您的位置:首页 > 产品设计 > UI/UE

UVA 11995 I Can Guess the Data Structure!

2014-04-03 15:44 696 查看
Description





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

#include<stack>
#include<queue>
using namespace std;
const int maxn = 1000+100;
int id[maxn],x[maxn],n;
bool isStack()
{
stack<int> s;
for(int i=0;i<n;i++)
{
if(id[i]==1) s.push(x[i]);
else
{
if(s.empty())  return false;
int val=s.top();  s.pop();
if(x[i]!=val) return false;
}
}
return true;
}
bool isQueue()
{
queue<int >q;
for(int i=0;i<n;i++)
{
if(id[i]==1) q.push(x[i]);
else
{
if(q.empty()) return false;
int val=q.front();  q.pop();
if(x[i]!=val) return false;
}
}
return true;
}
bool isPriority()
{
priority_queue<int > q;
for(int i=0;i<n;i++)
{
if(id[i]==1)
q.push(x[i]);
else
{
if(q.empty())
return false;
int val=q.top();
q.pop();
if(x[i]!=val) return false;
}
}
return true;
}
int main(){
while(scanf("%d",&n)!=EOF)
{
bool st=false,qu=false,pr=false;
for(int i=0;i<n;i++)
{
scanf("%d %d",&id[i],&x[i]);
}
st=isStack();
qu=isQueue();
pr=isPriority();
if(!st&&!qu&&!pr)
puts("impossible");
else if((!st&&qu&&pr)||(!qu&&st&&pr)||(!pr&&qu&&st)||pr&&qu&&st)
{
puts("not sure");
}
else if(st) puts("stack");
else if(qu) puts("queue");
else if(pr) puts("priority queue");
}
return 0;
}



                                            
内容来自用户分享和网络整理,不保证内容的准确性,如有侵权内容,可联系管理员处理 点击这里给我发消息
标签: