【hdu - 1022 （简单栈题目），，狂敲代码】

Train Problem I

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 11117    Accepted Submission(s): 4042


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
 
 
 
  1 // Project name : Defalut
  2 // File name    : main.cpp
  3 // Author       : iCoding
  4 // Date & Time  :
  5 // Email        : honi.linux@gmail.com
  6 
  7 #include <iostream>
  8 #include <stdio.h>
  9 #include <string>
 10 #include <cmath>
 11 #include <algorithm>
 12 #include <fstream>
 13 using namespace std;
 14 
 15 #define MAX_TRAIN_COUNT 15
 16 
 17 bool iFlagNoProblem;
 18 
 19 class Stack
 20 {
 21     public:
 22     char* a;
 23     int iTop;
 24     Stack();
 25     void pop();
 26     void push(int iNum);
 27     char getTop();
 28 };
 29 
 30 
 31 Stack::Stack()
 32 {
 33     a = new char[MAX_TRAIN_COUNT];
 34     iTop = -1;
 35     iFlagNoProblem = true;
 36 }
 37 
 38 void Stack::pop()
 39 {
 40     if (iFlagNoProblem)
 41     {
 42         if (iTop == -1)
 43         {
 44             iFlagNoProblem = false;
 45             return;
 46         }
 47         else
 48         {
 49             iTop--;
 50             return ;
 51         }
 52     }
 53     else
 54     {
 55         return ;
 56     }
 57 }
 58 
 59 char Stack::getTop()
 60 {
 61     if (iTop >= 0)
 62     {
 63         return a[iTop];
 64     }
 65     else
 66     {
 67         return '@';
 68     }
 69 }
 70 
 71 void Stack::push(int iNum)
 72 {
 73     if (iFlagNoProblem)
 74     {
 75         iTop++;
 76         a[iTop] = iNum;
 77     }
 78     else
 79     {
 80 
 81     }
 82 }
 83 
 84 
 85 int n;
 86 char iStrIn[MAX_TRAIN_COUNT];
 87 char iStrOut[MAX_TRAIN_COUNT];
 88 
 89 string iOperatorArr[MAX_TRAIN_COUNT * 2];
 90 int iOperatorTop;
 91 
 92 int iIndexStrIn;
 93 int iIndexStrOut;
 94 
 95 int main()
 96 {
 97     //freopen("in.dat", "r", stdin);
 98 
 99     while (scanf("%d", &n) != EOF)
100     {
101         scanf("%s%s", iStrIn, iStrOut);
102 
103         Stack iStack;
104         iOperatorTop = -1;
105         iIndexStrOut = 0;
106         for (iIndexStrIn = 0; iIndexStrIn < n; iIndexStrIn++)
107         {
108             iOperatorTop++;
109             iOperatorArr[iOperatorTop] = "in";
110             iStack.push(iStrIn[iIndexStrIn]);
111             while (iStack.getTop() == iStrOut[iIndexStrOut])
112             {
113                 iOperatorTop++;
114                 iOperatorArr[iOperatorTop] = "out";
115                 iStack.pop();
116                 iIndexStrOut++;
117             }
118         }
119 
120         if (iIndexStrOut != n)
121         {
122             iFlagNoProblem = false;
123         }
124         if (iFlagNoProblem)
125         {
126             cout << "Yes." << endl;
127             for (int i = 0; i <= iOperatorTop; i++)
128             {
129                 cout << iOperatorArr[i] << endl;
130             }
131             cout << "FINISH" << endl;
132         }
133         else
134         {
135             cout << "No." << endl;
136             cout << "FINISH" << endl;
137         }
138     }
139 
140     return 0;
141 }
142 
143 // end
144 // iCoding@CodeLab

 

posted @ 2012-08-06 13:52  ismdeep  阅读(412)  评论(0)  编辑  收藏  举报