分酒问题（DFS解法）

import java.util.ArrayList;
import java.util.LinkedList;
import java.util.Queue;
import java.util.Scanner;

//已知有3个容量分别为3kg，5kg和8kg且没有刻度的酒瓶3kg和5kg的瓶子均装满了酒。而8kg的瓶子为空。
//现要求仅用这3个酒瓶将这些酒均分为两个4kg，并分别装入5kg和8kg的瓶子中。

public class DispensingProblem {
    public static int N;  //酒瓶的数量
    public static Integer[] bottleMax;  //每个酒瓶最多装多少酒
    public static Integer[] bottleCurrent;  //每个酒瓶现在装了多少酒
    public static Integer[] bottleFinal;   //标注每个酒瓶的最终状态
    public static ArrayList<String> states;   //标注每一种状态，防止重复(每一种状态是一个向量，我用整型数组表示)
    public static ArrayList<String[]> paths;   //标注每一条路径
    public static int caseNum = 0;   //标注是第几种方法
    public static int minNum = 1000000;  //标注最少需要的步数
    public static void dfs(int n){
        String testS = String.valueOf(bottleCurrent[0]);
        for(int m =1;m<N;m++)
            testS+=String.valueOf(bottleCurrent[m]);
        boolean flag = true;
        for(int k = 0;k<N;k++)
            if(bottleCurrent[k]!=bottleFinal[k])
                flag = false;
        if(flag){
            caseNum ++ ;
            if(n <= minNum){
                if(n<minNum){
                    paths.clear();    
                }
                String[] tempS= new String[states.size()];
                for(int ll = 0;ll<states.size();ll++)
                    tempS[ll] = states.get(ll);
                paths.add(tempS);
                minNum = n;
            }
            System.out.println("第"+caseNum+"种方法：");
            for(int l=0;l<states.size();l++)
                System.out.println(states.get(l));
            System.out.println("总共需要移动"+n+"步");
            System.out.println("------------------------------------");
            return;
        }
        //找出当前可能的所有移动
        //数据不大，不需要优化
        //每个瓶子只能倒满或者倒空
        //注意要标注每一种状态，防止状态重复
        for(int i = 0 ; i < N;i++)
            for(int j = 0; j < N ;j++){
                if(i==j)
                    continue;
                //从i瓶把所有酒倒入j瓶
                int nI = bottleCurrent[i];
                int nJ = bottleCurrent[j];
                int temp = nI + nJ - bottleMax[j];
                if(temp<=0){
                    //能全倒进去
                    bottleCurrent[i] = 0;
                    bottleCurrent[j] = nI + nJ;
                    String s = String.valueOf(bottleCurrent[0]);
                    for(int m =1;m<N;m++)
                        s+=String.valueOf(bottleCurrent[m]);
                    if(!states.contains(s)){
                        states.add(s);
                        dfs(n+1);
                        //回溯
                        states.remove(states.indexOf(s));
                    }
                    //回溯
                    bottleCurrent[i] = nI;
                    bottleCurrent[j] = nJ;
                }
                else{
                    //不能全倒进去
                    bottleCurrent[i] = temp;
                    bottleCurrent[j] = bottleMax[j];
                    String s = String.valueOf(bottleCurrent[0]);
                    for(int m =1;m<N;m++)
                        s+=String.valueOf(bottleCurrent[m]);
                    if(!states.contains(s)){
                        states.add(s);
                        dfs(n+1);
                        //回溯
                        states.remove(states.indexOf(s));
                    }
                    //回溯
                    bottleCurrent[i] = nI;
                    bottleCurrent[j] = nJ;
                }
            }
        //找遍所有状态都不可行，则表明不能出现这种状态
        //System.out.println("不可能存在这种状态！");
    }
    public static void main(String[] args){
        Scanner sc = new Scanner(System.in);
        System.out.println("请输入瓶子个数");
        N = sc.nextInt();
        bottleMax = new Integer[N];
        bottleCurrent = new Integer[N];
        bottleFinal = new Integer[N];
        states = new ArrayList<String>();
        paths = new ArrayList<String[]>();
        System.out.println("请输入每个瓶子的容量");
        for(int i = 0 ;i < N; i ++){
            bottleMax[i] = sc.nextInt();
        }
        System.out.println("请输入每个瓶子当前有多少酒");
        for(int i = 0 ;i < N; i ++){
            bottleCurrent[i] = sc.nextInt();
        }
        System.out.println("请输入最终希望每个瓶子有多少酒");
        for(int i = 0 ;i < N; i ++){
            bottleFinal[i] = sc.nextInt();
        }
        String s = String.valueOf(bottleCurrent[0]);
        for(int i =1;i<N;i++)
            s+=String.valueOf(bottleCurrent[i]);
        states.add(s);
        dfs(0);
        System.out.println("******************************");
        System.out.println("最少需要"+minNum+"步");
        int index = 0;
        for(int i = 0;i<paths.size();i++){
            index++;
            System.out.println("第"+index+"种最短方法：");
            String[] temp = paths.get(i);
            for(int j = 0;j<temp.length;j++)
                System.out.println(temp[j]);
            System.out.println("-----------------------------------");
        }
        System.out.println("******************************");
    }
}