• Bridge over a rough river


    Description

    A group of N travelers (1 ≤ N ≤ 50) has approached an old and shabby bridge and wishes to cross the river as soon as possible. However, there can be no more than two persons on the bridge at a time. Besides it's necessary to light the way with a torch for safe crossing but the group has only one torch.

    Each traveler needs ti seconds to cross the river on the bridge; i=1, ... , N (ti are integers from 1 to 100). If two travelers are crossing together their crossing time is the time of the slowest traveler.

    The task is to determine minimal crossing time for the whole group.

    Input

    The input consists of two lines: the first line contains the value of N and the second one contains the values of ti (separated by one or several spaces).

    Output

    The output contains one line with the result.

    Sample Input

    4 6 7 6 5

    Sample Output

    29

    Source
    Northeastern Europe 2001, Western Subregion

    题目的意思是说,几个人过桥(有的题目是用船过河),因为桥很窄,每次只能有两个人通过,因为看不清,还要用一火把照明(总共只有一个火把),两个人同时过桥的时候,花费的时间为两个人中速度最慢的那个人花费的时间.现给出N个人的单独过桥时需花费的时间,求出这N个人全部过桥要花费多长时间...

    /*最佳方案构造:以下是构造N个人(N≥1)过桥最佳方案的方法:  
    1) 如果N=1、2,所有人直接过桥。  
    2) 如果N=3,由最快的人往返一次把其他两人送过河。  
    3) 如果N≥4,设A、B为走得最快和次快的旅行者,过桥所需时间分别为a、b;
     而Z、Y为走得最慢和次慢的旅行者,过桥所需时间分别为z、y。那么    
     当2b>a+y时,使用模式一将Z和Y移动过桥;    
     当2b<a+y时,使用模式二将Z和Y移动过桥;    
     当2b=a+y时,使用模式一将Z和Y移动过桥。
    这样就使问题转变为N-2个旅行者的情形,从而递归解决之。
         ……         
        A Z →          
        A ←          
        ……
    也就是“由A护送到对岸,A返回”,称作“模式一”。

        ……          
        ……  
    第n-2步:   A B →   
    第n-1步:    A ←    
    第n步:     Y Z →  
    第n+1步:    B ←          
        ……
    这个模式是“由A和B护送到对岸,A和B返回”,称作“模式二”。
    http://www.oursci.org/magazine/200204/020411-01.htm

