Pop Sequence（25 分）
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
AC代码：
//02-线性结构4 Pop Sequence（25 分）
#include<iostream>
#include<stdio.h>
using namespace std;
#define MAXSIZE 1000
typedef struct
{
int data[MAXSIZE];
int top;
}SqStack;
int InitStack(SqStack *s)
{
s->top=-1;
return 0;
}
int Push(SqStack *s,int e)
{
if(s->top==MAXSIZE-1)
return 1;
s->top++;
s->data[s->top]=e;
return 0;
}
int Pop(SqStack *s)
{
int e;
if(s->top==-1)
return 1;
e=s->data[s->top];
s->top--;
return e;
}
int main()
{
int m,n,k,i,h;
cin>>m>>n>>k;
SqStack s,t;
while(k--)
{
InitStack(&s);
InitStack(&t);
for(i=n-1;i>=0;i--)//注意i开始的位置，栈顶是最先出栈的元素
{
cin>>t.data[i];
}
t.top=n-1;
i=0;
while(i<=n)//出栈模拟
{
if(s.data[s.top]==t.data[t.top]&&s.top!=-1)//如果两个栈元素相同，说明压栈压到了该弹出的元素，弹出
{
Pop(&s);
Pop(&t);
}
else if(s.top<m-1&&i<n)//栈未满的情况下按顺序压栈
{
i++;
Push(&s,i);
}
else
break;
}
if(s.top==-1&&t.top==-1)//s栈能按t栈顺序，两个栈全部Pop空
cout<<"YES"<<endl;
else
cout<<"NO"<<endl;
}
return 0;
}
//注意栈T是反向存储的
//S栈按照从1开始的顺序压栈，反复的比较S和T的栈顶
//如果栈顶元素相同，就弹出，否则继续对S进行压栈（在栈未满的情况下）
//如果这个输入的pop序列是合法的，那么就能顺利的弹出两个栈的所有元素
//即两个栈的栈顶指针都指向-1；
//本题利用STL写会更简单，此处是为了熟悉stack结构