# 02-线性结构4 Pop Sequence

### 题目描述

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

### 代码实现

``````#include <stdio.h>
#include <stdbool.h>

typedef struct {
int data[1000];
int count;
} MyStack;

MyStack stack;

// 建立栈
void CreateStack()
{
stack.count = 0;
}

// Push
void PushStack(int num)
{
stack.data[stack.count] = num;
stack.count++;
}

// Pop
void PopStack()
{
stack.count--;
}

bool CheckPopInfo(int *num, int n, int m)
{
int cur = 1;
for (int i = 0; i < n; i++) {
// stack为空或者栈顶元素与num[i]不相等，则继续入栈
while (stack.count == 0 || stack.data[stack.count-1] != num[i]) {
if (cur > n || stack.count >= m) {
break;   // 过限
}
stack.data[stack.count] = cur++;
stack.count++;
}

// 判断是否符合，不等于num[i]或者stack是空说明是break跳出来的，不符合
if (stack.data[stack.count-1] != num[i] || stack.count == 0) {
return false;
}

// 继续查看下一个数
stack.count--;   // 出栈
}
return true;
}

int main()
{
int m;
int n;
int k;
scanf("%d%d%d", &m, &n, &k);
for (int j = 0; j < k; j++) {
int num[1000] = {0};
for (int i = 0; i < n; i++) {
scanf("%d", &num[i]);
}

CreateStack();
if (CheckPopInfo(num, n, m)) {
printf("YES\n");
} else {
printf("NO\n");
}
}

return 0;
}