【PAT甲】1051 Pop Sequence (25分)判断出栈顺序的合法性

problem

1051 Pop Sequence (25分)
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

给定一个大小为n的栈，按照1,2,3,…n的顺序依次入栈
给出k个长为m的出栈序列，判断是否合法

solution

直观的思路就是将入栈序列一个一个入栈，与出栈序列相比较，一样就出栈，不一样就继续入栈，当入栈序列和出栈序列都为空时，表示出栈顺序合法。

建立一个辅助栈把输入的第一个序列中的数字一个一个压入该辅助栈，并按照第二个序列的顺序从该栈中弹出数字。
(1)如果元素是栈顶的元素，则pop出来；
(2)如果不是栈顶元素，则根据入栈顺序将没入栈的元素push进栈，直到将该元素push栈中，然后将该元素pop出来；如果push完所有元素都没有找到该元素，那么这个出栈顺序是错误的。最后辅助栈为空栈，并且遍历完了出栈序列的最后一个元素（二者缺一不可），那么该序列就是一个弹出序列

#include<iostream>
#include<stack>
using namespace std;
int a[1010];int main(){int n, m, k;cin>>n>>m>>k;while(k--){for(int i = 1; i <= m; i++)cin>>a[i];int ok = 1, cur = 0;stack<int>s;for(int i = 1; i <= m; i++){if(s.empty())s.push(++cur);while(s.top()!=a[i]&&s.size()<=n)s.push(++cur);if(s.size()>n){ok = 0; break;}if(s.top()==a[i]){s.pop(); continue;}else{ok = 0; break;}}if(ok==1&&s.size()==0){cout<<"YES\n";}else{cout<<"NO\n";}}return 0;
}