02-线性结构4 Pop Sequence (25分) （栈）

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本文介绍了一个算法，用于判断给定的出栈序列是否为特定容量栈的有效出栈序列。通过实例解析了算法的工作原理，并提供了完整的C++实现代码。

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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
/*
 * 举个例子吧： 3 2 1 7 5 6 4是不可能的序列
   因为3第一个出栈  1,2必定还在栈中 321是可能的 但是7先进栈 4，5,6还在栈中 4先进栈 所以出栈顺序该是6 5 4故不可能
   从左往右扫描数组 得到t 维护MIN值 如果t比MIN大 则将MIN+1-t-1的值进栈 小的话 则出栈一个和t比较 不相等则不可能，相等则继续
   另外 栈中元素个数如果超过栈容量 则也不可能得到
*/
#include "iostream"
using namespace std;
#define MAXSIZE 1000
typedef int ElementType;
struct Node {
	ElementType data[MAXSIZE];
	int top;
};
typedef Node *ptrToNode;
typedef ptrToNode Stack;
/* 初始化栈
*/
void initStack(Stack *stack) {
	*stack = (Stack)malloc(sizeof(Node));
	(*stack)->top = -1;
}
/* 进栈
*/
bool push(Stack *stack, ElementType ele) {
	if ((*stack)->top == MAXSIZE - 1)
		return 0;
	(*stack)->top++;
	(*stack)->data[(*stack)->top] = ele;
	return 1;
}
/* 出栈
*/
ElementType pop(Stack *stack) {
	if ((*stack)->top == -1)
		return 0;
	return (*stack)->data[(*stack)->top--];
}
ElementType length(Stack *stack) {
	return (*stack)->top + 1;
}
int main() {
	int k, m, n;
	int a;
	int min = 0;
	Stack s;
	bool flag = true;
	cin >> k >> m >> n;
	while (n--) {
		flag = true;
		initStack(&s);
		int l = m;
		min = 0;
		while (l--) {
			cin >> a;
			if (a > min) {
				for (int i = min + 1; i < a; i++)  //先弹出的a，所以min-（a-1）的数 应按顺序在栈中
					push(&s, i);
				min = a;
			}
			if (length(&s) > k - 1)
				flag = false;
			if (a < min) {
				int b = pop(&s);
				if (a != b) {
					flag = false;
				}
			}
		}
		if (flag) cout << "YES" << endl;
		else cout << "NO" << endl;
	}
}