1.这道题是判断出栈队列是否合理。
2.采用了用栈来模拟情况。
3.当目前栈为空,或者栈不为空并且栈顶不等于目标值,并且队列中还有数值可以压入,栈的size小于最大容量,那么就一直循环执行压入操作,把123456789。。。队列中的值依次压入栈,直到跳出循环。
4.跳出循环后,判断栈顶是否等于目标值,不等于的话,这个sequence就是不合理的,如果等于,则把栈顶的值pop掉,再执行步骤3。
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
AC代码:
//#include<string> //#include<stack> //#include<unordered_set> //#include <sstream> //#include "func.h" //#include <list> #include <iomanip> #include<unordered_map> #include<set> #include<queue> #include<map> #include<vector> #include <algorithm> #include<stdio.h> #include<iostream> #include<string> #include<memory.h> #include<limits.h> #include<stack> using namespace std; int main(void) { int maxCap, maxSequence, querySum; cin >> maxCap >> maxSequence >> querySum; for (int k = 0; k < querySum; k++) { int idx = 1; stack<int> sta; int maxNum = maxCap;//栈里面能够存在的最大的值 vector<int> sequence(maxSequence); for (int i = 0; i < maxSequence; i++) scanf("%d", &sequence[i]); bool result = true; for (int i = 0; i < maxSequence; i++) { if (sequence[i]>maxNum) {//不合理 result = false; break; } while ((sta.empty() || (!sta.empty() && sta.top() != sequence[i])) && idx <= maxSequence && sta.size() <= maxCap) {// 如果栈为空,或者栈不为空但是头部不等于目标值,并且还有值可以压入,队列size小于最大容量 sta.push(idx++); } if (sta.top() != sequence[i]) {//如果经过上面的压入操作后,仍不满足要求,则不合理 result = false; break; } sta.pop(); maxNum++;//弹出了一个,可以压入更大的一个值 } if (result) cout << "YES" << endl; else cout << "NO" << endl; } return 0; }