UVa 11995:I Can Guess the Data Structure!（数据结构练习）

I Can Guess the Data Structure!

There is a bag-like data structure, supporting two operations:

1 x

Throw an element x into the bag.

2

Take out an element from the bag.

Given a sequence of operations with return values, you're going to guess the data structure. It is a stack (Last-In, First-Out), a queue (First-In, First-Out), a priority-queue (Always take out larger elements first) or something else that you can hardly imagine!

Input

There are several test cases. Each test case begins with a line containing a single integer n (1<=n<=1000). Each of the next n lines is either a type-1 command, or an integer 2 followed by an integer x. That means after executing a type-2 command, we get an element x without error. The value of x is always a positive integer not larger than 100. The input is terminated by end-of-file (EOF). The size of input file does not exceed 1MB.

Output

For each test case, output one of the following:

stack

It's definitely a stack.

queue

It's definitely a queue.

priority queue

It's definitely a priority queue.

impossible

It can't be a stack, a queue or a priority queue.

not sure

It can be more than one of the three data structures mentioned above.

Sample Input

6
1 1
1 2
1 3
2 1
2 2
2 3
6
1 1
1 2
1 3
2 3
2 2
2 1
2
1 1
2 2
4
1 2
1 1
2 1
2 2
7
1 2
1 5
1 1
1 3
2 5
1 4
2 4

Output for the Sample Input

queue
not sure
impossible
stack
priority queue

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Rujia Liu's Present 3: A Data Structure Contest Celebrating the 100th Anniversary of Tsinghua University
Special Thanks: Yiming Li
Note: Please make sure to test your program with the gift I/O files before submitting!

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　　数据结构练习。

　　这道题的题目是“猜猜数据结构”，题意就是给你一些输入输出数据，让你根据这些数据判断是什么数据结构。要猜的数据结构只有三种，栈（stack）、队列（queue）、优先队列（priority_queue）。输出有5种情况，前三种分别是确定了一种数据结构，第四种是三种数据结构都不符合，第五种是有2种或2种以上符合。

　　思路就是在程序中定义这三种数据结构，根据输入数据，产生各自的输出结果，分别与给定的输出输出对比。如果与测试数据不同，则这种数据结构不可能。最后记录下符合的数据结构的个数，分情况判断输出即可。

　　代码：

 #include <iostream>
 #include <queue>
 #include <stack>
 using namespace std;
 int main()
 {
     int i,n;
     while(cin>>n){
         queue <int> q;
         priority_queue <int> pq;
         stack <int> s;
         bool f[] = {};
         for(i=;i<=n;i++){
             int a,b;
             cin>>a>>b;
             if(a==){    //入
                 q.push(b);
                 pq.push(b);
                 s.push(b);
             }
             else{    //出
                 //依次对比
                 if(!f[] && !q.empty()){
                     int x = q.front();
                     q.pop();
                     if(x!=b) f[]=true;
                 }
                 else f[]=true;

                 if(!f[] && !pq.empty()){
                     int x = pq.top();
                     pq.pop();
                     if(x!=b) f[]=true;
                 }
                 else f[]=true;

                 if(!f[] && !s.empty()){
                     int x = s.top();
                     s.pop();
                     if(x!=b) f[]=true;
                 }
                 else f[]=true;
             }
         }
         //查找有几个符合输出
         int num=;
         for(i=;i<;i++)
             if(!f[i])
                 num++;
         if(num==)
             cout<<"impossible"<<endl;
         else if(num==){
             if(!f[])
                 cout<<"queue"<<endl;
             else if(!f[])
                 cout<<"priority queue"<<endl;
             else if(!f[])
                 cout<<"stack"<<endl;
         }
         else
             cout<<"not sure"<<endl;
     }
     return ;
 }

Freecode : www.cnblogs.com/yym2013