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  • hdu 1022 Train Problem I

    Train Problem I

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
    Total Submission(s): 26603    Accepted Submission(s): 10061


    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
     
    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
     
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    巧妙利用栈解题,列车进站的问题。可以进站马上出站,出站的顺序可以随意定。
     
    题意:有N辆火车,以序列1方式进站,判断是否能以序列2方式出栈。进站不一定是一次性进入,也就是说中途可以出站。
     
    附上代码:
     
     1 #include <iostream>
     2 #include <cstdio>
     3 #include <stack>
     4 using namespace std;
     5 int main()
     6 {
     7     int n,i,j,m,num;
     8     char str1[105],str2[105];
     9     int flag[105];
    10     while(~scanf("%d %s %s",&n,str1,str2))
    11     {
    12         stack <char> s;    //定义一个栈
    13         while(!s.empty())   //如果栈不为空,则必须将栈清空
    14             s.pop();
    15         i=j=num=0;
    16         while(j<n)
    17         {
    18             if(s.empty()||s.top()!=str2[j]&&i<n)  //如果栈是空的,或者当前要出来的车不是栈中的第一辆可出来的车,则存入当前这辆车
    19             {
    20                 s.push(str1[i]);
    21                 flag[num++]=1;   //并标记为1,表示进站
    22                 i++;
    23             }
    24             else
    25             {
    26                 if(s.top()==str2[j])  //如果此时栈中可以出站的车正好是要求出站的车,则马上从栈中出来
    27                 {
    28                     s.pop();
    29                     flag[num++]=0;  //出站车,标记为0
    30                     j++;
    31                 }
    32 
    33                 else     //如果i>n,则跳出循环
    34                     break;
    35             }
    36         }
    37         if(s.empty())   //如果此时栈是空的,则表示序列1的列车以序列2的顺序全部出去,因次输出Yes.,否则输出No.
    38         {
    39             printf("Yes.
    ");
    40             for(i=0; i<num; i++)
    41             {
    42                 if(flag[i])    //标记为1就进来一辆车,0则出去一辆车
    43                     printf("in
    ");
    44                 else
    45                     printf("out
    ");
    46             }
    47         }
    48         else
    49             printf("No.
    ");
    50         printf("FINISH
    ");
    51     }
    52     return 0;
    53 }
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  • 原文地址:https://www.cnblogs.com/pshw/p/4818302.html
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