1051. Pop Sequence

原题连接：https://www.patest.cn/contests/pat-a-practise/1051

题目：

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

这道题我参阅了 http://blog.csdn.net/whzyb1991/article/details/46663867 这篇博文的思路，稍有改动，并且我是用链式存储堆栈，而博文的作者是用顺序存储的方式，读者可以对比两种
方式的优缺点。
这道题并不是有多复杂，关键在于想出it is not a possible pop sequence of the stack的条件！我是苦于没有思路，才参阅了博主的文章。共勉！
我的代码：

#include<stdio.h>
#include<stdlib.h>
#include<stdbool.h>
#define Max 1000
typedef struct SNode{
    int Data;
    struct SNode *Next;
}Stack;

Stack *CreateStack()
{
    Stack *Ptrs=(Stack*)malloc(sizeof(struct SNode));
    Ptrs->Next=NULL;
    return Ptrs;
}

bool IsEmpty(Stack *Ptrs)
{
    return(Ptrs->Next==NULL);
}

void Push(Stack *Ptrs,int X)
{
    Stack *S;
    S=(Stack *)malloc(sizeof(struct SNode));
    S->Data=X;
    S->Next=Ptrs->Next;
    Ptrs->Next=S;
}

void Pop(Stack *Ptrs)
{
    if(IsEmpty(Ptrs)) return;
    Stack *FirstCell;
    FirstCell=Ptrs->Next;
    Ptrs->Next=FirstCell->Next;
    free(FirstCell);
}
/* 计算堆栈的长度 */
int Length(Stack *Ptrs)
{
    Stack *p;
    p=Ptrs->Next;
    int cnt =0;
    while(p)
    {
        cnt++;
        p=p->Next;
    }
    return cnt;
}
/* 检验每一line是否符合要求 */
int check_stack(int *v,int N,int M)
{
    int i=0;
    int num=1;
    Stack*ps;
    ps=CreateStack();
    Push(ps,0);  //先给栈中压入一个元素0
    while(i<N)   //审核每个已给元素
    {  /* 核心 */  //入栈条件；当前栈顶元素小于已给数字，并且（前提条件）栈内元素的总容量小于栈的容量
        while(ps->Next->Data<v[i] && Length(ps)<M+1)
            Push(ps,num++);     //压入数num后，由于"the order is 1,2,……N。故下一个压入的数必为num+1;
        if (ps->Next->Data==v[i])
        {
            Pop(ps);   //栈顶元素出栈
            i++;       //之后的栈顶元素和下一个已给元素继续比较
        }
        else
        {
            free(ps);
            return 0;}   //False!
    }
    free(ps);
    return 1;

}
int main()
{
    int M,N,K;
    scanf("%d %d %d",&M,&N,&K);

    int *v=(int *)malloc(sizeof(int)*N);
    int i,j;
    for(i=0;i<K;i++)
    {
        for (j=0;j<N;j++)
        {
            scanf("%d",v+j);
        }
        if(check_stack(v,N,M))printf("YES
");
        else printf("NO
");
    }
    free(v);
    return 0;

}