码迷,mamicode.com
首页 > 其他好文 > 详细

LightOJ 1212 <Double Ended Queue >

时间:2016-08-30 16:00:57      阅读:170      评论:0      收藏:0      [点我收藏+]

标签:

Description

A queue is a data structure based on the principle of ‘First In First Out‘ (FIFO). There are two ends; one end can be used only to insert an item and the other end to remove an item. A Double Ended Queue is a queue where you can insert an item in both sides as well as you can delete an item from either side. There are mainly four operations available to a double ended queue. They are:

  1. pushLeft(): inserts an item to the left end of the queue with the exception that the queue is not full.
  2. pushRight(): inserts an item to the right end of the queue with the exception that the queue is not full.
  3. popLeft(): removes an item from the left end of the queue with the exception that the queue is not empty.
  4. popRight(): removes an item from the right end of the queue with the exception that the queue is not empty.

Now you are given a queue and a list of commands, you have to report the behavior of the queue.

Input

Input starts with an integer T (≤ 20), denoting the number of test cases.

Each case starts with a line containing two integers n, m (1 ≤ n ≤ 10, 1 ≤ m ≤ 100), where n denotes the size of the queue and m denotes the number of commands. Each of the next m lines contains a command which is one of:

pushLeft x        pushes x (-100 ≤ x ≤ 100) in the left end of the queue

pushRight x      pushes x (-100 ≤ x ≤ 100) in the right end of the queue

popLeft             pops an item from the left end of the queue

popRight           pops an item from the right end of the queue

Output

For each case, print the case number in a line. Then for each operation, show its corresponding output as shown in the sample. Be careful about spelling.

Sample Input

1

3 8

pushLeft 1

pushLeft 2

pushRight -1

pushRight 1

popLeft

popRight

popLeft

popRight

Sample Output

Case 1:

Pushed in left: 1

Pushed in left: 2

Pushed in right: -1

The queue is full

Popped from left: 2

Popped from right: -1

Popped from left: 1

The queue is empty

模拟双向队列。因为数据规模小,可以直接开个大数组,从中间开始操作。
AC代码:
#include<stdio.h>
#include<stdlib.h>


int main()
{
    char s[20];
int num[600];
int a,b,n,m,T,k,c,judge;
while(scanf("%d",&T)!=EOF&&T!=0)
{
    for(int j=1;j<=T;j++)
        {
            printf("Case %d:\n",j);
            scanf("%d%d",&n,&m);
            judge=0;
            a=300;b=300;k=0;
            for(int i=1;i<=m;i++)
                {
                    getchar();
                    scanf("%s",s);
                    if(s[1]==o)
                        {
                            if(k==0)printf("The queue is empty\n");
                            else if(k==1&&s[3]==L){k--;printf("Popped from left: %d\n",num[a]);}
                            else  if(k==1&&s[3]==R){k--;printf("Popped from right: %d\n",num[a]);}
                            else  if(k>1&&s[3]==L){k--;printf("Popped from left: %d\n",num[a]);a++;}
                            else  if(k>1&&s[3]==R){k--;printf("Popped from right: %d\n",num[b]);b--;}
                        }
                    if(s[1]==u)
                        {
                            scanf("%d",&c);
                            if(k==n)printf("The queue is full\n");
                            if(k<n)
                                {
                                    if(k==0&&s[4]==L){k++;num[a]=c;printf("Pushed in left: %d\n",num[a]);}
                                    else  if(k==0&&s[4]==R){k++;num[a]=c;printf("Pushed in right: %d\n",num[a]);}
                                    else  if(k>0&&s[4]==L){k++;a--;num[a]=c;printf("Pushed in left: %d\n",num[a]);}
                                    else  if(k>0&&s[4]==R){k++;b++;num[b]=c;printf("Pushed in right: %d\n",num[b]);}

                                }
                        }
                }
        }
}
}

 

LightOJ 1212 <Double Ended Queue >

标签:

原文地址:http://www.cnblogs.com/914295860-jry/p/5822327.html

(0)
(0)
   
举报
评论 一句话评论(0
登录后才能评论!
© 2014 mamicode.com 版权所有  联系我们:gaon5@hotmail.com
迷上了代码!