使用邻接矩阵在迷宫图中寻找最短路径
Finding the shortest path in a maze graph using adjacency matrix
我在解决一个在迷宫图中寻找最短路径的特定问题时遇到了一些麻烦。可能,我被困住了,因为迷宫是由一个四维数组初始化的,比如
adjacent = new boolean[height][width][height][width];
第一对和第二对索引以 row/column 形式指定图中的位置。该图如下所示:
XXXXXXXXXXXXX
..........X.X
X.XXX.XXX.X.X
X.X.X...X.X.X
X.X.XXX.XXX.X
X...X.....X..
XXXXXXXXXXXXX
ArrayList 必须按照从头到尾的顺序保存路径中顶点的位置。
我已经写好了构造函数和连接方法;但是,我在寻找最短路径方法时遇到了麻烦。
这是我如何创建迷宫图的示例:
final int edges[][] = {{1, 0, 1, 1}, {1, 1, 1, 2}, {1, 1, 2, 1},
{1, 2, 1, 3}, {1, 3, 1, 4}, {1, 4, 1, 5}, {1, 5, 1, 6},
{1, 5, 2, 5}, {1, 6, 1, 7}, {1, 7, 1, 8}, {1, 8, 1, 9},
{1, 9, 2, 9}, {1, 11, 2, 11}, {2, 1, 3, 1}, {2, 5, 3, 5},
{2, 9, 3, 9}, {2, 11, 3, 11}, {3, 1, 4, 1}, {3, 3, 4, 3},
{3, 5, 3, 6}, {3, 6, 3, 7}, {3, 7, 4, 7}, {3, 11, 4, 11},
{4, 1, 5, 1}, {4, 3, 5, 3}, {4, 7, 5, 7}, {4, 11, 5, 11},
{5, 1, 5, 2}, {5, 2, 5, 3}, {5, 5, 5, 6}, {5, 6, 5, 7},
{5, 7, 5, 8}, {5, 8, 5, 9}, {5, 11, 5, 12}};
MazeGraph maze = new MazeGraph(13, 7);
for (int[] edge : edges)
maze.connect(new Location(edge[0], edge[1]), new Location(edge[2], edge[3]));
首先,这个
adjacent = new boolean[height][width][height][width];
与此矛盾:
The first and second pair of indices specify a location in the graph
in row/column formation.
是column/row,不是row/column。
Dijkstra's algorithm 应该为您的矩阵实施。引用:
Let the node at which we are starting be called the initial node. Let
the distance of node Y be the distance from the initial node to Y.
Dijkstra's algorithm will assign some initial distance values and will
try to improve them step by step.
Assign to every node a tentative distance value: set it to zero for
our initial node and to infinity for all other nodes.
Set the initial node as current. Mark all other nodes unvisited. Create a set of all the unvisited nodes called the unvisited set.
For the current node, consider all of its unvisited neighbors and calculate their tentative distances. Compare the newly calculated
tentative distance to the current assigned value and assign the
smaller one. For example, if the current node A is marked with a
distance of 6, and the edge connecting it with a neighbor B has length
2, then the distance to B (through A) will be 6 + 2 = 8. If B was
previously marked with a distance greater than 8 then change it to 8.
Otherwise, keep the current value.
When we are done considering all of the neighbors of the current node, mark the current node as visited and remove it from the
unvisited set. A visited node will never be checked again.
If the destination node has been marked visited (when planning a route between two specific nodes) or if the smallest tentative
distance among the nodes in the unvisited set is infinity (when
planning a complete traversal; occurs when there is no connection
between the initial node and remaining unvisited nodes), then stop.
The algorithm has finished.
Otherwise, select the unvisited node that is marked with the smallest tentative distance, set it as the new "current node", and go
back to step 3.
我在解决一个在迷宫图中寻找最短路径的特定问题时遇到了一些麻烦。可能,我被困住了,因为迷宫是由一个四维数组初始化的,比如
adjacent = new boolean[height][width][height][width];
第一对和第二对索引以 row/column 形式指定图中的位置。该图如下所示:
XXXXXXXXXXXXX
..........X.X
X.XXX.XXX.X.X
X.X.X...X.X.X
X.X.XXX.XXX.X
X...X.....X..
XXXXXXXXXXXXX
ArrayList 必须按照从头到尾的顺序保存路径中顶点的位置。
我已经写好了构造函数和连接方法;但是,我在寻找最短路径方法时遇到了麻烦。 这是我如何创建迷宫图的示例:
final int edges[][] = {{1, 0, 1, 1}, {1, 1, 1, 2}, {1, 1, 2, 1},
{1, 2, 1, 3}, {1, 3, 1, 4}, {1, 4, 1, 5}, {1, 5, 1, 6},
{1, 5, 2, 5}, {1, 6, 1, 7}, {1, 7, 1, 8}, {1, 8, 1, 9},
{1, 9, 2, 9}, {1, 11, 2, 11}, {2, 1, 3, 1}, {2, 5, 3, 5},
{2, 9, 3, 9}, {2, 11, 3, 11}, {3, 1, 4, 1}, {3, 3, 4, 3},
{3, 5, 3, 6}, {3, 6, 3, 7}, {3, 7, 4, 7}, {3, 11, 4, 11},
{4, 1, 5, 1}, {4, 3, 5, 3}, {4, 7, 5, 7}, {4, 11, 5, 11},
{5, 1, 5, 2}, {5, 2, 5, 3}, {5, 5, 5, 6}, {5, 6, 5, 7},
{5, 7, 5, 8}, {5, 8, 5, 9}, {5, 11, 5, 12}};
MazeGraph maze = new MazeGraph(13, 7);
for (int[] edge : edges)
maze.connect(new Location(edge[0], edge[1]), new Location(edge[2], edge[3]));
首先,这个
adjacent = new boolean[height][width][height][width];
与此矛盾:
The first and second pair of indices specify a location in the graph in row/column formation.
是column/row,不是row/column。
Dijkstra's algorithm 应该为您的矩阵实施。引用:
Let the node at which we are starting be called the initial node. Let the distance of node Y be the distance from the initial node to Y. Dijkstra's algorithm will assign some initial distance values and will try to improve them step by step.
Assign to every node a tentative distance value: set it to zero for our initial node and to infinity for all other nodes.
Set the initial node as current. Mark all other nodes unvisited. Create a set of all the unvisited nodes called the unvisited set.
For the current node, consider all of its unvisited neighbors and calculate their tentative distances. Compare the newly calculated tentative distance to the current assigned value and assign the smaller one. For example, if the current node A is marked with a distance of 6, and the edge connecting it with a neighbor B has length 2, then the distance to B (through A) will be 6 + 2 = 8. If B was previously marked with a distance greater than 8 then change it to 8. Otherwise, keep the current value.
When we are done considering all of the neighbors of the current node, mark the current node as visited and remove it from the unvisited set. A visited node will never be checked again.
If the destination node has been marked visited (when planning a route between two specific nodes) or if the smallest tentative distance among the nodes in the unvisited set is infinity (when planning a complete traversal; occurs when there is no connection between the initial node and remaining unvisited nodes), then stop. The algorithm has finished.
Otherwise, select the unvisited node that is marked with the smallest tentative distance, set it as the new "current node", and go back to step 3.