初始化地图

function initMaze(r,c){let row = new Array(2 * r + 1)for(let i = 0; i < row.length; i++){let column = new Array(2 * c + 1)row[i] = columnfor(let j = 0; j < column.length; j++){row[i][j] = 1}}for(let i = 0; i < r; i++){for(let j = 0; j < c; j++){row[2 * i + 1][2 * j + 1] = 0}}console.log(row)
}initMaze(3,3)

计算二维数组坐标位置

let arr = [[0,0,0],[0,0,0],[0,0,0]
]for(let i = 0; i < 9; i++){let row = Math.floor(i / 3)let column = i % 3arr[row][column] = false
}console.log(arr)

偏移量方向预制

let offset = [{x:-1,y:0},{x:1,y:0},{x:0,y:-1},{x:0,y:1}
]let x = 0
let y = 0for(let i = 0; i < offset.length; i++){x = x + offset[i].xy = y + offset[i].yconsole.log(x,y)
}

随机数公式

1.　　0-x之间的随机数：
Math.round(Math.random()*x);2.　　x至y之间的随机数
Math.round(Math.random()*(y-x)+x);3.　　1-x之间的随机数：
Math.ceil(Math.random()*x);

Prim算法

const INF = Number.MAX_SAFE_INTEGERfunction findMinKey(graph,key,visited){let min = INFlet minIndex//找到候选边中成本最小的节点for(let v = 0; v < graph.length; v++){if(!visited[v] && key[v] < min){min = key[v]minIndex = v}}return minIndex
}const prim = (graph) => {const visited = [],key = [],parent = [];let {length} = graph;for(let v = 0; v < length; v++){visited[v] = falsekey[v] = INF}//把到第一个顶点的权值初始化为0key[0] = 0parent[0] = -1//根节点不需要比for(let i = 0; i < length - 1; i++){//找到成本最小边的顶点let u = findMinKey(graph,key,visited)//标记下该顶点已被访问visited[u] = true;//以及该顶点到对应其他顶点是不是成本最小的//如果是那么就走到该顶点去for(let v = 0; v < length; v++){if(graph[u][v] && !visited[v] && graph[u][v] < key[v]) {parent[v] = ukey[v] = graph[u][v]}}}return parent
}const graph = [[0,2,4,0,0,0],[2,0,2,4,2,0],[4,2,0,0,3,0],[0,4,0,0,3,2],[0,2,3,3,0,2],[0,0,0,2,2,0]
]
const parent = prim(graph)
console.log('Edge    Weight')
for (let i = 1; i < graph.length; i++) {console.log(parent[i] + ' - ' + i + '   ' + graph[i][parent[i]]);
}

使用Prim算法生成迷宫

1. 生成2 * k + 1的迷宫，1表示墙，0表示路
2. 随机选一个顶点，在该顶点上下左右随机抽取一个位置，如果没有访问过而且没有越界就选这个点生成迷宫
3. 重复第2步

function roadmap(r,c){let map = []let rLen = 2 * r + 1let cLen = 2 * c + 1for(let i = 0; i < rLen; i++){map[i] = []for(let j = 0; j < cLen; j++){//生成0101的格式if((i ^ (i - 1)) === 1 && (j ^ (j - 1)) === 1){map[i][j] = 0}else{map[i][j] = 1}}}map[1][0] = 0;map[2 * r - 1][2 * c] = 0;return map
}const MathUtil = {randomInt(a = 0,b){if(typeof b === 'undefined'){return Math.floor(Math.random() * a)} else {return Math.floor(Math.random() * (b - a) + a) }}
}function maze(map){let row = map.length >> 1let col = map[0].length >> 1let size = row * collet notAccessed = new Array(size).fill(0)let accessed = []let cur = MathUtil.randomInt(0,size)let offsS = [-col,col,-1,1] //cur在notAccessed要走的偏移量let offsR = [-1,1,0,0] //cur在map中row要走的偏移量let offsC = [0,0,-1,1]//cur在map中col要走的偏移量accessed.push(cur)notAccessed[cur] = 1while(accessed.length < size){let tr = Math.floor(cur / row)let tc = cur % collet num = 0;let pos = -1//判断四周格子是不是可以走while(num++ < 4){let dir = MathUtil.randomInt(0,4)nr = tr + offsR[dir]nc = tc + offsC[dir]if(nr >= 0 && nc >= 0 && nr < row && nc < col && notAccessed[cur + offsS[dir]] === 0){pos = dirbreak }}if(pos < 0){//堵死的情况cur = accessed[MathUtil.randomInt(0,accessed.length)]}else{//可以走的情况tr = 2 * tr + 1tc = 2 * tc + 1map[tr + offsR[pos]][tc + offsC[pos]] = 0cur = cur + offsS[pos]notAccessed[cur] = 1accessed.push(cur)}}return map
}function geMaze(r,c){return maze(roadmap(r,c))
}