threejs 里面的3d管道的每个节点ID是唯一的,且对应x,y,z坐标。那么当需要从A点到B点的时候,可能出现有多条路径可走,此时便需要求出最短行走路径,因此用到一个寻路径算法。我们将问题简化如下:
var begId = 191; //起点ID var endId = 185; //终点ID //所有路径,不区分开始和结束节点的前后顺序 var allPaths = [[185,184],[186,185],[187,186],[188,187],[189,187],[191,189]]; var result = []; var tree_num = 1; var over_num = 0; /* pos => 起点坐标id target => 终点坐标id key => 返回的字符串,保存上一层递归完成的路径节点 len => 节点的id差(本例中没用到) pos0 => 原始起点,防止走回头路 */ function tree(pos, target, key, len, pos0) { var _index = 0; var others = []; var tmp = ‘‘; for(x in allPaths) { // one point is pos if (allPaths[x] [0] == pos || allPaths[x] [1] == pos) { var other = allPaths[x] [0] == pos ? allPaths[x] [1] : allPaths[x] [0]; if ( (pos > target && (other >= pos || other >= pos0 ) ) || (pos < target && (other <= pos || other <= pos0 ) ) ) continue; others.push(other); if (_index > 0) tree_num ++; _index ++; } } for(y in others) { other = others[y] ; len += other - pos; // other one is end? if (other == target) { result.push(key+target+",|"+len); over_num ++; continue; } // else if (other > endId) continue ? tree(other, target, key+other+",", len, pos0); } // all tree over ? if (tree_num == over_num) { // console.log(result); } } function get_short_path(a,b){ tree(a,b,"",0, a); var ret = result; result = []; var min = 100; var min_path_str = ‘‘; //求出最短的路径 for (x in ret) { var path_arr_str = ret[x].split(‘|‘)[0]; var path_arr = path_arr_str.split(","); var rank = path_arr.length; if (rank < min) { min = rank; min_path_str = path_arr_str; } } return min_path_str.slice(0,-1).split(","); } // tree(begId, "", 0); var search_arr = [[185,188],[191,184]]; for ( var x in search_arr) { var ret = get_short_path(search_arr[x][0], search_arr[x][1]); console.log(ret); }
时间: 2024-10-03 16:36:22