我的迷宫是一个包含两个维度的int数组,int maze[][]
包含0,1,START(2),GOAL(3)
。我想打印最短的路径。
我有一个功能,但它没有显示最短路径,而是显示到结尾的一条路径:
bool RenderThread::find_path(int x, int y)
{
int maze_size=mmaze->size*2;
if ( x < 0 || x > maze_size || y < 0 || y > maze_size ) return FALSE;
if ( toSolve1->maze_data[y][x] == G ) return TRUE;
if ( toSolve1->maze_data[y][x] != PATH && toSolve1->maze_data[y][x] != S ) return FALSE;
toSolve1->setRed(y,x);
if ( find_path(x, y - 1) == TRUE ) return TRUE;
if ( find_path(x + 1, y) == TRUE ) return TRUE;
if ( find_path(x, y + 1) == TRUE ) return TRUE;
if ( find_path(x - 1, y) == TRUE ) return TRUE;
toSolve1->setPath(y,x);
return FALSE;
}
答案 0 :(得分:4)
我会推荐A* search算法。
function A*(start,goal)
closedset := the empty set // The set of nodes already evaluated.
openset := {start} // The set of tentative nodes to be evaluated, initially containing the start node
came_from := the empty map // The map of navigated nodes.
g_score[start] := 0 // Cost from start along best known path.
// Estimated total cost from start to goal through y.
f_score[start] := g_score[start] + heuristic_cost_estimate(start, goal)
while openset is not empty
current := the node in openset having the lowest f_score[] value
if current = goal
return reconstruct_path(came_from, goal)
remove current from openset
add current to closedset
for each neighbor in neighbor_nodes(current)
if neighbor in closedset
continue
tentative_g_score := g_score[current] + dist_between(current,neighbor)
if neighbor not in openset or tentative_g_score <= g_score[neighbor]
came_from[neighbor] := current
g_score[neighbor] := tentative_g_score
f_score[neighbor] := g_score[neighbor] + heuristic_cost_estimate(neighbor, goal)
if neighbor not in openset
add neighbor to openset
return failure
function reconstruct_path(came_from, current_node)
if came_from[current_node] in set
p := reconstruct_path(came_from, came_from[current_node])
return (p + current_node)
else
return current_node
答案 1 :(得分:0)
如果我们假设您的迷宫是网格,并且墙壁被标记为无法访问的网格空间,则A *跳转点搜索是目前此搜索空间中最快的最佳算法。