I Can Guess the Data Structure! C++STL

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最新推荐文章于 2021-10-31 23:34:59 发布

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分类专栏： ACMSTL ACM栈

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通过一系列操作及返回值，挑战者需猜测背后的数据结构是栈、队列还是优先队列，亦或是其他难以想象的结构。此游戏考验参与者对数据结构特性的理解与应用。

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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 y

Take out an element y 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

Output for the Sample Input

queue
not sure
impossible
stack
priority queue

#include<vector>
#include<cstdio>
#include<stack>
#include<queue>
using namespace std;

struct Data_Struct
{
    //vc存放指令pair序列
    vector<pair<int,int> > vc;

    //构造函数
    Data_Struct(int n)
    {
        for(int i=0;i<n;i++)
        {
            int x,y;
            scanf("%d%d",&x,&y);
            vc.push_back(make_pair(x,y));
        }
    }

    //判断是否可能是一个栈
    bool is_stack()
    {
        stack<int> S;
        for(int i=0;i<vc.size();i++)
        {
            int x=vc[i].first, y=vc[i].second;
            if(x==1) S.push(y);
            else if(x==2)
            {
                if(S.empty() || S.top()!=y ) return false;
                S.pop();
            }
        }
        return true;
    }

    //判断是否可能是一个队列
    bool is_queue()
    {
        queue<int> Q;
        for(int i=0;i<vc.size();i++)
        {
            int x=vc[i].first, y=vc[i].second;
            if(x==1) Q.push(y);
            else if(x==2)
            {
                if(Q.empty() || Q.front()!=y ) return false;
                Q.pop();
            }
        }
        return true;
    }

    //判断是否可能是一个优先队列
    bool is_priority_queue()
    {//默认谁大谁出
        priority_queue<int> Q;
        for(int i=0;i<vc.size();i++)
        {
            int x=vc[i].first,y=vc[i].second;
            if(x==1) Q.push(y);
            else
            {
                if(Q.empty() || Q.top()!=y )return false;
                Q.pop();//top()返回优先级最高的一个
            }
        }
        return true;
    }

    //输出整合后的结果
    void solve()
    {
        int S=is_stack(), Q=is_queue(), PQ=is_priority_queue();

        if(S+Q+PQ==0) printf("impossible\n");
        else if(S+Q+PQ==1)
        {
            if(S) printf("stack\n");
            else if(Q) printf("queue\n");
            else if(PQ) printf("priority queue\n");
        }
        else if(S+Q+PQ>1)
            printf("not sure\n");
    }
};

int main()
{
    int n;
    while(scanf("%d",&n)==1)
    {
        Data_Struct ds(n);

        ds.solve();
    }
    return 0;
}