Pop Sequence栈的顺序实现

问题描述：

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

分析：

对于一个数组，判断他是否符合栈先进后出的顺序：数组元素从左到右，当前元素可以出栈的条件为：一、可以进栈 二、比他小的元素已经按顺序入栈。

代码：

#include<stdio.h>
#include<stdlib.h>

#define Max_Size 1000			//栈的存储空间初始分配量


typedef struct node{
	int top;					//指示栈最上面的元素 
	int cap;					//栈的空间 
	int Data[Max_Size];			//顺序栈 
}Stack;

int M, N, K;					//M:栈的容量 N：待判断数组元素个数 K：须要被检查的数组个数 
								//全局变量，做用域为文件做用域 

Stack *InitStack()				//	构造一个空栈 
{
	Stack *PtrS = (Stack *)malloc(sizeof(struct node));//sizeof(struct node) 
	PtrS -> top = -1;			//top初始值为-1表示空栈 
	PtrS -> cap = Max_Size; 
	return PtrS; 
} 

int Top(Stack* PtrS)			//获取栈顶元素 
{
	return PtrS -> Data[PtrS -> top];
}

void Push(Stack*PtrS, int ele)	//将元素ele压入栈中 
{
	if(PtrS -> top == PtrS -> cap-1){	//栈满 
		printf("FULL\n");
		return;
	}	
	PtrS -> top++;				//栈顶指针上移 
	PtrS -> Data[PtrS -> top] = ele;
} 

void Pop(Stack *PtrS)
{
	if(PtrS -> top==-1){		//栈空 
		printf("Empty\n");
		return; 
	} 
	PtrS -> top--;
} 

int Check_Stack(int v[])		//对于给定的数组，检查数字顺序是否符合栈的规则 
{
	int new_cap = M+1;			//栈的容量为M+1 
	Stack *ps = InitStack();	//新建一个空栈。对于数组中的一个元素，出栈的充分条件为：一、可以进栈 二、它前面的元素已经入栈 
	int idx = 0;				//标记被检查数组元素下标
	Push(ps, 0);				//初始元素	
	int num = 1;
			
	while(idx != N){			//直到检查到最后一个元素，退出循环 
		while(idx!=N && (ps->top+1)<new_cap && Top(ps) < v[idx])//一、还有待检查元素 二、栈中元素小于栈的容量 三、栈顶元素小于待检查元素 
			Push(ps, num++);	//将小于待检查元素的数字压入栈中，以后数字加一 
		if(Top(ps) == v[idx]){  //if(Data[top] == v[idx]) 
			Pop(ps);
			//printf("Pass");			//该元素经过检查 
			idx++; 
		}
		else
			return 0;
	}
	return 1; 
}

int main()
{
	scanf("%d %d %d", &M, &N, &K);
	int *v =(int *)malloc(sizeof(int)*N);//开辟一个N个int数组的空间，v指向数组首地址
	int i;
	for(;K!=0;--K){				//k--
		for(i=0; i != N; ++i)	//i<N
			scanf("%d",v+i);
		if(Check_Stack(v))
			printf("YES\n");
		else
			printf("NO\n");
	}	
	return 0;
}