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最新推荐文章于 2020-02-24 16:30:13 发布

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本文介绍了一种通过模拟栈操作来判断给定序列是否为合法出栈序列的方法。该方法适用于给定最大栈容量和一系列待验证的出栈序列的情况。通过对每一步栈操作的模拟，确保序列的有效性。

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题目：http://pat.zju.edu.cn/contests/pat-a-practise/1051

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：

可以模拟出其出栈序列的抽象的规律：可以向后跳跃移动，但是只能连续向前移动。当然，隐含的条件是不能超出栈容量。

用程序来模拟的话，可以这样：

用指针index指代当前还未被push进站的元素的位置，如果当前栈为空，push一个元素，index自增，如果当前栈元素小于出栈序列的当前元素，一直push到相等(index自增)，否则如果当前元素小于出栈序列的当前元素，输出NO，等于的话，continue下一次判断。

参考代码：

#include<cstdio>
#include<stack>
using namespace std;

int main()
{
    int M, N, K, arr[1001];
    scanf("%d %d %d", &M, &N, &K);
    for(int i = 0; i < K; ++i)
    {
        for(int j = 1; j <= N; ++j)
        {
            scanf("%d", arr+j);
        }
        stack<int> st;
        bool is_yes = true;
        int index = 1;
        for(int j = 1; j <= N && is_yes; ++j)
        {
            if(st.empty())
            {
                st.push(index++);
            }
            if(st.top() < arr[j])
            {
                while(st.top() < arr[j])
                    st.push(index++);
                if(st.size() > M)
                {
                    is_yes = false;
                    break;
                }
            }
            if(st.top() != arr[j])
            {
                is_yes = false;
                break;
            }
            st.pop();
        }
        is_yes ? printf("YES\n") : printf("NO\n");
    }
    return 0;
}