PAT 1051 Pop Sequence[栈][难]

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

题目大意：给出一个栈最大容量，并且给出最大的N，要求是1....N是这样顺序入栈，输入K个检查的序列，需要判读是不是可能的弹出序列。

//我看见其实以为自己会，其实是不会的，以前见过这种题的，但是应该当时也没理解吧。

代码转自:https://www.liuchuo.net/archives/2232

#include <iostream>
#include <stack>
#include <vector>
#include<cstdio>
using namespace std;
int main() {
    int m, n, k;
    scanf("%d %d %d", &m, &n, &k);
    for(int i = ; i < k; i++) {
        bool flag = false;
        stack<int> s;
        vector<int> v(n + );
        for(int j = ; j <= n; j++)
            scanf("%d", &v[j]);//读入要检验的序列。
        int current = ;//指向输入的序列。
        for(int j = ; j <= n; j++) {
            s.push(j);
            if(s.size() > m) break;
            while(!s.empty() && s.top() == v[current]) {
                s.pop();
                current++;
            }
        }
        if(current == n + ) flag = true;
        if(flag) printf("YES\n");
        else printf("NO\n");
    }
    return ;
}

//真的很厉害了，学习了。

1.使用一个current来指向当前的检验序列。

2.如果栈的大小已经大于了容量，那么就退出。

3.还有这个while循环是最关键的，只要top值等于当前current指向的，那么就弹出，并且指向下一个元素，非常厉害了。

//学习了，另一位大佬的代码思路也是相同的。