杭电1022Train Problem I



3 123 312
in
in
in
out
out
out
FINISH
No.
FINISH
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.".
思路:紫书上原题,题意是给你一个进站前的数列,判断这几个数能否通过类似入栈出栈的操作后得到题目所给的数列;
用栈来模拟就好。
ps:一定要注意输出格式!!!写完代码之后,愣是一直wa,开始是因为栈没清空,后又发现是输出格式不对!!!
血的教训,改了好半天!!!
#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <queue>
#include <stack>
#include <map>
#include <vector>
#include <string> #define PI acos((double)-1)
#define E exp(double(1))
#define N 1000000
using namespace std; int main (void)
{
int t,a,b,flag,c;
string s,e,code;
stack<int>sk;
while(scanf("%d",&t) == && t)
{
cin>>s>>e;
flag=;
a=b=c=;
while(b<t)
{
if(s[a] == e[b])
{
a++;b++;
code[c++]=;
code[c++]=;
}
else if(!sk.empty() && e[b] == sk.top())
{
sk.pop();b++;
code[c++]=;
}
else if(a < t)
{
sk.push(s[a++]);
code[c++]=;
}
else
{
flag = ;
break;
}
}
if(flag)
{
cout<<"Yes."<<endl;
for(int i = ;i< *t;i++)
if(code[i])
cout<<"in"<<endl;
else
cout<<"out"<<endl;
}
else
cout<<"No."<<endl;
cout<<"FINISH"<<endl;
while(!sk.empty())
sk.pop();
}
return ;
}
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