1051. Pop Sequence (25)

秒速五厘米 2022-05-31 12:07 169阅读 0赞

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

--------------------

题目大意：

给定一个栈，他最多可以保留M个数字。按1, 2, 3, ..., N的顺序进栈并随机出栈。你的任务是判断一个给定的数字序列是否是正确的出栈顺序。例如，如果M是5N是7，我们可以从堆栈中的到1, 2, 3, 4, 5, 6, 7，但不是3, 2, 1, 7, 5, 6, 4。  
输入规格：  
每个输入文件包含一个测试用例。对于每个测试用例，第一行包含3个数字（都不超过1000）：M（堆栈的最大容量），N（进栈队列的长度），K（要检查的出栈序列的数量）。然后是K行，每行包含一个有N个数字组成的序列。直线上得所有数字都用空格隔开。  
输出规范：  
对于每个打印序列，如果他确实是堆栈的一个可能的出栈序列，打印“YES”，否则打印“NO”。

代码：

#include<stdio.h>
    #include<stack>
    #include<algorithm>
    using namespace std;
    int main()
    {
        int i,j,n,m,k,t,l,flag;
        scanf("%d %d %d",&n,&m,&k);
        for(i=0;i<k;i++)
        {
            l=1;
            flag=0;
            stack<int> s;
            for(j=0;j<m;j++)
            {
    
                scanf("%d",&t);
                if(flag==0)
                {
                    while(s.empty()||s.top()!=t)
                    {
                        s.push(l);
                        if(s.size()>n)
                        {
                            flag=1;
                            break;
                        }
                        l++;
                    }
                    if(flag==0&&s.size()>=1&&s.top()==t)
                    {
                        s.pop();
                    }
                }
            }
            if(flag==0)
            {
                printf("YES\n");
            }
            else
            {
                printf("NO\n");
            }
        }
        return 0;
    }