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最新推荐文章于 2021-07-14 18:35:37 发布

最新推荐文章于 2021-07-14 18:35:37 发布 · 250 阅读

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本文介绍了一道经典的栈应用问题——列车车厢重组问题。通过分析问题背景与要求，给出了解题思路及详细的C++代码实现。该问题要求确定是否能通过栈（车站）将按特定顺序进入的车厢重新排列成所需的顺序。

一、题目描述

There is a famous railway station in PopPush City. Country there is incredibly hilly. The station was built in last century. Unfortunately, funds were extremely limited that time. It was possible to establish only a surface track. Moreover, it turned out that the station could be only a dead-end one (see picture) and due to lack of available space it could have only one track.
The local tradition is that every train arriving from the direction A continues in the direction B with coaches reorganized in some way. Assume that the train arriving from the direction A has N <= 1000 coaches numbered in increasing order 1, 2, …, N. The chief for train reorganizations must know whether it is possible to marshal coaches continuing in the direction B so that their order will be a1, a2, …, aN. Help him and write a program that decides whether it is possible to get the required order of coaches. You can assume that single coaches can be disconnected from the train before they enter the station and that they can move themselves until they are on the track in the direction B. You can also suppose that at any time there can be located as many coaches as necessary in the station. But once a coach has entered the station it cannot return to the track in the direction A and also once it has left the station in the direction B it cannot return back to the station.

这里写图片描述

二、输入

The input consists of blocks of lines. Each block except the last describes one train and possibly more requirements for its reorganization. In the first line of the block there is the integer N described above. In each of the next lines of the block there is a permutation of 1, 2, …, N. The last line of the block contains just 0.
The last block consists of just one line containing 0.

三、输出

The output contains the lines corresponding to the lines with permutations in the input. A line of the output contains Yes if it is possible to marshal the coaches in the order required on the corresponding line of the input. Otherwise it contains No. In addition, there is one empty line after the lines corresponding to one block of the input. There is no line in the output corresponding to the last “null” block of the input.

例如：
输入：
5
1 2 3 4 5
5 4 1 2 3
0
6
6 5 4 3 2 1
0
0
输出：
Yes
No

Yes

四、解题思路

1、问题分析
Station就是一个栈，A是进栈的顺序，B就是出栈的顺序。判断通栈，按A的输入顺序，能否实现B的输出顺序。这就是典型的栈的应用问题。

2、算法步骤
因为入栈的顺序是按1、2、3…n，所以只用一个计数变量enterData表示就可以，没操作完一次计数变量enterData++；

使用一个queue（outList）保存初栈的顺序。使用一个栈st来表示station。

1）++enterData进入栈st中，当enterData大于n时，返回No结果

2）outList队首数据出队，判断st是否为空，如果st为空enterData++，操作1

3）判断st栈顶元素是否与outList顶元素队首相等，如果相等进行步骤4，否则跳到1

4）st栈顶出栈，outList队首元素出队，判断5是否为true，否则进入3

5）判断outList是否为空，如果outList为空放回true

五、代码

#include<iostream>
#include<stack>
#include<queue>

using namespace std;

int main()
{
    int firstNum, number;
    while(cin >> number && number != 0)
    {
        while(cin >> firstNum && firstNum!=0)
        {
            queue<int> outList;
            outList.push(firstNum);
            for(int i = 1; i < number; i++)     //保存出tation的队列B
            {
                int enterData;
                cin >> enterData;
                outList.push(enterData);
            }
            int enterData = 1;  //进入tation的编号
            stack<int> st;

            while(enterData <= number && outList.size() > 0)    //不断循环入栈，初栈操作
            {
                if(enterData == outList.front())    //满足初栈
                {
                    outList.pop();
                    while(!st.empty() && st.top() == outList.front())
                    {
                        outList.pop();
                        st.pop();
                    }
                }else
                {
                    st.push(enterData);     //入栈
                }
                enterData++;
            }

            if(outList.empty()) cout << "Yes" << endl;
            else cout << "No" << endl;
        }
        cout<<endl;
    }

    return 0;

}