02-线性结构4 Pop Sequence (25分) （栈）

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 51 2 3 4 5 6 73 2 1 7 5 6 47 6 5 4 3 2 15 6 4 3 7 2 11 7 6 5 4 3 2

Sample Output:

YESNONOYES
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 1000typedef 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;}}