HDU 1022.Train Problem I【栈的应用】【8月19】
2015-08-19 14:09
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Train Problem I
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
Sample Output
栈的应用:给定1个出栈序列,看当前入栈序列能否通过入栈出栈达到这个出栈序列.代码如下:
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.".
栈的应用:给定1个出栈序列,看当前入栈序列能否通过入栈出栈达到这个出栈序列.代码如下:
#include<cstdio> #include<stack> using namespace std; int main(){ int n; char f1[1010],f2[1010];//f1记录入栈顺序,f2记录出栈顺序 while(scanf("%d%s%s",&n,f1,f2)!=EOF){ int g[1010],i=0,j=0,k=0;//g[]记录进出情况 stack<char> t; while(i<n){ if(t.empty()||t.top()!=f2[i]&&j<n){ t.push(f1[j++]); g[k++]=1;//1为in } else if(t.top()==f2[i]){ t.pop(); g[k++]=0;//0为out i++; } else break; } if(t.empty()){//栈t为空,则可以生成当前出栈序列 printf("Yes.\n"); for(int i=0;i<k;i++){ if(g[i]) printf("in\n"); else printf("out\n"); } printf("FINISH\n"); } else printf("No.\nFINISH\n"); } return 0; }
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