1051. Pop Sequence (25)

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本文介绍了一种通过栈数据结构来判断给定序列是否为合法弹出序列的方法。输入包括栈的最大容量、推入序列长度及待验证序列，通过算法验证序列的有效性，并给出每个序列的判断结果。

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

思路：考察栈的应用，输入是有序的！！

参考代码：

#include <iostream>
#include <stack>
using namespace std;
stack<int> s;
int M,N,K;
int main(){
    cin>>M>>N>>K;
    for(int i=0;i<K;i++){
        while(!s.empty()){
            s.pop();
        }
        int isRight = 1;
        int countL = 0;
        for(int j=0;j<N;j++){
            int input;
            cin>>input;
            while(countL<input){
                if(s.size()>M-1){
                    isRight = 0;
                    break;
                }
                s.push(countL++);
            }
            int topV = s.top()+1;
            if(topV==input){
                s.pop();
            }else{
                isRight = 0;
            }
        }
        if(isRight==1){
            cout<<"YES"<<endl;
        }
        if(isRight==0){
            cout<<"NO"<<endl;
        }
    }
    return 0;
}