*...*..D
.G..*.....
**...**.
.S....*.
........
...G**..
........
.G..*...

下面是二维数组的地方 S-来源 D-目标 G点一定要参观。自由路径*阻止路径你能帮助我这将是有效的算法找到最短伯提的长度在Java中只有水平和垂直运动是possiable

Here is 2d array where S- SourceD-DestinationG-Point must be visited." . "Free paths"*"Block PathsCan you help me which would be the efficient algorithm to find the length of shortest pathi in javaOnly Horizontal and Vertical movements are possiable

推荐答案

要找到从的最短距离启动指向地图中所有其他的点，你可以使用 BFS 。 样品code：

To find the shortest distance from start point to all other points in the map, you can use a BFS. Sample code:

public void visit(String []map , Point start){
        int []x = {0,0,1,-1};//This represent 4 directions right, left, down , up
        int []y = {1,-1,0,0};
        LinkedList<Point> q = new LinkedList();
        q.add(start);
        int n = map.length;
        int m = map[0].length();
        int[][]dist = new int[n][m];
        for(int []a : dist){
            Arrays.fill(a,-1);
        }
        dist[start.x][start.y] = 0;
        while(!q.isEmpty()){
            Point p = q.removeFirst();
            for(int i = 0; i < 4; i++){
                int a = p.x + x[i];
                int b = p.y + y[i];
                if(a >= 0 && b >= 0 && a < n && b < m && dist[a][b] == -1 && map[a].charAt(b) != '*' ){
                    dist[a][b] = 1 + dist[p.x][p.y];
                    q.add(new Point(a,b));
                }
            }
        }
    }

这个问题的第二条路径实际上是一个旅行商问题，所以你需要转换从原始图形的曲线图中只有G，D和S点，与重量中的每个边缘这个图是在原路径之间的最短路径。从那以后，如果G数小（小于17），可以使用动态规划和掩码来解决这个问题。

The second path of the problem is actually a traveling salesman problem, so you need to convert from your original graph to a graph which only contains G,D and S points, with the weight of each edge in this graph is the shortest path between them in original path. From that onward, if the number of G is small (less than 17) you can use dynamic programming and bitmask to solve the problem.