杭电1022——Train Problem I(栈的应用)
2016-01-07 20:07
531 查看
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.”.
主要算法:
扫描字符串2。如果当前字符与栈顶元素相等,则出栈,并且扫描字符串2中下一个字符;如果栈为空或者字符串2中扫描的当前字符与栈顶元素不相等,则字符串1中字符按顺序压入栈中,直到满足字符串2当前扫描字符和栈顶字符相等。再次,栈顶元素出栈,并且扫描字符串2中下一个字符。如此重复下去,如果最终栈为空,则输出Yes.如果栈顶元素和字符串2中当前扫描的字符不等,且字符串1中的元素已经全部进栈,则输出No.
注意:每个case完了之后要清空栈。
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.”.
主要算法:
扫描字符串2。如果当前字符与栈顶元素相等,则出栈,并且扫描字符串2中下一个字符;如果栈为空或者字符串2中扫描的当前字符与栈顶元素不相等,则字符串1中字符按顺序压入栈中,直到满足字符串2当前扫描字符和栈顶字符相等。再次,栈顶元素出栈,并且扫描字符串2中下一个字符。如此重复下去,如果最终栈为空,则输出Yes.如果栈顶元素和字符串2中当前扫描的字符不等,且字符串1中的元素已经全部进栈,则输出No.
注意:每个case完了之后要清空栈。
#include<stdio.h> #include<string.h> #define MAX 50 typedef struct{ char data[MAX]; int top; }Stack; void Initial(Stack *s); char getTop(Stack *s); int Push(Stack *s,char ch); int Pop(Stack *s); int main() { int n; char s1[MAX],s2[MAX]; char in_out[MAX][5]; Stack s; Initial(&s); while(scanf("%d%s%s",&n,s1,s2)!=EOF) { int i=0,j=0,k=0,flag=1; memset(in_out,0,MAX*5*sizeof(char)); Push(&s,s1[i++]); strcpy(in_out[k++],"in"); while(i<n||j<n) { while(s.top==0||s2[j]!=getTop(&s)) { if(i<n) { Push(&s,s1[i++]); strcpy(in_out[k++],"in"); } else { flag=0; break; } } if(flag==0) break; Pop(&s); strcpy(in_out[k++],"out"); j++; } if(s.top==0) { printf("Yes.\n"); for(i=0;i<k;i++) printf("%s\n",in_out[i]); printf("FINISH\n"); } else { printf("No.\nFINISH\n"); } s.top=0;//注意要将栈清空 } return 0; } void Initial(Stack *s) { s->top=0; } char getTop(Stack *s) { return s->data[s->top-1]; } int Push(Stack *s,char ch) { if(s->top>=MAX) return 0; else { s->data[s->top++]=ch; return 1; } } int Pop(Stack *s) { if(s->top>=MAX) return 0; else { s->data[--s->top]=0; return 1; } }
相关文章推荐
- IoC容器和 Dependency Injection模式 Inversion of Control Containers and the Dependency Injection pattern
- LeetCode Permutaions II
- SVN Checkout Failed的解决办法
- ubuntu 使用Perl NET::SMTP 发送Email auth 失败
- pthread_cond_wait
- There is no default constructor available in xxx错误引发
- 应用Btrfs管理磁盘,应用mdadm管理磁盘(softraid),使用raid卡管理磁盘
- 服务启动报错
- Binder&AIDL
- rails权限管理—devise+cancan+rolify
- 人工智能学习笔记(一)命题逻辑
- QWidget,QMainWindow和QDialog的区别 .
- 论机器人的环境感知与智主移动-兼谈基于微分几何的人工智能
- LeetCode Permutaions
- LintCode: Number of Airplanes in the Sky
- [已解决]使用matplotlib生成jpg报错:encoder jpeg not available
- 处于CLOSE_WAIT和TIME_WAIT状态连接的原因及解决
- Baidu地图的相关开发经验
- Communications link failure
- YTU 1007: Redraiment猜想