zoukankan      html  css  js  c++  java
  • HDOJ1548 A strange lift[BFS()]

    A strange lift

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
    Total Submission(s): 6056    Accepted Submission(s): 2238


    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 is on floor A,and you want to go to floor B,how many times at least he havt 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
     
    Recommend
    8600
     
     
     
     
     
     
    code:
     1 #include <iostream>   
     2 #include <iomanip>   
     3 #include <fstream>   
     4 #include <sstream>   
     5 #include <algorithm>   
     6 #include <string>   
     7 #include <set>   
     8 #include <utility>   
     9 #include <queue>   
    10 #include <stack>   
    11 #include <list>   
    12 #include <vector>   
    13 #include <cstdio>   
    14 #include <cstdlib>   
    15 #include <cstring>   
    16 #include <cmath>   
    17 #include <ctime>   
    18 #include <ctype.h> 
    19 using namespace std;
    20 
    21 int n,a,b;
    22 int vst[202];
    23 int table[202][202];
    24 
    25 struct node
    26 {
    27     int floor;
    28     int step;
    29 }Node;
    30 
    31 void bfs()
    32 {
    33     int i;
    34     node pre,last;
    35     pre.floor=a;
    36     pre.step=0;
    37     vst[a]=1;
    38     bool flag=false;
    39     queue<node>Que;
    40     Que.push(pre);
    41     while(!Que.empty())
    42     {
    43         pre=Que.front();
    44         Que.pop();
    45         if(pre.floor==b)
    46         {
    47             flag=true;
    48             break;
    49         }
    50         if(pre.step>=n)
    51             break;
    52         for(i=1;i<=n;i++)
    53         {
    54             last.floor=i;
    55             last.step=pre.step+1;
    56             if(table[pre.floor][last.floor]&&vst[last.floor]==0)
    57             {
    58                 vst[last.floor]=1;
    59                 Que.push(last);
    60             }
    61         }
    62     }
    63     if(!flag)
    64     {
    65         printf("-1\n");
    66         return;
    67     }
    68     printf("%d\n",pre.step);
    69 }
    70 
    71 int main()
    72 {
    73     int i;
    74     int temp;
    75     while(~scanf("%d",&n),n)
    76     {
    77         scanf("%d%d",&a,&b);
    78         memset(table,0,sizeof(table));
    79         memset(vst,0,sizeof(vst));
    80         for(i=1;i<=n;i++)
    81         {
    82             scanf("%d",&temp);
    83             if((i-temp)>0)
    84                 table[i][i-temp]=1;
    85             if((i+temp)<=n)
    86                 table[i][i+temp]=1;
    87         }
    88         if(a==b)
    89         {
    90             printf("0\n");
    91             continue;
    92         }
    93         bfs();
    94     }
    95     return 0;
    96 }
  • 相关阅读:
    NOIP 2011 提高组 计算系数(vijos 1739)(方法:二项式定理)
    NOIP 2012 提高组 借教室(vijos 1782) 总览
    NOIP 2012 提高组 借教室(vijos 1782) 线段树85分打法
    NOIP 2011 提高组 铺地毯(vijos 1736)(方法:纯枚举)
    获取指定时间的前一天、后一天及当前时间的前一周、前一个月
    input file禁用手机本地文件选择,只允许拍照上传图片
    给定一个时间,获取该时间所在周的周一及周日
    Fiddler手机抓包软件简单使用将h5效果显示在手机
    解决Linux服务器更换IP后,ssh连接被拒绝问题
    解决Hadoop启动报错:File /opt/hadoop/tmp/mapred/system/jobtracker.info could only be replicated to 0 nodes, instead of 1
  • 原文地址:https://www.cnblogs.com/XBWer/p/2615909.html
Copyright © 2011-2022 走看看