HDU 1022.Train Problem I【栈的应用】【8月19】
Train Problem I
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.".
栈的应用:给定1个出栈序列,看当前入栈序列是否能通过入栈出栈达到这个出栈序列.代码例如以下:
#include<cstdio>
#include<stack>
using namespace std;
int main(){
int n;
char f1[1010],f2[1010];//f1记录入栈顺序,f2记录出栈顺序
while(scanf("%d%s%s",&n,f1,f2)!=EOF){
int g[1010],i=0,j=0,k=0;//g[]记录进出情况
stack<char> t;
while(i<n){
if(t.empty()||t.top()!=f2[i]&&j<n){
t.push(f1[j++]);
g[k++]=1;//1为in
}
else if(t.top()==f2[i]){
t.pop();
g[k++]=0;//0为out
i++;
}
else break;
}
if(t.empty()){//栈t为空,则能够生成当前出栈序列
printf("Yes.\n");
for(int i=0;i<k;i++){
if(g[i]) printf("in\n");
else printf("out\n");
}
printf("FINISH\n");
}
else printf("No.\nFINISH\n");
}
return 0;
}
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