hdu 1022 Train Problem I(栈的应用+STL)
Train Problem I
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 20521 Accepted Submission(s): 7712
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 railway). Print a line contains "FINISH" after each test case. More details in the Sample Output.
3 123 321
3 123 312
Yes.
in
in
in
out
out
out
FINISH
No.
FINISHFor the first Sample Input, we let train 1 get in, then train 2 and train 3.HintHint
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.".
栈的水题一道。就直接对进出栈进行模拟。和之前做的那道题类似,用来练练STL还是不错的,直接用STL水过。
http://blog.csdn.net/whjkm/article/details/38455291 能够看看之前的那道题。
用来vector 和 stack容器。用vector来存储字符串;
#include <cstdio>
#include <cstring>
#include <string>
#include <iostream>
#include <stack>
#include <vector>
using namespace std;
int main()
{
int n,i,j;
string str1,str2;
while(cin>>n>>str1>>str2)
{
stack<int>s;
vector<string>v;
for(i=0,j=0;i<n&&j<n;i++)
{
s.push(str1[i]);//把元素进栈
v.push_back("in");//压进vector容器
while(s.top()==str2[j])//进行匹配
{
if(!s.empty())
{
s.pop();
v.push_back("out");
}
j++;
if(s.empty()) break;
}
}
if(j==n)
{
printf("Yes.\n");
for(i=0; i< v.size();i++)
cout<< v[i] <<endl;
printf("FINISH\n");
}
else
printf("No.\nFINISH\n");
}
return 0;
}
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