Swift之深度优先搜索和广度优先搜索

作者: 我系哆啦 | 来源:发表于2016-07-24 00:07 被阅读193次

    深度优先搜索

    深度优先搜索所遵循的搜索策略是尽可能“深”地搜索图。在深度优先搜索中,对于最新发现的顶点,如果它还有以此为起点而未探测到的边,就沿此边继续汉下去。当结点v的所有边都己被探寻过,搜索将回溯到发现结点v有那条边的始结点。这一过程一直进行到已发现从源结点可达的所有结点为止。如果还存在未被发现的结点,则选择其中一个作为源结点并重复以上过程,整个进程反复进行直到所有结点都被发现为止

    深度优先搜索基本模型

    void dfs(int step)
    {
    判定边界
    尝试每一种可能 for(int i = 0;i <n;i++)
    {
    继续下一步 dfs(stp+1)
    }
    返回
    }

    理解深度优先搜索的关键在于解决“当下该如何做”。至于“下一步如何做”则与“当下该如何做”是一样的。

    栗子一

    求解:□□□ + □□□ = □□□ 将数字1-9分别填入9个□中,没个数字只能使用一次是的等式成立。
    <pre>
    let count:Int = 9
    var book:[Int] = Array.init(repeatElement(0, count: count + 1))
    var num:[Int] = Array.init(repeatElement(0, count: count + 1))

    func dfs(step:Int) {

    //输出一种排列
    if step == (count + 1) {
        if ((num[1]*100 + num[2]*10 + num[3] + num[4]*100 + num[5]*10 + num[6]) == (num[7]*100 + num[8]*10 + num[9])) {
            print(num)
            return
        }
    }
    
    for i in 1...count {
        
        if book[i] == 0 {  //book[i] = 0 表示i没有使用,1表示已经用过了
            num[step] = i
            book[i] = 1
            
            dfs(step: step+1) //通过函数的递归来实现
            book[i] = 0  //回收
        }
    }
    

    }

    dfs(step: 1)
    </pre>

    栗子二

    在迷宫中求解起点到某一终点的最短步数
    <pre>
    var sumStep:Int = 99999
    let maze = [[0,0,1,0],
    [0,0,0,0],
    [0,0,1,0],
    [0,1,0,0],
    [0,0,0,1]]
    var book = Array.init(repeatElement([0,0,0,0,], count: 5))
    let endPoint = (2,3)

    func dfs(point: (x:Int,y:Int),step:Int) {

    if point == endPoint {
        if step < sumStep {
            sumStep = step
        }
        return
    }
    
    let next = [(0,1),  //向右走
                (1,0),  //向下走
                (0,-1), //向左走
                (-1,0)] //向上走
    
    for index in next.indices {
        
        let newPoint = (point.x + next[index].0,point.y + next[index].1) //新坐标点
        if (newPoint.0 < 0) || (newPoint.0 > 3) || (newPoint.1 < 0) || (newPoint.1 > 4) {
            continue
        }
        
        if (maze[newPoint.1][newPoint.0] == 0) &&  (book[newPoint.1][newPoint.0] == 0){
            
            book[newPoint.1][newPoint.0] = 1
            dfs(point: newPoint, step: step + 1)
            book[newPoint.1][newPoint.0] = 0
        }
    }
    

    }

    dfs(point: (0,0), step: 0)
    print(sumStep) //7
    </pre>

    广度优先搜索

    广度优先搜索是一种层层递进的算法,又译作宽度优先搜索或横向优先搜索,简称BFS,是一种图形搜索,是从根节点开始,沿着树的宽度遍历树的节点如果所有节点均被访问,则算法中止。广度优先搜索实现一般采用open-closed表。

    上面深度优先搜索的栗子二换成广度优先搜索代码如下:

    <pre>
    let maze = [[0,0,1,0],
    [0,0,0,0],
    [0,0,1,0],
    [0,1,0,0],
    [0,0,0,1]]
    var book = Array.init(repeatElement([0,0,0,0,], count: 5))
    //链表节点
    struct note {
    var x:Int = 0
    var y:Int = 0
    var f:Int = 0
    var step:Int = 0
    }

    func bfs(maze:[Array<Int>], startPoint: (x:Int,y:Int),endPoint: (x:Int,y:Int)) -> Int {

    var que:[note] = Array.init(repeatElement(note(), count: 1000))
    var head = 0
    var tail = 0
    
    let next = [(0,1),  //向右走
                (1,0),  //向下走
                (0,-1), //向左走
                (-1,0)] //向上走
    
    //往队列插入迷宫入口坐标
    que[tail].x = startPoint.x
    que[tail].y = startPoint.y
    que[tail].f = 0
    que[tail].step = 0
    tail+=1
    
    var flag = 0 //用来标记是都到达endPoint
    while head < tail {
        
            for index in next.indices {
                let nextPoint = (que[head].x + next[index].0,que[head].y + next[index].1) //新坐标点
                if (nextPoint.0 < 0) || (nextPoint.0 > 3) || (nextPoint.1 < 0) || (nextPoint.1 > 4) {
                    continue
                }
                
                if (maze[nextPoint.1][nextPoint.0] == 0) &&  (book[nextPoint.1][nextPoint.0] == 0){
                    book[nextPoint.1][nextPoint.0] = 1
                    
                    que[tail].x = nextPoint.0
                    que[tail].y = nextPoint.1
                    que[tail].f = head
                    que[tail].step = que[head].step + 1
                    tail+=1
                }
                
                if nextPoint == endPoint {
                    flag = 1
                    break;
                }
        }
        
        if flag == 1 {
            break
        }
        head+=1
    }
    
    return que[tail-1].step
    

    }

    let step = bfs(maze: maze, startPoint: (0,0), endPoint: (2,3))
    print(step) //7
    </pre>

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