#1051. Pop Sequence【栈 + 模拟】

2021-11-07 13:34:54 阅读：203 来源： 互联网

标签：1051 sequence int Sequence pop stk Pop stack numbers

Problem Description：

Given a stack which can keep M M M numbers at most. Push N N N numbers in the order of 1 , 2 , 3 , … , N 1, 2, 3, \ldots, N 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 M M is 5 and N N 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 M M (the maximum capacity of the stack), N N N (the length of push sequence), and K K K (the number of pop sequences to be checked). Then K K K lines follow, each contains a pop sequence of N N 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

Problem Analysis:

本题是一个栈模拟题，给定 k k k 组弹出序列，对于每一组弹出序列，我们模拟栈压入弹出的过程进行判断即可。时间复杂度为 O ( n 2 ) O(n^2) O(n2)

Code

#include <iostream>
#include <algorithm>
#include <stack>

using namespace std;

const int N = 1010;

int n, m, k;
int a[N];

bool check()
{
    stack<int> stk;
    for (int i = 1, j = 0; i <= n; i ++ )
    {
        stk.push(i);
        if (stk.size() > m) return false;
        
        while (stk.size() && stk.top() == a[j])
        {
            stk.pop();
            j ++ ;
        }
    }
    return stk.empty();
}

int main()
{
    cin >> m >> n >> k;
    while (k -- )
    {
        for (int i = 0; i < n; i ++ ) scanf("%d", &a[i]);
        if (check()) puts("YES");
        else puts("NO");
    }

    return 0;
}

标签：1051,sequence,int,Sequence,pop,stk,Pop,stack,numbers
来源： https://blog.csdn.net/geraltofrivia123/article/details/121190367

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