解题报告——Train Problem I(栈应用)
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Train Problem I
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 17889 Accepted Submission(s): 6686
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
非常经典的栈的应用问题。
题目没什么好说的,十分基础,只是要注意栈不为空和进出栈的操作正确。
但是有一个问题让我郁闷了好久,也导致我一下午一直处于wrong answer的状态。
就是当我用scanf(“%1d”,)(就是每次读入长度为1的整数) 时,对应的数据类型必须是char而不能是int型,这个在本地是没有任何错误的,也是没有办法发现的,这真是费了我好大功夫才找出来的(十几次的对比提交测试才把问题锁定)。
#include<stdio.h>char a[1000]; char b[1000];int s[1000];int c[3000];int main(){ int i,n; int tb,tc,top,ok; while(scanf("%d",&n)!=EOF) { for(i=0;i<n;i++) scanf("%1d",&a[i]); for(i=0;i<n;i++) scanf("%1d",&b[i]); ok=1; top=0,i=0,tb=0,tc=0; for(;tb<n;) { if(a[i]==b[tb]) { c[tc++]=1; c[tc++]=-1; tb++; i++; } else if(top&&s[top-1]==b[tb]) { c[tc++]=-1; tb++; top--; } else if(i<n) { s[top++]=a[i++]; c[tc++]=1; } else { ok=0; break; } } if(ok==0) printf("No.\n"); else { printf("Yes.\n"); for( i=0;i<tc;i++) { if(c[i]==1) printf("in\n"); else printf("out\n"); } } printf("FINISH\n"); } return 0;}
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