HDU1022 FLY

本文探讨了一个经典的火车调度问题,通过栈数据结构实现火车进出站的顺序调整,确保最多9辆火车能够按照特定顺序离开车站。文章提供了一段C++代码示例,详细展示了如何使用栈来解决这个问题,同时给出了输入输出样例。

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原来的博客https://blog.youkuaiyun.com/Sabrina_Martin/article/details/81292887

Problem Description

As the new term comes, the Ignatius Train Station is very busy nowadays. A lot of student want to get back to school by train(because the trains in the Ignatius Train Station is the fastest all over the world v). But here comes a problem, there is only one railway where all the trains stop. So all the trains come in from one side and get out from the other side. For this problem, if train A gets into the railway first, and then train B gets into the railway before train A leaves, train A can’t leave until train B leaves. The pictures below figure out the problem. Now the problem for you is, there are at most 9 trains in the station, all the trains has an ID(numbered from 1 to n), the trains get into the railway in an order O1, your task is to determine whether the trains can get out in an order O2.

Input

The input contains several test cases. Each test case consists of an integer, the number of trains, and two strings, the order of the trains come in:O1, and the order of the trains leave:O2. The input is terminated by the end of file. More details in the Sample Input.

Output

The output contains a string “No.” if you can’t exchange O2 to O1, or you should output a line contains “Yes.”, and then output your way in exchanging the order(you should output “in” for a train getting into the railway, and “out” for a train getting out of the railway). Print a line contains “FINISH” after each test case. More details in the Sample Output.

Sample Input

3 123 321
3 123 312

Sample Output

Yes.
in
in
in
out
out
out
FINISH
No.
FINISH

Hint Hint

For the first Sample Input, we let train 1 get in, then train 2 and train 3.
So now train 3 is at the top of the railway, so train 3 can leave first, then train 2 and train 1.
In the second Sample input, we should let train 3 leave first, so we have to let train 1 get in, then train 2 and train 3.
Now we can let train 3 leave.
But after that we can’t let train 1 leave before train 2, because train 2 is at the top of the railway at the moment.
So we output “No.”.
我感觉写得很好的程序

#include<iostream>
#include<algorithm>
#include<vector>
#include<list>
#include<queue>
#include<set>
#include<cmath>
#include<iomanip>
#include<stack>
#include<map>
#define N 100000
#define LL long long int
using namespace std;
int a[1000];
int main(){
    string str,aim;
    int n;
    while(cin>>n>>str>>aim){
        stack<char> s;
        int now=0;
        int p=0;
        for(int i=0;i<n;i++){
            s.push(str[i]);
            a[p++]=0;
            while(!s.empty()&&s.top()==aim[now]){    
                a[p++]=1;
                s.pop();
                now++;
            }
        }
        if(now!=n){
            cout<<"No."<<endl;
        }
        else{
            cout<<"Yes."<<endl;
            for(int i=0;i<p;i++){
                if(a[i]==0)
                    cout<<"in"<<endl;
                else cout<<"out"<<endl;
            }
        }
        cout<<"FINISH"<<endl;
    }
}
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