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  • Prim算法生成迷宫

    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] = column
        for(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 % 3
        arr[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 = 0
    
    for(let i = 0; i < offset.length; i++){
        x = x + offset[i].x
        y = y + offset[i].y
    
        console.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_INTEGER
    
    function findMinKey(graph,key,visited){
        let min = INF
        let 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] = false
            key[v] = INF
        }
        //把到第一个顶点的权值初始化为0
        key[0] = 0
        parent[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] = u
                    key[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 + 1
        let cLen = 2 * c + 1
    
        for(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 >> 1
        let col = map[0].length >> 1
        let size = row * col
        let 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] = 1
    
        while(accessed.length < size){
            let tr = Math.floor(cur / row)
            let tc = cur % col
            let 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 = dir
                    break 
                }
            }
    
            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 + 1
                tc = 2 * tc + 1
                
                map[tr + offsR[pos]][tc + offsC[pos]] = 0
                cur = cur + offsS[pos]
                notAccessed[cur] = 1
    
                accessed.push(cur)
            }
        }
        
        return map
    }
    
    function geMaze(r,c){
        return maze(roadmap(r,c))
    }
    
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  • 原文地址:https://www.cnblogs.com/pluslius/p/10878282.html
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