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 ;
}
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