zoukankan      html  css  js  c++  java
  • UVA-11995

    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 isa stack (Last-In, First-Out), a queue (First-In, First-Out), a priority-queue (Always take out largerelements first) or something else that you can hardly imagine!InputThere 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 xis always a positive integer not larger than 100. The input is terminated by end-of-file (EOF).OutputFor 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 mentionedabove.Sample Input61 11 21 32 12 22 361 11 21 32 32 22 121 12 241 21 12 12 271 21 51 11 32 51 42 4Sample Outputqueuenot sureimpossiblestackpriority queue


    题解:分别定义 stack、queue、priority_queue 判断这个操作序列是不是符合上述结构。

    AC代码为:


    #include<stdio.h>  
    #include<stack>  
    #include<queue>  
    using namespace std;  
    int main()  
    {  
        int n, i, x, y, f[4];  
        while(~scanf("%d",&n))  
        {  
            stack<int> s;  
            queue<int> q;  
            priority_queue<int, vector<int>, less<int> > pq;  
            for(i=0;i<3;i++) 
    f[i]=1;  
            for(i=0;i<n;i++)  
            {  
                scanf("%d%d",&x,&y);  
                if(x == 1)  
                {  
                    s.push(y);  
                    q.push(y);  
                    pq.push(y);  
                }  
                else  
                {  
                    if(!s.empty())  
                    {  
                        if(s.top() != y)  
                            f[0] = 0;  
                        s.pop();  
                    }  
                    else  
                        f[0] = 0;  
      
                    if(!q.empty())  
                    {  
                        if(q.front() != y)  
                            f[1] = 0;  
                        q.pop();  
                    }  
                    else  
                        f[1] = 0;  
      
                    if(!pq.empty())  
                    {  
                        if(pq.top() != y)  
                            f[2] = 0;  
                        pq.pop();  
                    }  
                    else  
                        f[2] = 0;  
                }  
            }  
            int num = 0;  
            for(i = 0; i < 3; i++)  
                if(f[i] == 1)  
                    num++;  
            if(num == 0)  
                printf("impossible ");  
            else if(num > 1)  
                printf("not sure ");  
            else  
            {  
                if(f[0] == 1)  
                    printf("stack ");  
                else if(f[1] == 1)  
                    printf("queue ");  
                else  
                    printf("priority queue ");  
            }  
        }  
        return 0;  
    }  


  • 相关阅读:
    (笔试题)镇长选举
    (笔试题)最小的非“重复的数”
    ( 笔试题)只出现一次的数
    (算法)二叉树中两个结点的最近公共父结点
    (笔试题)区间最大重叠
    (剑指Offer)面试题61:按之字形顺序打印二叉树
    (算法)Partition方法求数组第k大的数
    (剑指Offer)面试题60:把二叉树打印成多行
    整理一些不错的、网上好评的电影、电视、视频等资源地址
    个人网站/博客,建站好的域名和网站供应商网站整理
  • 原文地址:https://www.cnblogs.com/csushl/p/9386628.html
Copyright © 2011-2022 走看看