向队列ac代码：

#include <iostream>
#include <cstring>
#include <deque>
using namespace std;
deque <int>qu;
int main()
{
  int t, temp, dir;
  char com[1005];
  while (scanf("%d", &t)!=EOF){
    dir = 1;
    for (int i=1; i<=t; i++){
      scanf("%s", com);
      if (strcmp(com, "toFront") == 0){
        scanf("%d", &temp);
        if (dir > 0)
          qu.push_front(temp);
        else
          qu.push_back(temp);
      }
      if (strcmp(com, "push_back") == 0){
        scanf("%d", &temp);
        if (dir > 0)
          qu.push_back(temp);
        else
          qu.push_front(temp);
      }
      if (strcmp(com, "front") == 0){
        if (qu.empty())
          printf("No job for Ada?\n");
        else{
          if (dir > 0){
            printf("%d\n", qu.front());
            qu.pop_front();         
          }
          else{
            printf("%d\n", qu.back());
            qu.pop_back();
          }
        }
      }
      if (strcmp(com, "back") == 0){
        if (qu.empty())
          printf("No job for Ada?\n");
        else{
          if (dir > 0){
            printf("%d\n", qu.back());
            qu.pop_back();          
          }
          else{
            printf("%d\n", qu.front());
            qu.pop_front();
          }
        }       
      }
      if (strcmp(com, "reverse") == 0)
        dir = -dir;
    }
    qu.clear();
  }
return 0;
}

模拟ac代码：

#include <iostream>
#include <cstring>
using namespace std;
int a[1000005];
int main()
{
  int t, front, rear, dir, temp, sum;
  char com[1005];
  while (scanf("%d", &t)!=EOF){
    dir = 1;
    sum = 0;
    front = rear = 1000005 / 2;
    for (int i=1; i<=t; i++){
      scanf("%s", com);
      if (strcmp(com, "toFront") == 0){
        scanf("%d", &temp);
        if (sum){
          front -= dir;         
          a[front] = temp;
        }
        else
          a[front] = temp;      
        sum++;
      }
      if (strcmp(com, "push_back") == 0){
        scanf("%d", &temp);
        if (sum){
          rear += dir;
          a[rear] = temp;
        }
        else
          a[rear] = temp;
        sum++;
      }
      if (strcmp(com, "front") == 0){
        if (!sum)
          printf("No job for Ada?\n");
        else{
          printf("%d\n", a[front]);
          sum--;
          if (sum)
            front += dir;         
        }
      }
      if (strcmp(com, "back") == 0){
        if (!sum)
          printf("No job for Ada?\n");
        else{
          printf("%d\n", a[rear]);
          sum--;
          if (sum)
            rear -= dir;  
        }         
      }
      if (strcmp(com, "reverse") == 0 && sum){
        dir = -dir;
        temp = rear;
        rear = front;
        front = temp;
      }
    }
  }
return 0;
}