HDU 1022.Train Problem I【栈的应用】【8月19】

标签：栈   应用   acm   hdu   

Train Problem I

As the new term comes, the Ignatius Train Station is very busy nowadays. A lot of student want to get back to school by train(because the trains in the Ignatius Train Station is the fastest all over the world ^v^). But here comes 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.

3 123 321
3 123 312

Yes.
in
in
in
out
out
out
FINISH
No.
FINISH

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.".

#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，则可以生成当前出栈序列
            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;
}

HDU 1022.Train Problem I【栈的应用】【8月19】

标签：栈   应用   acm   hdu   

原文地址：http://blog.csdn.net/a995549572/article/details/47779781