思维体操: HDU1022Train Problem I
Train Problem I
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 38390 Accepted Submission(s): 14423
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.



More details in the Sample Input.
for a train getting out of 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.".
RunId : 21242780 Language : G++ Author : hnustwanghe
Code Render Status : Rendered By HDOJ G++ Code Render Version 0.01 Beta
#include<iostream>
#include<string>
#include<cstdio>
#include<vector>
#include<stack>
using namespace std;
int main(){
int n;
while(scanf("%d",&n)==1){
if(n == 0){
printf("Yes.\nFINISH\n");
continue;
}
stack<int> S;
string O1,O2;
cin >> O1 >> O2;
int pos = 0;
int a[50],cur = 0;
for(int i=0;i<O1.length();i++){
S.push(O1[i]-'0');
a[cur++] = 1;///入栈
while(!S.empty() && S.top()== O2[pos]-'0' && pos <O2.length()){
S.pop();
pos++;
a[cur++] = 2;///出栈
}
}
if(pos == O2.length() && S.empty()){
printf("Yes.\n");
for(int i=0;i<cur;i++){
printf("%s\n",a[i]<2?"in":"out");
}
printf("FINISH\n");
}
else
printf("No.\nFINISH\n");
}
}
#include<iostream>
#include<string>
#include<cstdio>
#include<vector>
#include<stack>
using namespace std; int main(){
int n;
while(scanf("%d",&n)==1){
if(n == 0){
printf("Yes.\nFINISH\n");
continue;
}
stack<int> S;
string O1,O2;
cin >> O1 >> O2;
int pos = 0;
int a[50],cur = 0;
for(int i=0;i<O1.length();i++){
S.push(O1[i]-'0');
a[cur++] = 1;///入栈
while(!S.empty() && S.top()== O2[pos]-'0' && pos <O2.length()){
S.pop();
pos++;
a[cur++] = 2;///出栈
}
}
if(pos == O2.length() && S.empty()){
printf("Yes.\n");
for(int i=0;i<cur;i++){
printf("%s\n",a[i]<2?"in":"out");
}
printf("FINISH\n");
}
else
printf("No.\nFINISH\n");
}
}
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