递归（SRM144D2）

Problem Statement

You work for an electric company, and the power goes out in a rather large apartment complex with a lot of irate tenants. You isolate the problem to a network of sewers underneath the complex with a step-up transformer at every junction in the maze of ducts. Before the power can be restored, every transformer must be checked for proper operation and fixed if necessary. To make things worse, the sewer ducts are arranged as a tree with the root of the tree at the entrance to the network of sewers. This means that in order to get from one transformer to the next, there will be a lot of backtracking through the long and claustrophobic ducts because there are no shortcuts between junctions. Furthermore, it's a Sunday; you only have one available technician on duty to search the sewer network for the bad transformers. Your supervisor wants to know how quickly you can get the power back on; he's so impatient that he wants the power back on the moment the technician okays the last transformer, without even waiting for the technician to exit the sewers first.
You will be given three int[]'s: fromJunction, toJunction, and ductLength that represents each sewer duct. Duct i starts at junction (fromJunction[i]) and leads to junction (toJunction[i]). ductlength[i] represents the amount of minutes it takes for the technician to traverse the duct connecting fromJunction[i] and toJunction[i]. Consider the amount of time it takes for your technician to check/repair the transformer to be instantaneous. Your technician will start at junction 0 which is the root of the sewer system. Your goal is to calculate the minimum number of minutes it will take for your technician to check all of the transformers. You will return an int that represents this minimum number of minutes.
Definition

Class:
PowerOutage
Method:
estimateTimeOut
Parameters:
int[], int[], int[]
Returns:
int
Method signature:
int estimateTimeOut(int[] fromJunction, int[] toJunction, int[] ductLength)
(be sure your method is public)


Constraints
-
fromJunction will contain between 1 and 50 elements, inclusive.
-
toJunction will contain between 1 and 50 elements, inclusive.
-
ductLength will contain between 1 and 50 elements, inclusive.
-
toJunction, fromJunction, and ductLength must all contain the same number of elements.
-
Every element of fromJunction will be between 0 and 49 inclusive.
-
Every element of toJunction will be between 1 and 49 inclusive.
-
fromJunction[i] will be less than toJunction[i] for all valid values of i.
-
Every (fromJunction[i],toJunction[i]) pair will be unique for all valid values of i.
-
Every element of ductlength will be between 1 and 2000000 inclusive.
-
The graph represented by the set of edges (fromJunction[i],toJunction[i]) will never contain a loop, and all junctions can be reached from junction 0.
Examples
0)


{0}
{1}
{10}
Returns: 10
The simplest sewer system possible. Your technician would first check transformer 0, travel to junction 1 and check transformer 1, completing his check. This will take 10 minutes.
1)


{0,1,0}
{1,2,3}
{10,10,10}
Returns: 40
Starting at junction 0, if the technician travels to junction 3 first, then backtracks to 0 and travels to junction 1 and then junction 2, all four transformers can be checked in 40 minutes, which is the minimum.
2)


{0,0,0,1,4}
{1,3,4,2,5}
{10,10,100,10,5}
Returns: 165
Traveling in the order 0-1-2-1-0-3-0-4-5 results in a time of 165 minutes which is the minimum.
3)


{0,0,0,1,4,4,6,7,7,7,20}
{1,3,4,2,5,6,7,20,9,10,31}
{10,10,100,10,5,1,1,100,1,1,5}
Returns: 281
Visiting junctions in the order 0-3-0-1-2-1-0-4-5-4-6-7-9-7-10-7-8-11 is optimal, which takes (10+10+10+10+10+10+100+5+5+1+1+1+1+1+1+100+5) or 281 minutes.
4)


