HDU 1022 Train Problem I

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最新推荐文章于 2024-07-21 14:30:00 发布

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分类专栏： hdu ACM 水题128题文章标签： output input exchange integer each string

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本文探讨了一个经典的列车调度问题，通过栈的数据结构实现列车进出站的顺序调整。文章提供了两种编程实现方式，一种使用C++标准模板库(STL)，另一种则采用C语言结合数组模拟栈的操作。

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Train Problem I

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 11821 Accepted Submission(s): 4332

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

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.".

Author

Ignatius.L

分析：很典型的堆栈的题目，若图方便直接用STL或者用数组模拟出一个堆栈都可以

代码：

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

stack<char>s;
string stra,strb;
int i,j,k,n;
int result[9];

int main()
{
while(cin>>n>>stra>>strb)
{
i=j=k=0;
s.push(stra[i]);
result[k++]=1;
while(i<n&&j<n)
{
if(s.size()&&s.top()==strb[j])
{
s.pop();
result[k++]=0;
j++;
}
else
{
i++;
if(i==n)
break;
s.push(stra[i]);
result[k++]=1;
}
}
while(!s.empty())
s.pop();
if(i==n)
cout<<"No."<<endl;
else
{
cout<<"Yes."<<endl;
for(i=0;i<k;i++)
{
if(result[i])
cout<<"in"<<endl;
else
cout<<"out"<<endl;
}
}
cout<<"FINISH"<<endl;
}
return 0;
}

代码2：C语言数组模拟

#include<stdio.h>
#include<string.h>

int main()
{
    int n;
    char a1[10];
    char a2[10];
    int b[10];
    int c[1000];

    while(scanf("%d",&n)!=EOF)
    {
        memset(c,0,sizeof(c));
        int i=0,j=0,k=0,m=0;
            getchar();
            scanf("%s",a1);
            scanf("%s",a2);
        b[0]=0;
        while(j<n)
        {
            if(b[0]==0)
            {
                m=0;
                b[m]=a1[i]-'0';
                c[k++]=1;
                i++;
                m++;
            }
            else
            {
                if(b[m-1]==(a2[j]-'0'))
                {
                    b[m-1]=0;
                    m--;
                    j++;
                    k++;
                }
                else
                {
                    if(i==n)
                        break;
                    b[m]=a1[i]-'0';
                    c[k++]=1;
                    i++;
                    m++;
                }
            }
        }
        if(b[0]==0)
        {
            printf("Yes.\n");
            for(i=0;i<k;i++)
            {
                if(c[i])
                    printf("in\n");
                else
                    printf("out\n");
            }
            printf("FINISH\n");
        }
        else
        {
            printf("No.\n");
            printf("FINISH\n");
        }
    }
    return 0;
}