Train Problem I

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 29144    Accepted Submission(s): 11062


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 HintHint 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.".【程式碼如下】#include <iostream> #include <stdio.h> #include <stack> #define max 100//巨集定義  與int const max=100;不同，巨集定義是定義一個數，const是定義一個固定變數 using namespace std; int main() {     stack<char>s;//很明顯，這個題要用棧做     int n, i, j, k, result[max];//result用來記錄是出棧還是入棧，進棧標記為1，出棧標記為0     char str1[max], str2[max];//兩個字串，用字元陣列表示     while(cin>>n>>str1>>str2)//cin>>str可以用來輸入字元陣列，下次別說不會啦！     {         i=0, j=0, k=1;         s.push(str1[0]);//為了防止棧空，壓一個進去（棧的題目應該都這樣吧）         result[0] = 1;//記錄進來了一個         while(i<n && j<n)         {             if(s.size() && s.top()==str2[j])//s.size表示用判斷棧不為空，棧頂元素要與序列2的第一個字元一樣，這是             {//如果棧頂元素和序列2當前元素相等，則彈棧，序列2集團向後移一位                 j++;                 s.pop();                 result[k++] = 0;//第一個元素就可以出棧了，然後做k++             }             else             {//否則從序列1中取當前元素壓入棧中                 if(i==n) break;//如果全部都壓進來了，還是彈不出去，就說明失敗了                 s.push(str1[++i]);//i<n，就壓棧！                 result[k++] = 1;             }         }         if(i==n)//如果i=n表示棧頂元素元素不等於序列2當前元素，且序列1中元素都已經入過棧，判斷不能得到和序列2一樣的答案            cout<<"No."<<endl;         else         {//輸出入棧方式             cout<<"Yes."<<endl;             for(i=0; i<k; i++)                 if(result[i])                     cout<<"in"<<endl;                 else                     cout<<"out"<<endl;         }         cout<<"FINISH"<<endl;     }     return 0; } 【分析】題解盡在程式碼中。。。