1051. Pop Sequence (25)

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

这个题刚开始完全没有思路，后来看了别人写的博客，发现此题并不难。打个比喻，如果要检查一辆车是否合格，肯定是每个零件都要检查一遍（这个比喻不一定恰当），同样，检查一个序列是否可由条件生成，那么就检查一下这个序列的每个元素是否可生成。所以问题划归为检查一个数。

算法的定义是有限个有序的步骤解决一个问题，所以很多时候，走出第一步就成功了一般。

具体代码如下：

/*************************************************************************
> File Name: 1051.c
> Author:
> Mail:
> Created Time: 2016年12月22日 星期四 23时39分48秒
************************************************************************/

#include<stdio.h>
#include <malloc.h>

int main()
{
int M, N, K;
int popEle, pushEle;
int i, j, k = 1;
int rear = -1;
int flag = 1;

scanf("%d%d%d", &M, &N, &K);
int* Stack = (int*)malloc(M * sizeof(int));

for (i = 0; i < K; i++) {
flag = 1;
rear = -1;
k = 1;
for (j = 0; j < N; j++) { //inspect j_th element
scanf("%d", &popEle);
if (rear != -1 && Stack[rear] == popEle) { //find it in the stack
rear -= 1;
} else { //find in the numbers not pushed into stack
while (k != popEle && k <= N && rear < M-2) { //rear < M-2, not M-1
Stack[++rear] = k;
++k;
}
if (k != popEle) {
flag = 0;
} else {
++k;
}
}
}
if (flag == 1) {
printf("YES\n");
} else {
printf("NO\n");
}
}
return 0;
}

编程之路，进步，再进一步！