1051 Pop Sequence

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.

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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

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题目大意

有个容量限制为MAXSIZE的栈，分别把1，2，3，…，K⼊栈

然后给你N个系列的出栈顺序，问这些出栈顺序是否可能存在分别输出YES or NO

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思路

按照要求进⾏模拟即可

box存入出栈顺序

OP()封装判断运算

OP中的flag表示当前推送的数字，flag==K+1时表示推送完成

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C/C++ 

#include<bits/stdc++.h>
using namespace std;
list<int> box;
int N,K,MAXSIZE,num;
bool OP();
int main()
{
    cin >> MAXSIZE >> K >> N;
    while (N--){
        for(int z=0;z<K;z++){
            cin >> num;
            box.push_back(num);
        }
        puts((OP()?"YES":"NO"));
        box.clear();
    }
    return 0;
}

bool OP() {
    stack<int> s;
    int flag = 1;
    while (!box.empty()){
        if(!s.empty() && s.top()==box.front()) {
            s.pop();
            box.pop_front();
        }
        else{
            if(s.size()==MAXSIZE || flag==K+1) return false;
            s.push(flag++);
        }
    }
    return true;
}

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