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  • A1051. Pop Sequence (25)

    Given a stack which can keep M numbers at most. Push N numbers in the order of 1, 2, 3, ..., N and pop randomly. You are supposed to tell if a given sequence of numbers is a possible pop sequence of the stack. For example, if M is 5 and N is 7, we can obtain 1, 2, 3, 4, 5, 6, 7 from the stack, but not 3, 2, 1, 7, 5, 6, 4.

    Input Specification:

    Each input file contains one test case. For each case, the first line contains 3 numbers (all no more than 1000): M (the maximum capacity of the stack), N (the length of push sequence), and K (the number of pop sequences to be checked). Then K lines follow, each contains a pop sequence of N numbers. All the numbers in a line are separated by a space.

    Output Specification:

    For each pop sequence, print in one line "YES" if it is indeed a possible pop sequence of the stack, or "NO" if not.

    Sample Input:

    5 7 5
    1 2 3 4 5 6 7
    3 2 1 7 5 6 4
    7 6 5 4 3 2 1
    5 6 4 3 7 2 1
    1 7 6 5 4 3 2
    

    Sample Output:

    YES
    NO
    NO
    YES
    NO
    
    #include <stdio.h>
    #include <stdlib.h>
    #include <iostream>
    #include <string.h>
    #include <math.h>
    #include <algorithm>
    #include <string>
    #include <stack> 
    using namespace std;
    
    
    int main(){
        stack<int> st;
        int m,n,k;
        scanf("%d %d %d",&m,&n,&k);
        int a[n];
        while(k--)
        {
            //初始化栈 
            while(!st.empty())
            {
                st.pop();
            } 
            for(int i=0;i<n;i++)
            {
                scanf("%d",&a[i]);
            }
            int current=0;
            bool flag=true;
            for(int i=1;i<=n;i++)
            {
                st.push(i);
                if(st.size()>m)
                {
                    flag=false;
                    break;
                }
                while(!st.empty()&&st.top()==a[current])
                {
                    st.pop();
                    current++;
                }
            }
            
            if(st.empty()==true&&flag==true)
            printf("YES
    ");
            else
            printf("NO
    ");
            
        }
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/ligen/p/4313317.html
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