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


    title: A strange lift
    tags: [广搜,最短路径]

    题目链接

    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
    

    题意

    有一栋楼,它有电梯,但是每层楼只有两个按钮,UP 和down ,这层楼只能上x层或者下x层,层数大于等于1,当然小于楼层高度。

    分析

    简单的广搜问题,由于在最短路分类中找到的这个题目,就用对短路来做了。

    代码

    #include<bits/stdc++.h>
    #define inf 0x3f3f3f3f
    #define MAX1 1050
     int road[MAX1][MAX1];
    using namespace std;
    int dij(int n,int s,int e)
    {
        int vis[n+1]= {0};
        int dis[n+1];
        for (int i=1; i<=n; i++)
            dis[i]=road[s][i];
        vis[s]=1;
        int count1=1;
        int op=s;
        while (count1<n)
        {
            int min1=inf;
            for (int i=1; i<=n; i++)
            {
                if (vis[i]==0&&dis[i]<min1)
                {
                    min1=dis[i];
                    op=i;
                }
            }
            vis[op]=1;
            count1++;
            for (int i=1; i<=n; i++)
            {
                if (vis[i]==0&&dis[i]>dis[op]+road[op][i])
                    dis[i]=dis[op]+road[op][i];
            }
        }
        return dis[e];
    }
    
    int main()
    {
    
        int n,a,b;
        while (scanf("%d",&n),n)
        {
            scanf("%d%d",&a,&b);
            int f[MAX1];
    
            for (int i=1; i<=n; i++)
                for (int j=1; j<=n; j++)
                    if (i==j)road[i][j]=0;
                    else road[i][j]=inf;
            for (int i=1; i<=n; i++)
            {
                int a;
                scanf("%d",&a);
                if (i-a>=1)
                    road[i][i-a]=1;
                if (i+a<=n)
                    road[i][i+a]=1;
            }
            int s=dij(n,a,b);
            if(s==inf)s=-1;
            printf("%d
    ",s);
    
        }
    
        return 0;
    }
    /**
    
    */
    
    
  • 相关阅读:
    查看crontab的日志记录定位定时任务问题
    Latex 表格内公式换行方法
    C语言中qsort函数用法
    7 种常用的排序算法-视觉直观感受
    Ubuntu下如何安装YouCompleteMe插件
    Linux下非root用户如何安装软件
    系统进化树-原理介绍及软件使用
    LaTeX 页眉页脚的设置
    TEXshade教程- 多重比对着色软件包
    easyUI自带的时间插件日期选择、月份选择、时间选择的使用
  • 原文地址:https://www.cnblogs.com/dccmmtop/p/6708381.html
Copyright © 2011-2022 走看看