{0,0,0,0,0}
{1,2,3,4,5}
{100,200,300,400,500}
Returns: 2500
div2级别1100分题，我没有做出来，拿到论坛上跟大家讨论了一下，其中bigbug9002给出了一个很好的解答，学到不少东西以下是他的代码：

import java.util.*;
public class PowerOutage{
    static int min=Integer.MAX_VALUE;  //最小路径
    static int junctionCount=0;        //总的顶点数
    static List<Integer> minPath=null;   //路径。
    public static void main(String[] args){
        /*int[] fromJunction={0,1,0};
        int[] toJunction={1,2,3};
        int[] ductLength={10,10,10};
        int[] fromJunction={0};
        int[] toJunction={1};
        int[] ductLength={10};
        int[] fromJunction={0,0,0,1,4};
        int[] toJunction={1,3,4,2,5};
        int[] ductLength={10,10,100,10,5};*/
        int[] fromJunction={0,0,0,1,4,4,6,7,7,7,8};
        int[] toJunction={1,3,4,2,5,6,7,8,9,10,11};
        int[] ductLength={10,10,100,10,5,1,1,100,1,1,5};
        System.out.println("Min Time:"+estimateTimeOut(fromJunction,toJunction,ductLength));
        System.out.println(""+minPath);
    }
    public static int estimateTimeOut(int[] fromJunction, int[] toJunction, int[] ductLength) {
        Set<Integer> jun=new TreeSet<Integer>();
        for(int i=0;i<fromJunction.length;i++){
            jun.add(fromJunction[i]);
        }
        for(int i=0;i<fromJunction.length;i++){
            jun.add(toJunction[i]);
        }
        junctionCount=jun.size();  //上面的两个循环就是为了得到总的节点数（顶点数）
        int start=0;
        List<Integer> path=new ArrayList<Integer>();     
        path.add(0);
        int[] flag=new int[junctionCount];
        flag[0]=1;
        travel(start,fromJunction,toJunction,ductLength,path,0);
        return min;
    }
    /*@param start 开始遍历的顶点。
    * @param count 计数遍历过的顶点数
    * @param path  存放路径
    * @param cost  路径长
    * @param plag  标识数组，遍历过的标识为1
    * 递归调用
    */
    static void  travel(int start,int[] from,int[] to,int[] ductLength,List<Integer> path,int cost){
        if(transferAll(path)){         //得到结果
            if(cost<min){                 //如果比最小的路径长还小
                min=cost;
                minPath=new ArrayList<Integer>(path);
                return ;
            }
        }
        List <Integer> adj=new ArrayList<Integer>();      //把所有的和start邻接的点及路径长都放在adj中.
        for(int i=0;i<from.length;i++){
            if(from[i]==start){
                 adj.add(to[i]);
                 adj.add(ductLength[i]);     
            }
        }
        for(int i=0;i<to.length;i++){
            if(to[i]==start){
                 adj.add(from[i]);
                 adj.add(ductLength[i]);
            }
        }

        Iterator<Integer> it=adj.iterator();
        while(it.hasNext()){
            int next=it.next();
            if(!isCycle(path,start,next)){                 //下一步不会构成环
                //path=new ArrayList<Integer>(path);
                path.add(next);                            //加入路径
                travel(next,from,to,ductLength,path,cost+it.next());
                path.remove(path.size()-1);
            }else{
                it.next();
            }
        }
    }
    static boolean isCycle(List<Integer> path,int one,int nextone){  //从one 到nextone是不是走过,并且是环的一部分.
        Object[] pathArray=path.toArray();
        if(pathArray.length<2) return false;
        int p=0,q=p+1;
        while(q<pathArray.length){
            if((Integer)(pathArray[p])==one&&(Integer)(pathArray[q])==nextone) return true;
            p=q;
            q++;
        }
        return false;
    }
    static boolean transferAll(List<Integer> path){  //在路径中是不是包含了全部的顶点
        Set<Integer> jun=new HashSet<Integer>(path);
        if(jun.size()==junctionCount) return true;
        return false;
    }

}

从两点总结我的收获：

1. 递归，学了几年编程看到递归心里还是没底。这段代码主要是借助递归完成对树的深度遍历，其中start邻接点数组的建立感觉很有意思，把权值也加入了进去，在递归的函数调用时通过travel(next,from,to,ductLength,path,cost+it.next())来跳到下一个节点。如果不这样设置的话应该要另外建一个数组或将原数组改为二维，肯定会增加复杂度。

另外，递归过程中加入了对圈的判定，也是比较严谨。

2. 使用Hashset来储存路径避免重复现象，感觉很有用。