此为《算法的兴趣》读书笔记，我用javascript从新完成算法。

敝人鄙见

此题作者完成得过于庞杂，我将初始状况定义为：[3,3,0,0,true]，释义：顺次示意，彼岸僧人数目、彼岸妖怪数目、彼岸僧人数目、彼岸妖怪数目、船在彼岸否。有了以上定义，完全可以将这个题目算作与上一章桶等分水谁人题目是一样的题目，两岸是两个“桶”，僧人和妖怪是”水”，”水”在两个”桶“中返来倒，末了悉数倒到彼岸谁人桶中。

题目形貌

有三个僧人和三个妖怪要应用唯一的一条小船过河，这条小船一次只能载两个人，同时，无论是在河的两岸照样在船上，只需妖怪的数目大于僧人的数目，妖怪们就会将僧人吃掉。如今须要挑选一种过河的部署，保证僧人和妖怪都能过河且僧人不能被妖怪吃掉。

变量定义

var states = [[3,3,0,0,true]];          //初值，递次为：当地僧人、妖怪;对岸僧人、妖怪、船在彼岸        
var IsLocal = true;                     //是不是在彼岸，是为真，在对岸为假

检测搭船部署是不是可行（倒水要领合理？）

function CanTakeDumpAction(curr,local,from,to){
    //检测船上，僧人数目大于即是妖怪或许僧人为零且总数为1或2
    if((from >= to || from === 0 && to > 0) && (from + to <= 2) && (from + to > 0)){
        if(local){            //彼岸与彼岸是差别的
            //船过岸后，两岸都要满足要么僧人为0,要么僧人数目大于即是妖怪
            if((curr[0] >= from && curr[1] >= to && (curr[0] - from == 0 || curr[0] - from >= curr[1] - to)) && (curr[2] + from == 0 || curr[2] + from >= curr[3] + to)){
                return true;
            }
        }else{
            if((curr[2] >= from && curr[3] >= to && (curr[2] - from == 0 || curr[2] - from >= curr[3] - to)) && (curr[0] + from == 0 || curr[0] + from >= curr[1] + to)){
                return true;
            }
        }
    }
    return false;
}

船到岸后（过河）操纵（倒水）

function DumpWater(curr,local,from,to){
    var next = curr.slice();       
    if(local){        //彼岸与彼岸有差别的操纵
        next[0] -= from;
        next[1] -= to;
        next[2] += from;
        next[3] += to;
    }else{
        next[0] += from;
        next[1] += to;
        next[2] -= from;
        next[3] -= to;
    }
    next[4] = !local    //船到对岸
    return next;
}

检测状况是不是涌现过

这个函数是保证不会进入死循环。

function IsStateExist(state){
    for(var i = 0; i < states.length; i++){
        if(state[0] == states[i][0] && state[1] == states[i][1] && state[2] == states[i][2] && state[3] == states[i][3] && state[4] == states[i][4]){
            return true;
        }
    }
    return false;
}

状况搜刮主函数

(function SearchState(states,local){
    var curr = states[states.length - 1];              //取初始状况
    if(curr[2] == 3 && curr[3] == 3){                  //找到解   
        var rs = ''
        states.forEach(function(al){
            rs += al.join(',') + ' -> ';
        });
        console.log(rs.substr(0,rs.length - 4))
    }

    for(var i = 0; i < 3; i++){                         //i示意搭船的僧人数目，0~2
        for(var j = 0; j < 3; j++){                     //j示意搭船的妖怪数目，0~2
            if(CanTakeDumpAction(curr,local,i,j)){      //搭船部署合理
                var next = DumpWater(curr,local,i,j);   //过河
                if(!IsStateExist(next)){       
                    states.push(next);
                    SearchState(states,!local);
                    states.pop();
                }
            }
        }
    }
})(states,IsLocal);

四组效果

3,3,0,0,true -> 3,1,0,2,false -> 3,2,0,1,true -> 3,0,0,3,false -> 3,1,0,2,true -> 1,1,2,2,false -> 2,2,1,1,true -> 0,2,3,1,false -> 0,3,3,0,true -> 0,1,3,2,false -> 0,2,3,1,true -> 0,0,3,3,false
3,3,0,0,true -> 3,1,0,2,false -> 3,2,0,1,true -> 3,0,0,3,false -> 3,1,0,2,true -> 1,1,2,2,false -> 2,2,1,1,true -> 0,2,3,1,false -> 0,3,3,0,true -> 0,1,3,2,false -> 1,1,2,2,true -> 0,0,3,3,false
3,3,0,0,true -> 2,2,1,1,false -> 3,2,0,1,true -> 3,0,0,3,false -> 3,1,0,2,true -> 1,1,2,2,false -> 2,2,1,1,true -> 0,2,3,1,false -> 0,3,3,0,true -> 0,1,3,2,false -> 0,2,3,1,true -> 0,0,3,3,false
3,3,0,0,true -> 2,2,1,1,false -> 3,2,0,1,true -> 3,0,0,3,false -> 3,1,0,2,true -> 1,1,2,2,false -> 2,2,1,1,true -> 0,2,3,1,false -> 0,3,3,0,true -> 0,1,3,2,false -> 1,1,2,2,true -> 0,0,3,3,false