02-线性结构4 Pop Sequence （25 分）

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<iostream>
#include<stdio.h>
#include<algorithm>
#include<string.h>
int book[1001],a[1001];

int main()
{
    int m,n,k,count1=0,i,j,t,q,count2=0;
    scanf("%d%d%d",&m,&n,&k);
    while(k--)
    {
        for(i=0;i<n;i++)
        {
            scanf("%d",&a[i]);
        }
        for(i=0;i<n-1;i++)
        {
            if(a[i]>a[i+1])
            {
                for(q=1;q<a[i];q++)
                {
                    if(!book[q])
                        count2++;
                }
                if(count2>=m)
                {
                    count1=1;
                    break;
                }
                for(j=a[i]-1;j>a[i+1];j--)
                {
                    if(!book[j])
                    {
                        count1=1;
                        break;
                    }
                }
                book[a[i]]=1;
                count2=0;
            }
            else
            {
                for(q=1;q<a[i];q++)
                {
                    if(!book[q])
                    count2++;
                }
                for(j=a[i]+1;j<a[i+1];j++)
                {
                    if(!book[j])
                    {
                        count2++;
                    }
                }
                if(count2>=m)
                {
                    count1=1;
                    break;
                }
                book[a[i]]=1;
                count2=0;
            }
        }
        if(!count1)
                printf("YES\n");
        else
                printf("NO\n");
        count1=0;
        count2=0;
        memset(book,0,sizeof(book));
    }
    return 0;
}

来源：CSDN

作者：poptox

链接：https://blog.csdn.net/qq_41514794/article/details/82942993

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线性结构