D. Train Problem I

Time Limit:1000ms

Case Time Limit:1000ms

Memory Limit:32768KB

64-bit integer IO format:%I64d      Java class name: Main

SubmitStatus PID: 5246

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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.".

解题思路：该题为模拟题，知识点为栈和队列的应用，可以用数组模拟。

最多9辆火车，编号为1,2,3....9，进出站方式入图（先进后出）。给出一个n，表示要进出站得火车数，给出火车进站顺序表，给出火车出站顺序表，问按给出入站表入站的火车能否以出站表顺序出站，如果可以，输出“YES.”以及出入情况信息（in，out），否则输出“NO.”，每组测试案例输出的最后一行以“FINISH”结束。

用3个队列及1个站来模拟，分别存储，火车进站顺序，火车出站顺序，站内火车情况（出入顺序要求情况）。输入火车进、出站顺序表，用字符数组做中介存储，后存入到对应队列。依题意得，火车进出站的次数最多为2*n次（每辆火车进站一次，出站一次），开始就循环结束判断条件纠结了很久，后面才想到。因为火车最多9辆，所以，这个地方不会导致程序运行超时。每次循环，判断出来，如果站内有火车可以按照给定的顺序出站，则让其出站（用队列保存火车出入情况），否则，若有火车未入栈，则让火车入站。最后如果火车进出站情况存储队列中有2*n个数据（所有的火车都进过站并且出来了，按题目给定的顺序）则输出“YES.”和火车出入情况及顺序（直接输出出入情况存储队列），否则输出“NO.”，其余按题意输出即可。

#include<stdio.h>#include<string.h>    //使用String串#include<iostream>    //使用c++输出流#include<stack>    //使用栈#include<queue>    //使用队列using namespace std;int main(){    int n,x,i;    string st;  //输出时内容暂存    while(scanf("%d",&n)!=EOF)    {        queue<int>  in,out;   //存储需要入站,出站的火车编号        queue<string> sign;    //存储火车存储站的状态（in，out，有序）        stack<int > s;   //存储现在车站中火车进出顺序        char a[20];   //存储输入时，火车进、出站顺序，后面会转存到队列        scanf("%s",a);   //读入火车入站顺序        for(i=0;i<n;i++)in.push(a[i]-48);    //将火车入站顺序转存到入站队列        scanf("%s",a);            for(i=0;i<n;i++)out.push(a[i]-48);        x=in.front();   //读取第一辆进站火车的车号        s.push(x);    //放入栈（站中）        in.pop();        sign.push("in");    //记录有一辆火车进站        int m=2*n-1;    //最多处理2*n次（进n次，出n次）        while(m--)        {            if(!out.empty()&&!s.empty()&&out.front()==s.top())     //若站（栈）内有火车，出站队列中有火车序号，且要处理的为同一辆火车，则处理这辆火车（出战）            {                out.pop();   //该火车可以出站                s.pop();     //该火车出站                sign.push("out");    //记录有火车出站            }            else if(!in.empty())    //若处于出站口的火车不能出站，且还有火车没有进站，则让下一辆火车进站            {                x=in.front();   //提取即将进站火车信息                in.pop();    //该火车将入站                s.push(x);   //火车入站                sign.push("in");     //记录有1辆火车入站            }        }        if(sign.size()==n*2)   //若火车出入站的次数为2*n次（每辆火车进一次站，出一次站）        {            printf("Yes.\n");    //火车以给出顺序进站，能以给出顺序出站            for(i=0;i<2*n;i++)            {                st=sign.front();                cout<<st<<endl;                sign.pop();            }        }        else printf("No.\n");   //火车以给出顺序进站，不能以给出顺序出站        printf("FINISH\n");    }    return 0;}