修道士和野人问题
休闲时刻看看神经网络方面的书，发现了修道士和野人的问题，不禁勾引起我写算法的欲望，曾经的三只大老虎三只小老虎过河问题、人狼羊白菜过河问题、汉诺塔、哈夫曼等等各种算法瞬间在脑海中约隐约现，修道士和野人问题我以前好像没有解开，中午吃饭的时候在脑海中重新构造思路，下午耗了点时间把它干掉。（算法不在代码里，而在思想中；所以尽量不要看我的代码，而要仔细分析我写的思路）
题目：
设有３个修道士和３个野人来到河边， 打算用一条船从河的左岸渡到河的右岸。但该船每次只能装载两个人， 在任何岸边野人的数目都不得超过修道士的人数， 否则修道士就会被野人吃掉。假设野人服从任何一种过河安排， 请问如何规划过河计划才能把所有人都安全地渡过河去。
首先考虑总共有（3+1）*（3+1）= 16 种不同的状态(因为左岸可以有0，1，2，3个传教士，左岸可以有0，1，2，3个野人)，所以可以考虑使用穷举法。
使用如下C#程序语言：
int MaxNum = 3; for (int monk = MaxNum; monk >= 0; monk--) { for (int savage = MaxNum; savage >= 0; savage--) { Console.Write("{{" + monk + "," + savage + "},{" + (MaxNum - monk) + "," + (MaxNum - savage) + "}} "); } Console.Write("\n"); }
生成16种状态图↓↓↓↓↓↓↓↓↓↓↓
状态图含义：
{a,b}：a，左岸修道士数量；b，左岸野人数量。
--------仅考虑左岸传教士和野蛮人数量(所有状态图)------------------------ {3,3} {3,2} {3,1} {3,0} {2,3} {2,2} {2,1} {2,0} {1,3} {1,2} {1,1} {1,0} {0,3} {0,2} {0,1} {0,0}
其中{3,3}是起始状态图；{0,0}是终止状态图。
去除在任何岸边野人的人数超过修道士的人数，保留所有可满足的状态图：
--------仅考虑左岸传教士和野蛮人数量(符合要求状态图)---------------------- {3,3} {3,2} {3,1} {3,0} {2,2} {1,1} {0,3} {0,2} {0,1} {0,0}
可知，左岸修道士和野人的总数是先减少、再增加、再减少、再增加...直到左岸的总人数为零(这个由船行动的方向决定的，请仔细思考)，所求的最终状态图 {0,0}，而且每次增减人数最多为2个（条件提示：该船每次只能装载两个人），由此可知左岸总人数的增减趋势是： -1或-2、+1或+2、-1或-2、+1或+2...。
左岸人头数作为参照：
--------仅考虑左岸传教士和野蛮人数量（分组）------------------------ heads num：6 {3,3} heads num：5 {3,2} heads num：4 {3,1} {2,2} heads num：3 {3,0} {0,3} heads num：2 {1,1} {0,2} heads num：1 {0,1} heads num：0 {0,0}
将船的停靠岸考虑在内，其中L、R分别表示船在左岸或右岸：
--------仅考虑左岸传教士和野蛮人数量,以及船的状态------------------------
heads num：6 {3,3,L}
heads num：5 {3,2,L} {3,2,R}
heads num：4 {3,1,L} {3,1,R} {2,2,L} {2,2,R}
heads num：3 {3,0,L} {3,0,R} {0,3,L} {0,3,R}
heads num：2 {1,1,L} {1,1,R} {0,2,L} {0,2,R}
heads num：1 {0,1,L} {0,1,R}
heads num：0 {0,0,R}
此时，初始状态图为 {3,3,L}，终止状态图为{0,0,R}。
其实讲到这里，几乎答案已经很明显了。上面已经提到：左岸总人数的增减趋势是： -1或-2、+1或+2、-1或-2、+1或+2...，假设每种状态图只能使用一次，按照-1或-2、+1或+2的循环模式的增长趋势，可得到如下几种走法：
--------可能的路径----------------------------------------
{3,3,L} {3,1,R} {3,2,L} {3,0,R} {3,1,L} {1,1,R} {2,2,L} {0,2,R} {0,3,L} {0,1,R} {0,2,L} {0,0,R}
{3,3,L} {3,1,R} {3,2,L} {3,0,R} {3,1,L} {1,1,R} {2,2,L} {0,2,R} {0,3,L} {0,1,R} {1,1,L} {0,0,R}
{3,3,L} {2,2,R} {3,2,L} {3,0,R} {3,1,L} {1,1,R} {2,2,L} {0,2,R} {0,3,L} {0,1,R} {0,2,L} {0,0,R}
{3,3,L} {2,2,R} {3,2,L} {3,0,R} {3,1,L} {1,1,R} {2,2,L} {0,2,R} {0,3,L} {0,1,R} {1,1,L} {0,0,R}
using System; using System.Collections.Generic; using System.Linq; using System.Text; using System.Threading.Tasks; namespace ConsoleApplication1 { public enum State { Left, Right } public class RiverBank { public State state = State.Left; public int monk = 0; public int savage = 0; public bool used = false; } public class Item { public int index = 0; public List<RiverBank> list = new List<RiverBank>(); } class Program { const int MaxNum = 3; static State state = State.Left; public State GetState { get { State r = state; state = (state == State.Left ? State.Right : State.Left); return r; } } static void Add(List<Item> list, int index, params RiverBank[] objs) { foreach (Item i in list) { if (i.index == index) { foreach (RiverBank ii in objs) { bool flag = true; for (int iii = 0; iii < i.list.Count; iii++) { if (i.list[iii].monk == ii.monk && i.list[iii].monk == ii.monk && i.list[iii].state == ii.state) { flag = false; break; } } if (flag) { i.list.Add(ii); } } return; } } Item item = new Item(); item.index = index; item.list.AddRange(objs); list.Add(item); } static void WriteDivInfo(String msg) { Console.WriteLine("\n--------" + msg.PadRight(45, '-')); } static void WriteInfo(List<RiverBank> riverBankList) { foreach (var i in riverBankList) { Console.Write("{" + i.monk + "," + i.savage + "," + (i.state == State.Right ? "R" : "L") + "} "); } Console.WriteLine(); } static void Main(string[] args) { List<Item> list = new List<Item>(); WriteDivInfo("仅考虑左岸传教士和野蛮人数量(所有状态图)"); for (int monk = MaxNum; monk >= 0; monk--) { for (int savage = MaxNum; savage >= 0; savage--) { Console.Write("{" + monk + "," + savage + "}\t"); } Console.Write("\n"); } WriteDivInfo("仅考虑左岸传教士和野蛮人数量(符合要求状态图)"); for (int monk = MaxNum; monk >= 0; monk--) { for (int savage = MaxNum; savage >= 0; savage--) { if (monk == 0 || monk == MaxNum || monk == savage) { int index = monk + savage; //RiverBank p = new RiverBank { monk = monk, savage = savage }; //Add(list, index, p); if (index > 0) { RiverBank p = new RiverBank { monk = monk, savage = savage, state = State.Left }; Add(list, index, p); } if (index < 2 * MaxNum) { RiverBank p = new RiverBank { monk = monk, savage = savage, state = State.Right }; Add(list, index, p); } Console.Write("{" + monk + "," + savage + "}\t"); } else { Console.Write(" \t"); } } Console.Write("\n"); } WriteDivInfo("仅考虑左岸传教士和野蛮人数量,以及船的状态"); foreach (Item item in list) { Console.Write("heads num：" + item.index + "\t"); foreach (RiverBank i in item.list) { //Console.Write("{" + i.monk + "," + i.savage + "} "); Console.Write("{" + i.monk + "," + i.savage + "," + (i.state == State.Right ? "R" : "L") + "} "); } Console.Write("\n"); } list.Sort((a, b) => { return a.index < b.index ? 1 : -1; }); WriteDivInfo("可能的路径"); List<RiverBank> usedList = new List<RiverBank>(); list[0].list[0].used = true; usedList.Add(list[0].list[0]); Cycle(ref list, ref usedList); Console.ReadKey(); } private static void Cycle(ref List<Item> list, ref List<RiverBank> usedList) { RiverBank item = usedList[usedList.Count - 1]; State state = item.state; int monk = item.monk; int savage = item.savage; for (int i = 1; i <= 2; i++) {//±1或±2 for (int j = 0; j <= i; j++) {//每次变化的传教士数 int k = i - j;//每次变化的野人数 int value = (state == State.Left ? -1 : 1); int m = monk + j * value; int s = savage + k * value; if (m == 0 && s == 0 && state == State.Left) { list[6].list[0].used = true; usedList.Add(list[6].list[0]); WriteInfo(usedList); list[6].list[0].used = false; usedList.Remove(list[6].list[0]); } else if (m >= 0 && s >= 0 && m <= MaxNum && s <= MaxNum) { foreach (var it in list) { foreach (var v in it.list) { if (v.used == false && v.state != state && v.monk == m && v.savage == s) { v.used = true; usedList.Add(v); Cycle(ref list, ref usedList); v.used = false; usedList.Remove(v); } } } } } } } } }
