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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
Sample Output
Yes. in in in out out out FINISH No. FINISH
Sample Output
Yes. in in in out out out FINISH No. FINISH
题目意思是：
有一个车站 3 123 3是有三辆火车进站 123 分别是1 2 3号车 321是车辆开出顺序也就是列车必须是3号车出站后2号车才能出站之后是 1 号，当第 1 号辆进来的时候因为第一辆出战必须是3号车所以此时1号不能走，驶入2号，2号也不能走，驶入3号吗，3号可以走了，接着2号走，然后是1号走。
4 1234 2134 此时开出的顺序就不一样了。1号进来不能走，2号进来正好出车顺序是2134 所以 此时2号可以走了，2号走完正好1号走，1 号走完，3号进来，3号接着就出去了，最后是4号进4号出
这是在进站的时候就有出站情况的数据
/* 4 1234 2134
Yes.
int
int
out
out
int
out
int
out
FINISH */
#include<cstdio> #include<cstring> #include<iostream> using namespace std; int main() { int n; char c[15],g[15],zhan[15]; int cg[20]; while(cin>>n>>c>>g) { memset(cg,0,sizeof(cg)); int i=-1,j=0,k=-1,x=-1; zhan[++k]=c[++i];// 先把第一个进来的车存入栈 cg[++x]=0;// 进栈用 0 标记（便于一会输出出栈进栈顺序） while(c[i]!='\0') { while(k>=0&&zhan[k]==g[j])// 比如 进栈 123 出栈 321 可以在纸上演示一遍就明白这个循环了 { cg[++x]=1; --k; ++j; } cg[++x]=0; zhan[++k]=c[++i]; } if(i==n&&j==n) { printf("Yes.\n"); for(i=0;i<x;i++) puts(cg[i]?"out":"in"); } else puts("No."); puts("FINISH"); } return 0; }
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