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
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:有一个类似“包包”的数据结构,支持两种操作。1x:把元素x放进包包;2:从包包中拿出一个元素。给出一系列的操作以及返回值,你的任务是猜猜这个“包包”到底是什么。它可能是一个栈,队列,优先队列或者其他什么奇怪的东西。 分析:只需要依次判断输入是否可能是栈,队列或优先队列,然后中合起来即可。注意到题目中说的“无错的返回”,因此在执行pop操作的时候要调用一下empty(),否则可能会异常退出。 代码如下:
#include<cstdio>
#include<queue>
#include<stack>
#include<cstdlib>
using namespace std; const int maxn = + ;
int n, t[maxn], v[maxn]; int check_stack() {
stack<int> s;
for(int i = ; i < n; i++) {
if(t[i] == ) {
if(s.empty()) return ;
int x = s.top(); s.pop();
if(x != v[i]) return ;
}
else s.push(v[i]);
}
return ;
} int check_queue() {
queue<int> s;
for(int i = ; i < n; i++) {
if(t[i] == ) {
if(s.empty()) return ;
int x = s.front(); s.pop();
if(x != v[i]) return ;
}
else s.push(v[i]);
}
return ;
} int check_pq() {
priority_queue<int> s;
for(int i = ; i < n; i++) {
if(t[i] == ) {
if(s.empty()) return ;
int x = s.top(); s.pop();
if(x != v[i]) return ;
}
else s.push(v[i]);
}
return ;
} int main() {
while(scanf("%d", &n) == ) {
for(int i = ; i < n; i++) scanf("%d%d", &t[i], &v[i]);
int s = check_stack();
int q = check_queue();
int pq = check_pq();
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 I Can Guess the Data Structure!(ADT)的更多相关文章
- 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!(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! [STL应用]
11995 - I Can Guess the Data Structure! Time limit: 1.000 seconds Problem I I Can Guess the Data Str ...
- 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. ...
- 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!
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 ...
- 高级数据结构学习笔记 / Data Structure(updating)
树状数组 查询操作:O(logn) 修改操作:O(logn) #define lowbit(x) (x & -x) int tr[N]; // 树状数组 // 添加c个大小为x的数值 vo ...
- uva-11995 - I Can Guess the Data Structure!(栈,优先队列,队列,水题)
11995 - I Can Guess the Data Structure! There is a bag-like data structure, supporting two operation ...
随机推荐
- java 小结3 hashcode和equals I/o问题
我需要把星期天看的一些东西记录下来,要不然会忘记. hashCode.equals: 1)每个java对象都有hashCode和equals方法. java的终极类是object类,那么object类 ...
- 【Java基础】一个有意思的泛型方法Arrays.asList(T... a)
总结 利用Arrays.asList方法返回的List是不允许add和remove的,这种list的长度不可变,因为底层依然是写数组. Arrays.asList的返回值是调用是传入T类型的List, ...
- 伪分布配置完成启动jobtracker和tasktracker没有启动
检查logs目录下的hadoop-root-jobtracker日志文件 2014-02-26 19:56:06,782 FATAL org.apache.hadoop.mapred.JobTrack ...
- Robocopy是微软Windows Server 2003资源工具包中众多多用途的实用程序之一(它是基于强大的拷贝程序
Robocopy是微软Windows Server 2003资源工具包中众多多用途的实用程序之一(它是基于强大的拷贝程序).没错,Robocopy的功能是拷贝文件,你也许会觉得无聊并且要翻阅下一篇文章 ...
- A Tour of Go Arrays
The type [n]T is an array of n values of type T. The expression var a [10]int declares a variable a ...
- mongodb的查询操作符
本文地址:http://www.cnblogs.com/egger/archive/2013/05/04/3059374.html 欢迎转载 ,请保留此链接! 官方参考: http://docs. ...
- snowflake算法(java版)
转自:http://www.cnblogs.com/haoxinyue/p/5208136.html 1. 数据库自增长序列或字段 最常见的方式.利用数据库,全数据库唯一. 优点: 1)简单,代码方 ...
- 安装J2EE的SDK报错:could not find the required version of the Java(TM)2 Runtime Environment in '(null)'的解决办法。
国内私募机构九鼎控股打造APP,来就送 20元现金领取地址:http://jdb.jiudingcapital.com/phone.html内部邀请码:C8E245J (不写邀请码,没有现金送)国内私 ...
- android访问asset目录下的资源
android提供了AssetManager来访问asset目录下的资源, 在activity中通过getAssets()获取AssetManager 常用的api如下: 1.列举路径下的资源Stri ...
- oracle并行模式(Parallel)
1. 用途 强行启用并行度来执行当前SQL.这个在Oracle 9i之后的版本可以使用,之前的版本现在没有环境进行测试.也就是说,加上这个说明,可以强行启用Oracle的多线程处理功能.举例的话,就 ...