HDU - 1022 Train Problem I STL 压栈

最新推荐文章于 2019-06-08 23:13:15 发布

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最新推荐文章于 2019-06-08 23:13:15 发布

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分类专栏： HDU 文章标签： ACM STL 栈

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本文介绍了一个火车调度问题，探讨了如何使用数组模拟栈来验证一组火车能否按照特定顺序离开车站。通过具体的输入输出示例，文章详细展示了算法的实现过程。

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Train Problem I

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 41471 Accepted Submission(s): 15510

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

通过火车进站前的顺序判断是否能改变为所给的顺序，火车需先进站然后再出站，可用STL中的stack实现，这里用数组模拟压栈

#include<stdio.h>
#include<string.h>
int main()
{
	int n;
	char s1[20]={0},s2[20]={0},s3[20];
	int s1_in[20],s2_in[20];
	while(~scanf("%d %s %s",&n,s1,s2))	
	{
		memset(s3,0,sizeof(s3));//需对数组进行清空
		
		int s[20]={0};
		int i1=0,i2=0,i3=0,i=0;
		//scanf("%s" "%s",s1,s2);
		for(i1=0;i1<n;i1++)//将字符转换为数字
		{
			s1_in[20]=s1[i]-'0';
			s2_in[20]=s2[i]-'0';
			
		}
		for(i1=0;i1<n;i1++)
		{
			s3[i3]=s1[i1];//火车进站
			i3++;
			i++;
			while(s3[i3-1]==s2[i2]&&i2<n)
			{
				i3--;
				s[i]=1;
				i2++;
				i++;
			}
		}
		//printf("%d %d %d %d\n",i1,i2,i3,i);
		if(i2==n)//火车全部出站
		{
			printf("Yes.\n");
			for(i1=0;i1<i;i1++)
			{
				if(s[i1]==1)
				{
					printf("out\n");
				}
				else
				printf("in\n");
			}
			printf("FINISH\n");
		}
		else
		{
			printf("No.\n");
			printf("FINISH\n");
		}
	}
	return 0;
}