zoukankan      html  css  js  c++  java
  • HDU 2548 A strange lift

    A strange lift

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
    Total Submission(s): 29442    Accepted Submission(s): 10641


    Problem Description
    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.
     
    Output
    For 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
     
    #include <iostream>
    #include <queue>
    using namespace std;
    int main()
    {
        int a[205];
        int v[205];
        int n,s,z;
        queue<int>q;
        while(cin>>n&&n)
        {
            int f=0;
            cin>>s>>z;
            while(!q.empty()) q.pop();
            int i;
            for(i=1;i<=n;i++)
            {
                cin>>a[i];
                v[i]=-1;
            }
            q.push(s);
            v[s]=0;
            while(!q.empty())
            {
                int x,step;
                x=q.front();q.pop();
                step=v[x];
                if(x==z) {cout<<v[x]<<endl;f=1;break;}
                if(x+a[x]<=n&&v[x+a[x]]==-1) 
                {
                    v[x+a[x]]=step+1;
                    q.push(x+a[x]);
                }
                if(x-a[x]>=1&&v[x-a[x]]==-1)
                {
                    v[x-a[x]]=step+1;
                    q.push(x-a[x]);
                }
            }
            if(!f) cout<<-1<<endl;
        }
        return 0;
    }
  • 相关阅读:
    MMORPG 游戏服务器端设计转载
    OpenFileMapping失败 原因ERROR_FILE_NOT_FOUND
    让程序在崩溃时体面的退出转
    [c\C++]线程函数的比较
    VS2005 2008 2010字体颜色配置方案
    用文件映射(File Mapping)实现进程间内存共享
    [Debug]通过LeakDiag查看内存泄漏
    线程函数的设计以及MsgWaitForMultipleObjects函数的使用要点 转
    VC++ 中使用内存映射文件处理大文件
    透视投影
  • 原文地址:https://www.cnblogs.com/caiyishuai/p/8411122.html
Copyright © 2011-2022 走看看