题目:
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 Nnumbers. 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
胡乱分析:
补档.。。。。题目简单翻译下,给你一个容量为Capacity的堆栈,按照顺序push进Num个数,问是否可以pop出的指定数组,思路全在备注里了2333
代码:
#include <iostream>
#include <stack>
using namespace std;
int main() {
int Capacity, Num, cnt;//Capacity堆栈容量 Num待检查的元素的个数 cnt组数
cin >> Capacity >> Num >> cnt;
while (cnt--) {
int *a = new int[Num + 1]; //存放带检查元素的数组(下标从1开始)
bool flag = true; //一个检查堆栈是否溢出的标记
stack<int> S; //一个模拟用的堆栈
for (int i = 1; i <= Num; i++) {
cin >> a[i];
}
int p = 1; //一个滑动的指针
for (int i = 1; i <= Num; i++) { //元素按照顺序入栈
S.push(i);
if (S.size() > Capacity) { //如果超出堆栈容量
flag = false; break; //改变flag跳出循环
}
while (!S.empty() && S.top() == a[p]) {//顶端元素为a[p]
S.pop(); //弹出这个元素
p++; //p指向数组下一个元素
}
}
if (p - 1 == Num && flag) { //如果整个数组均被检查完&&没有溢出堆栈容量
cout << "YES" << endl;
}
else {
cout << "NO" << endl;
}
}
}