• UVA 11995


    I Can Guess the Data Structure!

    Time limit: 1.000 seconds

    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
    

     考查基本数据结构定义,实现判断即可。

    /*
    * @author  Panoss
    */
    #include<iostream>
    #include<cstdio>
    #include<cstring>
    #include<string>
    #include<algorithm>
    #include<vector>
    #include<ctime>
    #include<stack>
    #include<queue>
    #include<list>
    using namespace std;
    #define DBG 0
    #define fori(i,a,b) for(int i = (a); i < (b); i++)
    #define forie(i,a,b) for(int i = (a); i <= (b); i++)
    #define ford(i,a,b) for(int i = (a); i > (b); i--)
    #define forde(i,a,b) for(int i = (a); i >= (b); i--)
    #define forls(i,a,b,n) for(int i = (a); i != (b); i = n[i])
    #define mset(a,v) memset(a, v, sizeof(a))
    #define mcpy(a,b) memcpy(a, b, sizeof(a))
    #define dout  DBG && cerr << __LINE__ << " >>| "
    #define checkv(x) dout << #x"=" << (x) << " | "<<endl
    #define checka(array,a,b) if(DBG) { 
        dout << #array"[] | " << endl; 
        forie(i, a, b) cerr << "[" << i << "]=" << array[i] << " |" << ((i - (a)+1) % 5 ? " " : "
    "); 
    if (((b)-(a)+1) % 5) cerr << endl; 
    }
    #define redata(T, x) T x; cin >> x
    #define MIN_LD -2147483648
    #define MAX_LD  2147483647
    #define MIN_LLD -9223372036854775808
    #define MAX_LLD  9223372036854775807
    #define MAX_INF 18446744073709551615
    inline int  reint() { int d; scanf("%d", &d); return d; }
    inline long relong() { long l; scanf("%ld", &l); return l; }
    inline char rechar() { scanf(" "); return getchar(); }
    inline double redouble() { double d; scanf("%lf", &d); return d; }
    inline string restring() { string s; cin >> s; return s; }
    
    struct Cmd
    {
        int cmd;
        int x;
    };
    
    vector<Cmd> C;
    
    bool is_stack()
    {
        stack<int> S;
        fori(i,0,C.size())
        {
            if(C[i].cmd == 1)
                S.push(C[i].x);
            else
            {
                if(S.empty()) return false;
                int value = S.top();
                S.pop();
                if(value != C[i].x) return false;
            }
        }
        return true;
    }
    
    bool is_queue()
    {
        queue<int> Q;
        fori(i,0,C.size())
        {
            if(C[i].cmd == 1)
                Q.push(C[i].x);
            else
            {
                if(Q.empty()) return false;
                int value = Q.front();
                Q.pop();
                if(value != C[i].x) return false;
            }
        }
        return true;
    }
    
    bool is_pri_queue()
    {
        priority_queue<int> Q;
        fori(i,0,C.size())
        {
            if(C[i].cmd == 1)
                Q.push(C[i].x);
            else
            {
                if(Q.empty()) return false;
                int value = Q.top();
                Q.pop();
                if(value != C[i].x) return false;
            }
        }
        return true;
    }
    
    int main()
    {
        int n;
        while(scanf("%d",&n)==1)
        {
            C.clear();
            forie(i,1,n)
            {
                Cmd command;
                scanf("%d%d",&command.cmd,&command.x);
                C.push_back(command);
            }
            bool is_st = is_stack();
            bool is_qu = is_queue();
            bool is_pri_qu = is_pri_queue();
            if(!is_st&&!is_qu&&!is_pri_qu) puts("impossible");
            else if(is_st&&is_qu&&is_pri_qu) puts("not sure");
            else if((is_st&&is_qu&&!is_pri_qu)||(is_st&&!is_qu&&is_pri_qu)||(!is_st&&is_qu&&is_pri_qu)) puts("not sure");
            else if(is_st&&!is_qu&&!is_pri_qu) puts("stack");
            else if(!is_st&&is_qu&&!is_pri_qu) puts("queue");
            else if(!is_st&&!is_qu&&is_pri_qu) puts("priority queue");
        }
    
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/Panoss/p/3815461.html
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