uva 11995 判别数据类型

Problem I

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

题目大意:根据数据的操作<1>插入<2>删除判别是哪一种数据类型,有栈、队列、优先队列几种类型,如三种
都不符合输出impossbile,符合两种以上输出not sure,只符合一种的对应输出它的类型名称。

#include <iostream>
#include <queue>
#include <stack>
#include <cstdio>
#define Max 1010
using namespace std;

int N;
int op[Max],num[Max];
stack<int> S;
queue<int> Q;
priority_queue<int> PQ;
int check_stack()
{
    int i,t;
    while(!S.empty()) S.pop();
    for(i=0;i<N;i++)
    {
        if(op[i]==2)
        {
            if(S.empty()) return 0;
            t=S.top();S.pop();
            if(t!=num[i]) return 0;
        }
        else S.push(num[i]);
    }
    return 1;
}
int check_queue()
{
    int i,t;
    while(!Q.empty()) Q.pop();
    for(i=0;i<N;i++)
    {
        if(op[i]==2)
        {
            if(Q.empty()) return 0;
            t=Q.front();Q.pop();
            if(t!=num[i]) return 0;
        }
        else Q.push(num[i]);
    }
    return 1;
}
int check_priorityqueue()
{
    int i,t;
    while(!PQ.empty()) PQ.pop();
    for(i=0;i<N;i++)
    {
        if(op[i]==2)
        {
            if(PQ.empty()) return 0;
            t=PQ.top();PQ.pop();
            if(t!=num[i]) return 0;
        }
        else PQ.push(num[i]);
    }
    return 1;
}
int main()
{
    int i;
    int s,q,pq;
    while(scanf("%d",&N)!=EOF)
    {
        for(i=0;i<N;i++) scanf("%d %d",op+i,num+i);
        s=check_stack();
        q=check_queue();
        pq=check_priorityqueue();
        if(!s && !q && !pq) printf("impossible\n");
        else if(s && !q && !pq) printf("stack\n");
        else if(!s && q && !pq) printf("queue\n");
        else if(!s && !q && pq) printf("priority queue\n");
        else printf("not sure\n");
    }
    return 0;
}

 


posted on 2014-01-26 13:49  雄..  阅读(645)  评论(1编辑  收藏  举报

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