UVA 11995 I Can Guess the Data Structure
UVA 11995 I Can Guess the Data Structure!
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
4Universidad de Valladolid OJ: 11995 – I Can Guess the Data Structure! 2/2
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
11995
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).
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
4Universidad de Valladolid OJ: 11995 – I Can Guess the Data Structure! 2/2
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
/* Author: 2486 Memory: 0 KB Time: 18 MS Language: C++ 4.8.2 Result: Accepted */ //将所有的STL分别进行判断就可以了 #include <iostream> #include <string> #include <algorithm> #include <stack> #include <queue> #include <cstdio> #include <cstring> using namespace std; int n,cmd,dig; bool isstack,isque,isprique; stack<int>sta; queue<int>que; priority_queue<int>prique; int main() { #ifndef ONLINE_JUDGE freopen("D://imput.txt","r",stdin); #endif // ONLINE_JUDGE while(~scanf("%d",&n)) { while(!sta.empty())sta.pop(); while(!que.empty())que.pop(); while(!prique.empty())prique.pop(); isstack=true,isque=true,isprique=true; for(int i=0; i<n; i++) { scanf("%d%d",&cmd,&dig); if(isstack) { if(cmd==1) { sta.push(dig); } else { if(sta.empty()) { isstack=false; while(!sta.empty())sta.pop(); } else { int v=sta.top(); if(v==dig) { sta.pop(); } else { isstack=false; while(!sta.empty())sta.pop(); } } } } if(isque) { if(cmd==1) { que.push(dig); } else { if(que.empty()) { isque=false; while(!que.empty())que.pop(); } else { int v=que.front(); if(v==dig) { que.pop(); } else { isque=false; while(!que.empty())que.pop(); } } } } if (isprique) { if(cmd==1) { prique.push(dig); } else { if(prique.empty()) { isprique=false; while(!prique.empty())prique.pop(); } else { int v=prique.top(); if(v==dig) { prique.pop(); } else { isprique=false; while(!prique.empty())prique.pop(); } } } } } if(isstack+isque+isprique==1) { if(isstack)printf("stack"); else if(isque)printf("queue"); else printf("priority queue"); } else if(isstack+isque+isprique==0) { printf("impossible"); } else { printf("not sure"); } printf("\n"); } return 0; }