1051. Pop Sequence

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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 =;
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=;
int num=;
Stack*ps;
ps=CreateStack();
Push(ps,); //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+)
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 ;} //False!
}
free(ps);
return ; }
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=;i<K;i++)
{
for (j=;j<N;j++)
{
scanf("%d",v+j);
}
if(check_stack(v,N,M))printf("YES\n");
else printf("NO\n");
}
free(v);
return ; }