出栈序列（Pop Sequence）详解

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.

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.

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

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

YES
NO
NO
YES
NO

我们以出栈序列元素为研究对象，一个一个判断之。先看出栈序列中第一个元素，第一个元素出栈，意味着小于它的整数曾经都已经入栈 ，也就是说出栈序列中下一个已出栈元素其值必然不小于栈顶元素（小于的都还在栈里面，不可能出栈了）所以不满足这个条件的出栈序列排除之；下一个已出栈元素其值大于栈顶元素，就让小于它的整数继续入栈，如果超出栈的容量大小，也排除之。如果所有元素都满足我们的要求，那么这个出栈序列就是正确的。

#include
using namespace std;

int main()
{
    int M,N,K;
    cin>>M>>N>>K;
    int input;
    int j;
    int arr[1000];//用来存放随机出栈序列
    //有K个需要判断的随机出栈序列
    for(int i=0;i>input;
            arr[i]=input;
        }
        //初始化一个堆栈
        int stack[1000];
        int top=-1;
        //栈中第一个元素设置为1
        int num=1;
        stack[++top]=num;
        //用来检查出栈序列中的每一个元素
        for(j=0;jstack[top])
            {
                stack[++top]=++num;
            }
            if(top>=M)
                break;
            if(stack[top]==arr[j])
                    top--;//注意此时num的值是没有减少的，意味着小于下次入栈元素的整数不包括已经出栈的数
            //应对出栈序列是栈底的情况
            if(top<0)
                stack[++top]=++num;
        }
        if(j==N)
            cout<<"YES"<