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：

 1 #include <iostream>

 2 #include <queue>

 3 #include <stack>

 4 using namespace std;  5 int main()  6 {  7     int i,n;  8     while(cin>>n){  9         queue <int> q; 10         priority_queue <int> pq; 11         stack <int> s; 12         bool f[3] = {0}; 13         for(i=1;i<=n;i++){ 14             int a,b; 15             cin>>a>>b; 16             if(a==1){    //入

17  q.push(b); 18  pq.push(b); 19  s.push(b); 20  } 21             else{    //出 22                 //依次对比

23                 if(!f[0] && !q.empty()){ 24                     int x = q.front(); 25  q.pop(); 26                     if(x!=b) f[0]=true; 27  } 28                 else f[0]=true; 29                 

30                 

31                 if(!f[1] && !pq.empty()){ 32                     int x = pq.top(); 33  pq.pop(); 34                     if(x!=b) f[1]=true; 35  } 36                 else f[1]=true; 37 

38                 if(!f[2] && !s.empty()){ 39                     int x = s.top(); 40  s.pop(); 41                     if(x!=b) f[2]=true; 42  } 43                 else f[2]=true; 44  } 45  } 46         //查找有几个符合输出

47         int num=0; 48         for(i=0;i<3;i++) 49             if(!f[i]) 50                 num++; 51         if(num==0) 52             cout<<"impossible"<<endl; 53         else if(num==1){ 54             if(!f[0]) 55                 cout<<"queue"<<endl; 56             else if(!f[1]) 57                 cout<<"priority queue"<<endl; 58             else if(!f[2]) 59                 cout<<"stack"<<endl; 60  } 61         else 

62             cout<<"not sure"<<endl; 63  } 64     return 0; 65 }

 

Freecode : www.cnblogs.com/yym2013