      1 import java.util.HashMap;
      2 import java.util.Iterator;
      3 import java.util.Map;
      4 
      5 public class BridgeAcross {
      6 
      7     static Map humanInfo;
      8 
      9     public BridgeAcross() {
     10         Map<String, Integer> humanInfo = new HashMap<String, Integer>();
     11         humanInfo.put("name1"1);
     12         humanInfo.put("name2"2);
     13         humanInfo.put("name3"5);
     14         humanInfo.put("name4"7);
     15         this.humanInfo = humanInfo;
     16     }
     17 
     18     public static Map getHumanInfo() {
     19         return humanInfo;
     20     }
     21 
     22     public void setHumanInfo(Map humanInfo) {
     23         this.humanInfo = humanInfo;
     24     }
     25 
     26     public static int minTime(int[] times) {
     27         int[] time_sorted = sort(times);
     28         return search(time_sorted);
     29     }
     30 
     31     public static int search(int[] times) {
     32         int n = times.length;
     33         if (n == 1)
     34             return times[0];
     35         else if (n == 2) {
     36             System.out.println("-->" + getHuman(times[0]) + "+"
     37                     + getHuman(times[1]));
     38             return times[1];
     39         } else if (n == 3) {
     40             System.out.println("-->" + getHuman(times[0]) + "+"
     41                     + getHuman(times[1]));
     42             System.out.println("<--" + getHuman(times[0]));
     43             System.out.println("-->" + getHuman(times[0]) + "+"
     44                     + getHuman(times[2]));
     45             return (times[0+ times[1+ times[2]);
     46         } else {
     47             int[] time_temp = remainingTime(times);
     48             if (2 * times[1> times[0+ times[n - 2]) {
     49                 System.out.println("-->" + getHuman(times[0]) + "+"
     50                         + getHuman(times[1]));
     51                 System.out.println("<--" + getHuman(times[0]));
     52                 System.out.println("-->" + getHuman(times[0]) + "+"
     53                         + getHuman(times[n - 1]));
     54                 System.out.println("<--" + getHuman(times[0]));
     55                 System.out.println("-->" + getHuman(times[0]) + "+"
     56                         + getHuman(times[n - 2]));
     57                 System.out.println("<--" + getHuman(times[0]));
     58                 return 2 * times[0+ times[n - 1+ times[n - 2]
     59                         + search(time_temp);
     60             } else {
     61                 System.out.println("-->" + getHuman(times[0]) + "+"
     62                         + getHuman(times[1]));
     63                 System.out.println("<--" + getHuman(times[0]));
     64                 System.out.println("-->" + getHuman(times[n - 1]) + "+"
     65                         + getHuman(times[n - 2]));
     66                 System.out.println("<--" + getHuman(times[1]));
     67                 // System.out.println("-->" + getHuman(times[0]) + "+" +
     68                 // getHuman(times[1]));
     69                 return 2 * times[1+ times[n - 1+ times[0]
     70                         + search(time_temp);
     71             }
     72         }
     73     }
     74 
     75     public static int[] sort(int[] times) {
     76         int temp = 0;
     77         for (int k = 0; k < times.length - 1; k++) {
     78             for (int i = 0; i < times.length - 1; i++) {
     79                 if (times[i + 1< times[i]) {
     80                     temp = times[i + 1];
     81                     times[i + 1= times[i];
     82                     times[i] = temp;
     83                 }
     84             }
     85         }
     86         return times;
     87     }
     88 
     89     public static int[] remainingTime(int[] times) {
     90         int[] sub_time = new int[times.length - 2];
     91         for (int i = 0; i < sub_time.length; i++) {
     92             sub_time[i] = times[i];
     93         }
     94         return sub_time;
     95     }
     96 
     97     public static String getHuman(int time) {
     98         Map map = getHumanInfo();
     99         Iterator iterator = map.entrySet().iterator();
    100         while (iterator.hasNext()) {
    101             Map.Entry emtry = (Map.Entry) iterator.next();
    102             int value = (Integer) emtry.getValue();
    103             if (time == value) {
    104                 return (String) emtry.getKey();
    105             }
    106         }
    107         return "";
    108     }
    109 
    110     public static int[] getTimes() {
    111         int[] times = new int[getHumanInfo().size()];
    112         int i = 0;
    113         Iterator iterator = getHumanInfo().entrySet().iterator();
    114         while (iterator.hasNext()) {
    115             Map.Entry emtry = (Map.Entry) iterator.next();
    116             times[i++= (Integer) emtry.getValue();
    117         }
    118         return times;
    119     }
    120 
    121     /**
    122      * @param args
    123      */
    124     public static void main(String[] args) {
    125         BridgeAcross bridge = new BridgeAcross();
    126         int[] times = bridge.getTimes();
    127         System.out.println(minTime(times));
    128     }
    129 
    130 }
    131 
     

  • 相关阅读:
    Javascript对象原型prototype和继承
    Javascript(js)使用function定义构造函数
    phpExcel中文帮助手册
    php curl_init函数用法
    nginx启动,重启,关闭命令
    Nginx配置文件详细说明
    Nginx 简单的负载均衡配置示例
    MySQL数据库的同步配置
    ucenter home 视频增加缩略图
    web分词程序和算法
  • 原文地址:https://www.cnblogs.com/kelin1314/p/1907574.html
Copyright © 2020-2023  润新知