hdu1022-Train Problem I

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 3213 123 312

 

Sample Output

Yes.inininoutoutoutFINISHNo.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.".

代码实现：

#include<iostream>
#include<cstring>
#include<stack>
using namespace std;
int main()
{
    int n,po[100],s[100],i,j,k,mark,sum,f,g,in[100];
    char a[100],b[100];
    stack<char> q;
    while(cin>>n>>a>>b)
    {
        while(!q.empty())
        {
            q.pop();
        }
        f=1;//标记结果,0表示no,1表示yes
        sum=0;//标记有多少步操作
        memset(po,-1,sizeof(po));//标记操作的数组，1进 0出
        memset(s,0,sizeof(s));//标记每个数是否在栈中，0不在 1在
        memset(in,0,sizeof(in));//标记每个数是否已经入过栈，入过栈为1
//        for(i=0;i<10;i++)
//        {
//            cout<<s[i]<<" ";
//        }
        for(i=0; i<n; i++)
        {
            //cout<<1;
//            cout<<b[i]<<endl;
//            cout<<s[b[i]-'0']<<endl;
            g=0;
            if(s[b[i]-'0']==0)
            {
                //cout<<1;
                for(j=0; j<n; j++) //找b[i]在a字符串里的位置
                {
                    if(a[j]==b[i])
                    {
                        mark=j;
                        //cout<<mark;
                        g=1;
                        break;
                    }
                }
                if(g==0)
                {
                    f=0;
                    break;
                }
                for(j=0; j<=mark; j++)
                {
                    if(in[a[j]-'0']==0)
                    {
                        in[a[j]-'0']=1;
                        q.push(a[j]);
                        s[a[j]-'0']=1;
                        po[sum++]=1;
                    }
                }
                s[a[mark]-'0']=0;
                po[sum++]=0;
                q.pop();
            }
            else
            {
                //cout<<2;
                if(q.top()==b[i])
                {
                    s[b[i]-'0']=0;
                    po[sum++]=0;
                    q.pop();
                }
                else
                {
                    f=0;
                    break;
                }
            }
        }
        if(f==0)
        {
            cout<<"No.\nFINISH\n";
        }
        else
        {
            cout<<"Yes.\n";
            for(i=0; i<sum; i++)
            {
                if(po[i]==1)
                {
                    cout<<"in\n";
                }
                else
                {
                    cout<<"out\n";
                }
            }
            cout<<"FINISH\n";
        }
        while(!q.empty())
        {
            q.pop();
        }
    }
    return 0;
}

这道题目做着好费劲

1.总思路为：遍历b数组，判断每个字符是否在栈中，如果不在，在a字符串中把b[i]及b[i]之前没有入过栈的字符入栈，这里不是把a字符串中把b[i]及b[i]之前的字符入栈，前面已经出栈的和入栈还没有出栈的，就不需要再重复入栈了，然后弹出b[i]；如果在，看栈顶的字符是不是b[i]，如果是，弹出，如果不是，输出no

2.a,b数组中存的是字符，用整型数组标记的时候，记得减去‘0’