版权声明：抱最大的希望，为最大的努力，做最坏的打算。 https://blog.csdn.net/qq_37748451/article/details/86604680
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题意：
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。
主要想通过这个题复习一下常用的数据结构；
栈：后进先出，只允许在一端进行操作
队列：先进先出，类似于排队
优先队列：基于堆的完全二叉树 用来插入元素或者删除最大元素来使用的
#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[3] = {0};
for(i=1;i<=n;i++){
int a,b;
cin>>a>>b;
if(a==1){ //入
q.push(b);
pq.push(b);
s.push(b);
}
else{ //出
//依次对比
if(!f[0] && !q.empty()){
int x = q.front();
q.pop();
if(x!=b) f[0]=true;
}
else f[0]=true;
if(!f[1] && !pq.empty()){
int x = pq.top();
pq.pop();
if(x!=b) f[1]=true;
}
else f[1]=true;
if(!f[2] && !s.empty()){
int x = s.top();
s.pop();
if(x!=b) f[2]=true;
}
else f[2]=true;
}
}
//查找有几个符合输出
int num=0;
for(i=0;i<3;i++)
if(!f[i])
num++;
if(num==0)
cout<<"impossible"<<endl;
else if(num==1){
if(!f[0])
cout<<"queue"<<endl;
else if(!f[1])
cout<<"priority queue"<<endl;
else if(!f[2])
cout<<"stack"<<endl;
}
else
cout<<"not sure"<<endl;
}
return 0;
}