02-线性结构4 Pop Sequence（25 分）

最新推荐文章于 2024-04-05 17:23:19 发布

peterchen96

最新推荐文章于 2024-04-05 17:23:19 发布

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本题来源于中国大学MOOC的数据结构课程，由浙江大学陈越教授讲解。给定一个最大容量为M的栈，将1到N的数字顺序压入栈中，然后随机弹出。任务是判断给出的K个弹出序列是否可能从栈中产生。输入包含M、N和K，输出每个序列是否可能。示例输入和输出展示了不同情况下的判断结果。

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题目来源：中国大学MOOC-陈越、何钦铭-数据结构-2018春
作者: 陈越
单位: 浙江大学

问题描述：
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

解答：模拟栈的运行，把1-N依次放入栈中，模拟，看能不能顺利全部弹出栈。

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

stack<int> moni;
int main()
{
    int MaxSize,N,testN;
    cin>>MaxSize>>N>>testN;
    int test[1001];
    for(int j=1; j<=testN; j++)
    {
        while(!moni.empty())
        {
            moni.pop();
        }
        for(int i=1; i<=N; i++)
            cin>>test[i];
        int i=1,temp=1;
        bool flag=true;
        while(i<=N)
        {
            moni.push(i);
            if(moni.size()>MaxSize)
            {
                flag=false;
                break;
            }
            while(!moni.empty()&&test[temp]==moni.top())
            {
                moni.pop();
                temp++;
            }
            i++;
        }
        if(!moni.empty())
            flag=false;
        if(flag)
            cout<<"YES"<<endl;
        else
            cout<<"NO"<<endl;
    }
    return 0;
}