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=;i<N;i++)
{
if(op[i]==)
{
if(S.empty()) return ;
t=S.top();S.pop();
if(t!=num[i]) return ;
}
else S.push(num[i]);
}
return ;
}
int check_queue()
{
int i,t;
while(!Q.empty()) Q.pop();
for(i=;i<N;i++)
{
if(op[i]==)
{
if(Q.empty()) return ;
t=Q.front();Q.pop();
if(t!=num[i]) return ;
}
else Q.push(num[i]);
}
return ;
}
int check_priorityqueue()
{
int i,t;
while(!PQ.empty()) PQ.pop();
for(i=;i<N;i++)
{
if(op[i]==)
{
if(PQ.empty()) return ;
t=PQ.top();PQ.pop();
if(t!=num[i]) return ;
}
else PQ.push(num[i]);
}
return ;
}
int main()
{
int i;
int s,q,pq;
while(scanf("%d",&N)!=EOF)
{
for(i=;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 ;
}
uva 11995 判别数据类型的更多相关文章
- [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 ...
- 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 ...
- 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 ...
- 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 ...
- 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 ...
- 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 ...
- 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 ...
- UVA - 11995 - I Can Guess the Data Structure! STL 模拟
There is a bag-like data structure, supporting two operations: 1 x Throw an element x into the bag. ...
- UVA - 11995 模拟
#include<iostream> #include<cstdio> #include<algorithm> #include<cstdlib> #i ...
随机推荐
- [习题]输入自己的生日(年/月/日)#2 -- 日历(Calendar)控件的时光跳跃,一次跳回五年、十年前?--TodaysDate属性、VisibleDate属性
原文出處 http://www.dotblogs.com.tw/mis2000lab/archive/2013/06/10/calendar_visibledate_birthday_dropdow ...
- 转载自infoq:MYSQL的集群方案
分布式MySQL集群方案的探索与思考 2016-04-29 张成远 “本文整理自ArchSummit微信大讲堂张成远线上群分享内容 背景 数据库作为一个非常基础的系统,任何一家互联网公司都会 ...
- UVA - 1658 Admiral (最小费用最大流)
最短路对应费用,路径数量对应流量.为限制点经过次数,拆点为边.跑一次流量为2的最小费用最大流. 最小费用最大流和最大流EK算法是十分相似的,只是把找增广路的部分换成了求费用的最短路. #include ...
- python基础一 day13 生成器
#生成器函数# def generator():# print(1)# return 'a'## ret = generator()# print(ret) #只要含有yield关键字的函数都是生成器 ...
- 如何在Mac上放大
您是否发现有时自己眯眼盯着屏幕,希望屏幕上的东西只是“大”一点?无论您是否视力差,或只是想放大屏幕来看近景,这是很容易做到,只需要按一些按键.这篇文章将告诉您如何放大看浏览器或桌面的特写. 方法 ...
- ubuntu 16.04 安装node.js 8.x
引自 https://www.digitalocean.com/community/tutorials/how-to-install-node-js-on-ubuntu-16-04#how-to-in ...
- servlet上传多个文件(乱码解决)
首先,建议将编码设置为GB2312,并在WEB-INF\lib里导入:commons-fileupload-1.3.jar和commons-io-2.4.jar, 可百度下下载,然后你编码完成后,上传 ...
- iOS7.1企业版发布后用户通过sarafi浏览器安装无效的解决方案
关于iOS7.1企业版发布后,用户通过sarafi浏览器安装无效的解决方案: 通过测试,已经完美解决. 方案一: iOS7.1企业应用无法安装应用程序 因为证书无效的解决方案 http://blog. ...
- (50)zabbix API二次开发使用与介绍
zabbix API开发库 zabbix API请求和响应都是json,并且还提供了各种语法的lib库,http://zabbix.org/wiki/Docs/api/libraries,包含php. ...
- mysqldump指令说明
3种形式mysqldump [OPTIONS] database [tables]mysqldump [OPTIONS] -B | --databases [OPTIONS] DB1 [DB2 DB3 ...