zoukankan      html  css  js  c++  java
  • A strange lift HDU

    There is a strange lift.The lift can stop can at every floor as you want, and there is a number Ki(0 <= Ki <= N) on every floor.The lift have just two buttons: up and down.When you at floor i,if you press the button "UP" , you will go up Ki floor,i.e,you will go to the i+Ki th floor,as the same, if you press the button "DOWN" , you will go down Ki floor,i.e,you will go to the i-Ki th floor. Of course, the lift can't go up high than N,and can't go down lower than 1. For example, there is a buliding with 5 floors, and k1 = 3, k2 = 3,k3 = 1,k4 = 2, k5 = 5.Begining from the 1 st floor,you can press the button "UP", and you'll go up to the 4 th floor,and if you press the button "DOWN", the lift can't do it, because it can't go down to the -2 th floor,as you know ,the -2 th floor isn't exist. 
    Here comes the problem: when you are on floor A,and you want to go to floor B,how many times at least he has to press the button "UP" or "DOWN"? 

    Input  The input consists of several test cases.,Each test case contains two lines. 
      The first line contains three integers N ,A,B( 1 <= N,A,B <= 200) which describe above,The second line consist N integers k1,k2,....kn. 
      A single 0 indicate the end of the input.OutputFor each case of the input output a interger, the least times you have to press the button when you on floor A,and you want to go to floor B.If you can't reach floor B,printf "-1".

    Sample Input

    5 1 5
    3 3 1 2 5
    0

    Sample Output

    3
    思路:不过是迷宫换成电梯的BFS
    AC Code:
    #include<iostream>
    #include<cstring>
    #include<queue>
    using namespace std;
    int INF=0x3f3f3f3f;
    int N,A,B;
    int a[213];
    int vis[213];
    int bfs(){
        if(A>N||B>N) return -1;
        queue<int> q;
        q.push(A);
        vis[A]=0;
        while(!q.empty() ){
            int p=q.front() ;
            q.pop() ;
            if(p==B) return vis[p];
            int up,down;
            up=p+a[p];
            down=p-a[p];
            if(up<=N&&up>=1&&vis[up]==INF){
                vis[up]=vis[p] +1;
                q.push(up); 
            }
            if(down>=1&&down<=N&&vis[down]==INF){
                vis[down]=vis[p]+1;
                q.push(down); 
            }
        }
        return -1;
    }
    int main(){
        while(cin>>N>>A>>B&&A){
            for(int i=1;i<=N;i++) cin>>a[i];
            memset(vis,INF,sizeof(vis));
            cout<<bfs()<<endl;
        }
    }
  • 相关阅读:
    好的学习资源
    对paper有用的idea
    斜杠青年
    简书随笔
    点云专业英文单词
    通过 UDP 发送数据的简单范例
    简单的聊天时范例(客户端)
    键盘输入
    简单的传输文件范例
    编写serversocket简单示例1
  • 原文地址:https://www.cnblogs.com/astonc/p/9910740.html
Copyright © 2011-2022 走看看