hdu 1022 栈的应用

本文探讨了一个经典的列车调度问题,即如何通过单轨铁路正确地安排列车进出站的顺序。问题中考虑了最多9辆列车,并提供了输入输出示例。文章还提供了一段C++代码实现,用于验证列车是否能按照指定的顺序离开车站。

摘要生成于 C知道 ,由 DeepSeek-R1 满血版支持, 前往体验 >

Problem Description
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.

Input
The input contains several test cases. Each test case consists of an integer, the number of trains, and two strings, the order of the trains come in:O1, and the order of the trains leave:O2. The input is terminated by the end of file. More details in the Sample Input.

Output
The output contains a string “No.” if you can’t exchange O2 to O1, or you should output a line contains “Yes.”, and then output your way in exchanging the order(you should output “in” for a train getting into the railway, and “out” for a train getting out of the railway). Print a line contains “FINISH” after each test case. More details in the Sample Output.

Sample Input
3 123 321
3 123 312

Sample Output
Yes.
in
in
in
out
out
out
FINISH
No.
FINISH

题解:

参考了紫书中的代码。
用string wrong answer好几次。换char 加了个条件A<=n水过

代码:

#include <iostream>
#include <stack>
#include <vector>
#include <string>
#include <cstdio>
using namespace std;
const int maxn = 1000+10;

char orig[maxn],target[maxn];

vector<string> vec;

int main()
{
    int n;
    while(cin>>n)
    {
        scanf("%s %s",orig+1,target+1);
        stack<char> s;

        int A=1,B=1;
        bool ok=1;

        while(B<=n)
        {
            if(orig[A]==target[B]) {A++;B++;vec.push_back("in");vec.push_back("out");}
            else if(!s.empty()&&s.top()==target[B]) {s.pop();B++;vec.push_back("out");}
            else if((orig[A]-'0')<=n&&A<=n) {s.push(orig[A]);A++;vec.push_back("in");}
            else {ok=0;break;}
        }
        if(ok)
        {
            printf("Yes.\n");
            for(int i=0;i<vec.size();i++)
            cout<<vec[i]<<endl;
            printf("FINISH\n");
        }
        else
        {
            printf("No.\n");
            printf("FINISH\n");
        }

        vec.clear();


    }
    return 0;
}
评论
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值