Train Problem(栈的应用)
Description

Input
Output
Sample Input
Sample Output
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.". 题目意思:给了火车进站和出站的顺序,问是否匹配,同时也要给出进出站的步骤。 解题思路:火车进出站实际上也模拟了栈的使用,我们可以对比两个字符串,s1串为进站顺序,s2串为出站顺序,对s1中每一个字符(火车)
比较s2的出站顺序,不相同则是进站,相同就要出站,以此类推,用一个数组标记进站和出站即可。 上代码:
#include<stdio.h>
#include<string.h>
#include<stack>
using namespace std;
int main()
{
int n,i,j,k,flag[];
char s1[],s2[];
while(scanf("%d %s%s",&n,s1,s2)!=EOF)
{
stack<char>s;
memset(flag,-,sizeof(flag));
j=;
k=;
for(i=;i<n;i++)
{
s.push(s1[i]);
flag[k++]=;
while(!s.empty()&&s.top()==s2[j])
{
s.pop();
flag[k++]=;
j++;
}
}
if(j==n)
{
printf("Yes.\n");
for(i=;i<k;i++)
{
if(flag[i]==)
printf("in\n");///1为进栈
else
printf("out\n");///0为出栈 }
}
else
printf("No.\n");
printf("FINISH\n");
}
return ;
}
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