HDU1022——Train Problem I（栈 stack）

Train Problem I

点击打开链接http://acm.hdu.edu.cn/showproblem.php?pid=1022

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.".
解题思路：
    这是一道简单的栈使用题。本题小龙人先用一个队列储存出栈顺序，然后依次进栈，若遇到栈顶元素与队列头元素一致，则立刻出栈（并出队列），再判断栈顶元素是否与队列头元素一致，若一致，循环以上步骤，若不一致，查询栈中元素是否有与队列头元素一致，若有，直接跳出输出NO，若没有，再依次进栈判断。最后有一个细节要处理好，当栈中元素出完后判断还有没有未进栈的火车，若没有，跳出输出YES。再进栈出栈同时别忘记记录，用map来储存是个不错的选择。小龙人希望同学看到解题思路后自己动手写代码，提高自己的敲代码能力。
代码实现：
#include <iostream>#include <cstdio>#include <queue>#include <vector>#include <stack>#include <map>#include <cstring>#include <cmath>#include <algorithm>using namespace std;const int maxn = 100000+10;int main(){    int n;    while( scanf("%d",&n)!=EOF )    {    stack<char>sta;    queue<char>que;    int i;    char train1[15] , train2[15];    scanf("%s",train1);    scanf("%s",train2);    for( i=0; i<n; i++ )        que.push(train2[i]);    map<int,string>mp;    int step = 0, flag = 0;    for( i=0; i<=n; i++ )    {        if(sta.empty())        {            if( i == n )                   // 注意判断有无未进站火车;                break;            sta.push(train1[i]);            mp[step++] = "in";            continue;        }        else        {            char q = que.front();            if( sta.top() == q )            {                sta.pop();                que.pop();                mp[step++] = "out";                i--;            }            else            {                for( int j=0; j<i-1; j++ )                {                    if( train1[j] == q )                    {                        flag = 1;                        break;                    }                }                if(flag == 1)                    break;                else                {                    sta.push(train1[i]);                    mp[step++] = "in";                }            }        }    }    if( flag == 1 )    {        printf("No.\n");    }    else    {        printf("Yes.\n");        for( i=0; i<step; i++)        {            cout << mp[i] <<endl;        }    }    printf("FINISH\n");    }    return 0;}