UVA 11995 STL 使用

There is a bag-like data structure, supporting two operations:

1 x

Throw an element x into the bag.

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

Output for the Sample Input

queue
not sure
impossible
stack
priority queue

题意:
判断是哪一种数据结构
题解：
STL <queue> <stack> priority_queue

 1 #include<iostream>
 2 #include<cstring>
 3 #include<cstdio>
 4 #include<algorithm>
 5 #include<queue>
 6 #include<stack>
 7 #include<map>
 8 #include<vector>
 9 using namespace std;
10 queue<int> q;
11 stack<int> s;
12 struct ss{
13   int f;    
14 friend bool operator < (ss a, ss b)
15 {return  a.f < b.f;}
16 };
17 priority_queue<ss> pq;
18 int aa,bb;
19 int n;
20 int ans,ans1,ans2,ans3;
21 int cun[1005][2];
22 int main()
23 {
24     while(scanf("%d",&n)!=EOF)
25 {
26     ans=0;ans1=0;ans2=0;ans3=0;
27     while(q.size()>0)
28     q.pop();
29     while(s.size()>0)
30     s.pop();
31     while(pq.size()>0)
32     pq.pop();
33     memset(cun,0,sizeof(cun));
34     for(int i=1;i<=n;i++)
35         scanf("%d %d",&cun[i][0],&cun[i][1]);
36     for(int i=1;i<=n;i++)
37     {
38         int aa=cun[i][0];
39         int bb=cun[i][1];
40         if(aa==1)
41         {
42            q.push(bb);
43            s.push(bb);
44            struct ss gg;
45            gg.f=bb; 
46            pq.push(gg);    
47         } 
48         else
49         {
50             ans++;
51             if(!q.empty()&&bb==q.front())
52             { ans1++; q.pop();}
53             if(!s.empty()&&bb==s.top())
54             {ans2++;s.pop();}
55             if(!pq.empty()&&bb==pq.top().f)
56             {ans3++;pq.pop();}    
57         }
58     }
59     //cout<<ans<<" "<<ans1<<" "<<ans2<<" "<<ans3<<" "<<endl;
60     int qu=0,st=0,pri=0;
61     if(ans==ans1)
62     qu=1;
63     if(ans==ans2)
64     st=1;
65     if(ans==ans3)
66     pri=1;
67     if (st + qu + pri > 1) printf("not sure
");
68     else if (st) printf("stack
");
69     else if (qu) printf("queue
");
70     else if (pri) printf("priority queue
");
71     else printf("impossible
");
72 
73 }
74     return 0;
75 }
76 /*
77 3
78 1 1
79 1 2
80 2 1
81 6
82 1 1
83 1 2
84 1 3
85 2 1
86 2 2
87 2 3
88 6
89 1 1
90 1 2
91 1 3
92 2 3
93 2 2
94 2 1
95 */