//判断同一条线能否连通,x轴相同或者y轴相同hasLine: function (p1, p2) {this.path = [];//同一点if (p1.x == p2.x && p1.y == p2.y) {return {status: false};}if (this.onlineY(p1, p2)) {var min = p1.y > p2.y ? p2.y : p1.y;min = min + 1;var max = p1.y > p2.y ? p1.y : p2.y;for (min; min < max; min++) {var p = {x: p1.x, y: min};if (!this.isEmpty(p)) {console.log('有障碍物p点………………');console.log(p);this.path = [];break;}this.path.push(p);}if (min == max) {return {status: true,data: this.path,dir: 'y' //y轴
                    };}this.path = [];return {status: false};}else if (this.onlineX(p1, p2)) {var j = p1.x > p2.x ? p2.x : p1.x;j = j + 1;var max = p1.x > p2.x ? p1.x : p2.x;for (j; j < max; j++) {var p = {x: j, y: p1.y};if (!this.isEmpty(p)) {console.log('有障碍物p点………………');console.log(p);this.path = [];break;}this.path.push(p);}if (j == max) {return {status: true,data: this.path,dir: 'x' //x轴
                    };}this.path = [];return {status: false};}return {status: false};
//2点是否有其中一点被全包围,若有，则返回trueisWrap: function (p1, p2) {//有一点为空，则条件不成立if (!this.isEmpty({x: p1.x, y: p1.y + 1}) && !this.isEmpty({x: p1.x,y: p1.y - 1}) && !this.isEmpty({x: p1.x - 1,y: p1.y}) && !this.isEmpty({x: p1.x + 1, y: p1.y})) {return true;}if (!this.isEmpty({x: p2.x, y: p2.y + 1}) && !this.isEmpty({x: p2.x,y: p2.y - 1}) && !this.isEmpty({x: p2.x - 1,y: p2.y}) && !this.isEmpty({x: p2.x + 1, y: p2.y})) {return true;}return false;}//两点在一条直线上，不能直线连接但是可以连通LineLink: function (p1, p2) {var pt0, pt1, pt2, pt3;//如果都在x轴，则自左至右扫描可能的路径，//每次构造4个顶点pt0, pt1, pt2, pt3，然后看他们两两之间是否连通if (this.onlineX(p1, p2)) {for (var i = 0; i < this.H; i++) {if (i == p1.y) {continue;}pt0 = p1;pt1 = {x: p1.x, y: i};pt2 = {x: p2.x, y: i};pt3 = p2;//如果顶点不为空，则该路不通。if (!this.isEmpty(pt1) || !this.isEmpty(pt2)) {continue;}if (this.hasLine(pt0, pt1).status && this.hasLine(pt1, pt2).status && this.hasLine(pt2, pt3).status) {this.drawLine(2, [pt0, pt3, pt1, pt2]);return [pt0, pt1, pt2, pt3];}}}//如果都在y轴，则自上至下扫描可能的路径，//每次构造4个顶点pt0, pt1, pt2, pt3，然后看他们两两之间是否连通if (this.onlineY(p1, p2)) {for (var j = 0; j < this.W; j++) {if (j == p1.x) {continue;}pt0 = p1;pt1 = {x: j, y: p1.y};pt2 = {x: j, y: p2.y};pt3 = p2;//如果顶点不为空，则该路不通。if (!this.isEmpty(pt1) || !this.isEmpty(pt2)) {continue;}if (this.hasLine(pt0, pt1).status && this.hasLine(pt1, pt2).status && this.hasLine(pt2, pt3).status) {this.drawLine(2, [pt0, pt3, pt1, pt2]);return [pt0, pt1, pt2, pt3];}}}}，//两点不在一条直线上，看是否可通curveLink: function (p1, p2) {var pt0, pt1, pt2, pt3;//特殊情况，先判断是否是一个转弯var spec1 = {x: p1.x, y: p2.y},spec2 = {x: p2.x, y: p1.y};if (this.isEmpty(spec1)) {if (this.hasLine(p1, spec1).status && this.hasLine(p2, spec1).status) {console.log('1个转弯');this.drawLine(1, [p1, p2, spec1]);return [p1, p2, spec1];}}if (this.isEmpty(spec2)) {if (this.hasLine(p1, spec2).status && this.hasLine(p2, spec2).status) {console.log('1个转弯');// console.table([pt0, spec2, pt3]);this.drawLine(1, [p1, p2, spec2]);return [p1, spec2, p2];}}//先纵向扫描可能的路径//同样，每次构造4个顶点，看是否可通for (var k = 0; k <= this.H; k++) {pt0 = p1;pt1 = {x: p1.x, y: k};pt2 = {x: p2.x, y: k};pt3 = p2;//2个交点都为空if (this.isEmpty(pt1) && this.isEmpty(pt2)) {//2个转弯if (this.hasLine(pt0, pt1).status && this.hasLine(pt1, pt2).status && this.hasLine(pt2, pt3).status) {console.log('2个转弯');this.drawLine(2, [pt0, pt3, pt1, pt2]);return [pt0, pt3, pt1, pt2];}}}//横向扫描所有可能的路径for (var k = 0; k <= this.W; k++) {pt0 = p1;pt1 = {x: k, y: p1.y};pt2 = {x: k, y: p2.y};pt3 = p2;//2个交点都为空if (this.isEmpty(pt1) && this.isEmpty(pt2)) {//2个转弯if (this.hasLine(pt0, pt1).status && this.hasLine(pt1, pt2).status && this.hasLine(pt2, pt3).status) {console.log('2个转弯');this.drawLine(2, [pt0, pt3, pt1, pt2]);return [pt0, pt3, pt1, pt2];}}}return false;}