PAT-A1051 Pop Sequence 题目内容及题解

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.

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

题目大意

题目给定了一个能容纳M个元素的栈和从1到N的入栈序列，要求判断题目给出的序列是否可能是给定栈的出栈序列。

解题思路

1. 模拟一个栈，逐个读取出栈序列，并作出判断，依次输出，并返回零值；
2. 给定栈空间可能为零（边界条件）；
3. 不论是否为出栈序列，必须将序列中所有数字读入，否则影响下一次判定。

代码

#include
#define maxn 1010
int main(){
    int N,M,K;
    int i;
    int stack[maxn],top,current,flag,num;
    scanf("%d%d%d",&M,&N,&K);
    stack[M]=-1;
    while(K--){
        top=-1;
        flag=1;
        num=1;
        if(M>0){
            stack[++top]=1;
        }else{
            printf("NO\n");
        }
        for(i=0;i

运行结果