11995 - I Can Guess the Data Structure! uva

最新推荐文章于 2019-12-08 22:31:42 发布

原创最新推荐文章于 2019-12-08 22:31:42 发布 · 451 阅读

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本文深入探讨了数据结构（如栈、队列、优先队列）及其操作原理，通过实例展示了如何通过序列操作来推断数据结构类型，并强调了正确处理残留数据的重要性。

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.

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

Output for the Sample Input

queue
not sure
impossible
stack
priority queue

本题考查了栈，队列，和优先队列3种ADT的概念，可直接调用STL即可

注意本题容易出错点为：每次判断完是否属于某种结构后，要对栈或队列清空，否则残留的数据将会对下次运算结果产生影响。

#include<iostream>
#include<cstring>
#include<queue>
#include<algorithm>
#include<cstdlib>
#include<cstdio>
#include<vector>
#include<stack>
using namespace std;
int main()
{
    int n,m;
    while(scanf("%d",&n)!=EOF){
        int a[1001],b[1001],t;
        bool flagv=false,flagq=false,flagp=false;
        queue<int> q;
        stack<int> v;
        priority_queue<int> p;
        for(int i = 0 ; i < n ; i++)
            scanf("%d %d",&a[i],&b[i]);
        for(int i = 0 ; i < n ; i++){
            if(1==a[i]){
                v.push(b[i]);
            }
            else {
                if(v.empty()!=true){
                    t=v.top();
                    v.pop();
                }
                else flagv=true;
                if(t != b[i]){
                    flagv=true;
                    break;
                }
            }
        }
        for(int i = 0 ; i < n ; i++){
            if(1==a[i])
                q.push(b[i]);
            else {
                if(q.empty()!=true){
                    t=q.front();
                    q.pop();
                }
                else flagq=true;
                if(t != b[i]){
                    flagq=true;
                    break;
                }
            }
        }
        for(int i = 0 ; i < n ; i++){
            if(1==a[i])
                p.push(b[i]);
            else {
                if(p.empty()!=true){
                    t=p.top();
                    p.pop();
                }
                else flagp=true;
                if(t != b[i]){
                    flagp=true;
                    break;
                }
            }
        }
        if(!flagv&&flagq&&flagp) {
          printf("stack\n");
          continue;
        }
        if(flagv&&!flagq&&flagp) {
          printf("queue\n");
          continue;
        }
        if(!flagp&&flagv&&flagq) {
          printf("priority queue\n");
          continue;
        }
        if(flagv && flagp && flagq)
            printf("impossible\n");
        else
            printf("not sure\n");
    }
    return 0;
}