B - I Can Guess the Data Structure!（建议使用栈队列和优先队列来模拟）

Posted 2021-02-07 liufei-

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

#include <iostream>
#include <queue>
#include <stack>
#include<cstdio>


using namespace std;

int main()
{
    int n;
    int i,x,y;
    while(cin>>n)
    {
        stack<int>s;
        queue<int>q;
        priority_queue<int>pq;


        int flag=0;     //判断输入符合几种情况
        int a[3]={1,1,1};   //三个数组元素，分别对应三种数据结构
        for(i=0;i<n;i++)
        {
            cin>>x>>y;
            if(x==1)     //将输入的数字分别放进三种数据结构里
            {
                s.push(y);
                q.push(y);
                pq.push(y);
            }
            else    //x=2的情况，也就是将每种数据结构元素拿出来和输入的数进行比较
            {
                if(!q.empty()&&a[0]==1)  //不为空且a=1
                {
                    int b=q.front();   //判断
                    q.pop();
                    if(y!=b)   //不相同，把a置零反之a=1
                        a[0]=0;
                }
                else a[0]=0;  //为空的情况a也是等于1


                if(!s.empty()&&a[1]==1)
                {
                    int b=s.top();
                    s.pop();
                    if(y!=b)
                        a[1]=0;
                }
                else a[1]=0;


                if(!pq.empty()&&a[2]==1)
                {
                    int b=pq.top();
                    pq.pop();
                    if(y!=b)
                        a[2]=0;
                }
                else a[2]=0;


            }
        }


            for(i=0;i<3;i++)  //计算有多少种数据结构满足输入条件
            {
                if(a[i]==1)
                    flag++;
            }



        if(flag==0)
                cout<<"impossible"<<endl;
            else if(flag==1)
            {
                if(a[0]==1)
                    cout<<"queue"<<endl;
                else if(a[1]==1)
                    cout<<"stack"<<endl;
                else if(a[2]==1)
                    cout<<"priority queue"<<endl;
            }
            else if(flag>1)
                cout<<"not sure"<<endl;
    }
    return 0;
}

大概思路：1的数字输入完之后，进入2数字判断，y与每种数据结构出来的元素进行判断。

顺，别忘记判断为空的情况