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).

******************************************************************************************************************

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

******************************************************************************************************************

Sample Output

queue

not sure

impossible

stack

priority queue

******************************************************************************************************************

题意:根据操作,判断属于那种数据结构。

思路:使用STL中对应函数,进行模拟,若矛盾则不符合。

#include<stdio.h>
#include<queue>
#include<stack>
using namespace std;
int main()
{
int n;
while (~scanf("%d", &n))
{
stack<int>st;
queue<int>qu;
priority_queue<int>pr;
int a = 1, b = 1, c = 1;
for (int i = 0; i < n; i++)
{
int x, y;
scanf("%d %d", &x, &y);
if (x == 1)//向其中加入数据
{
st.push(y);
qu.push(y);
pr.push(y);
}
else
{
if (st.empty())
{
a = b = c = 0;
continue;
}
if (a)
a = (st.top() == y);//判断顶部的数是否与y相等
if (b)
b = (qu.front() == y);
if (c)
c = (pr.top() == y);
st.pop();
qu.pop();
pr.pop();
}
}
if (a + b + c > 1)
printf("not sure\n");
else if (a)
printf("stack\n");
else if (b)
printf("queue\n");
else if (c)
printf("priority queue\n");
else
printf("impossible\n");
} return 0;
}

UVA - 11995 - I Can Guess the Data Structure! STL 模拟的更多相关文章

  1. [UVA] 11995 - I Can Guess the Data Structure! [STL应用]

    11995 - I Can Guess the Data Structure! Time limit: 1.000 seconds Problem I I Can Guess the Data Str ...

  2. UVA - 11995 I Can Guess the Data Structure!(模拟)

    思路:分别定义栈,队列,优先队列(数值大的优先级越高).每次放入的时候,就往分别向三个数据结构中加入这个数:每次取出的时候就检查这个数是否与三个数据结构的第一个数(栈顶,队首),不相等就排除这个数据结 ...

  3. UVa 11995:I Can Guess the Data Structure!(数据结构练习)

    I Can Guess the Data Structure! There is a bag-like data structure, supporting two operations: 1 x T ...

  4. UVA 11995 I Can Guess the Data Structure!(ADT)

    I Can Guess the Data Structure! There is a bag-like data structure, supporting two operations: 1 x T ...

  5. STL UVA 11995 I Can Guess the Data Structure!

    题目传送门 题意:训练指南P186 分析:主要为了熟悉STL中的stack,queue,priority_queue,尤其是优先队列从小到大的写法 #include <bits/stdc++.h ...

  6. UVa 11995 I Can Guess the Data Structure!

    做道水题凑凑题量,=_=||. 直接用STL里的queue.stack 和 priority_queue模拟就好了,看看取出的元素是否和输入中的相等,注意在此之前要判断一下是否非空. #include ...

  7. uva 11995 I Can Guess the Data Structure stack,queue,priority_queue

    题意:给你n个操做,判断是那种数据结构. #include<iostream> #include<cstdio> #include<cstdlib> #includ ...

  8. HDU 5929 Basic Data Structure 【模拟】 (2016CCPC东北地区大学生程序设计竞赛)

    Basic Data Structure Time Limit: 7000/3500 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Oth ...

  9. HDU 5929 Basic Data Structure(模拟 + 乱搞)题解

    题意:给定一种二进制操作nand,为 0 nand 0 = 10 nand 1 = 1 1 nand 0 = 1 1 nand 1 = 0 现在要你模拟一个队列,实现PUSH x 往队头塞入x,POP ...

随机推荐

  1. jQuery 动态标签生成插件

    前言: 最近对js的插件封装特别感兴趣,无耐就目前的技术想做到js的完全封装,还是有一定困难,就基于jQuery封装了一个小的插件,而且是基于对象级开发的,不是添加全局方法.高深的语法几乎没有,就有一 ...

  2. Elasticsearch6.3 使用jdbc连接

    Elasticsearch6.3开始执行sql,可以和使用数据库一样的CRUD进行操作elasticsearch,连接过程如下(安装下载Elasticsearch略): 一:项目中添加maven依赖 ...

  3. APScheduler API -- apscheduler.schedulers.base

    apscheduler.schedulers.base API class apscheduler.schedulers.base.BaseScheduler(gconfig={}, **option ...

  4. Strusts2笔记9--防止表单重复提交和注解开发

    防止表单重复提交: 用户可能由于各种原因,对表单进行重复提交.Struts2中使用令牌机制防止表单自动提交.以下引用自北京动力节点:

  5. DirectFB简介以及移植[一]【转】

    转自:https://blog.csdn.net/wavemcu/article/details/39251805 ****************************************** ...

  6. 解决Win7&Win8 64位下Source Insight提示未完整安装的问题[转]

    转自:http://www.cnblogs.com/sixiweb/p/3421533.html 网上的破解版的注册表文件都是针对32位系统的,所以在64位系统里运行根本无法破解.下面分别贴出这俩系统 ...

  7. 两个Bounding Box的IOU计算代码

    Bounding Box的数据结构为(xmin,ymin,xmax,ymax) 输入:box1,box2 输出:IOU值 import numpy as np def iou(box1,box2): ...

  8. Focal Loss笔记

    论文:<Focal Loss for Dense Object Detection> Focal Loss 是何恺明设计的为了解决one-stage目标检测在训练阶段前景类和背景类极度不均 ...

  9. Visual Studio 2017 for Mac

    Visual Studio 2017 for Mac Last Update: 2017/6/16 我们非常荣幸地宣布 Visual Studio 2017 for Mac 现已推出. Visual ...

  10. 用Max导出Unity3D使用的FBX文件流程注解(转载)

    http://www.cnblogs.com/wantnon/p/4564522.html 从max导出FBX到Unity,以下环节需要特别注意.1,单位设置   很多人在建模,动画的时候,默认的ma ...