我想找到闭合形状的总数。
在图像中,有 6 个 no of close polygons 。 我尝试了以下方法
for (NSInteger i = 0; i < [arrLinesInfo count]; i++) {
NSDictionary *dictLineInfo = [arrLinesInfo objectAtIndex:i];
startPoint = CGPointMake([[dictLineInfo valueForKey:@"line_start_point_x"] doubleValue], [[dictLineInfo valueForKey:@"line_start_point_y"] doubleValue]);
endPoint = CGPointMake([[dictLineInfo valueForKey:@"line_end_point_x"] doubleValue], [[dictLineInfo valueForKey:@"line_end_point_y"] doubleValue]);
[self isCircularRoute:startPoint withEndPoint:endPoint];
}
-(void) isCircularRoute:(CGPoint) lineStartPoint withEndPoint:(CGPoint) lineEndPoint
{
NSPredicate *pre= [NSPredicate predicateWithFormat:[NSString stringWithFormat:@"
(self.line_end_point_x == '%f' && self.line_end_point_y == '%f') OR
(self.line_start_point_x == '%f' && self.line_start_point_y == '%f') OR
(self.line_end_point_x == '%f' && self.line_end_point_y == '%f') OR
(self.line_start_point_x == '%f' && self.line_start_point_y == '%f')", lineStartPoint.x,
lineStartPoint.y,
lineStartPoint.x,
lineStartPoint.y,
lineEndPoint.x,
lineEndPoint.y,
lineEndPoint.x,
lineEndPoint.y]];
NSMutableArray *arrSamePointRef = [[arrLinesInfo filteredArrayUsingPredicate:pre] mutableCopy];
arrSamePointRef = [[arrSamePointRef filteredArrayUsingPredicate:[NSPredicate predicateWithFormat:[NSString stringWithFormat:@"
(self.line_start_point_x != '%f' && self.line_start_point_y != '%f') &&
(self.line_end_point_x != '%f' && self.line_end_point_y != '%f')", lineStartPoint.x
, lineStartPoint.y
, lineEndPoint.x
, lineEndPoint.y]]] mutableCopy];//[arrSamePointRef removeObject:dictLineInfo];
if(arrSamePointRef.count > 2){
totalPolygon = totalPolygon + 1;
}
NSLog(@"totalPolygon : ===== %tu", totalPolygon);
for (NSDictionary *dictSingleLine in arrSamePointRef) {
CGPoint newStartPoint = CGPointMake([[dictSingleLine valueForKey:@"line_start_point_x"] doubleValue], [[dictSingleLine valueForKey:@"line_start_point_y"] doubleValue]);
CGPoint newEndPoint = CGPointMake([[dictSingleLine valueForKey:@"line_end_point_x"] doubleValue], [[dictSingleLine valueForKey:@"line_end_point_y"] doubleValue]);
[self isCircularRoute:newStartPoint withEndPoint:newEndPoint];
}
}
这是无限循环。
我在数组中有所有起点和终点对象。 如下所示的数组对象
[ { "point_start_lbl" : "a", "point_end_lbl" : "b", "line_start_point_x" : 200, "line_start_point_y" : 10, "line_end_point_x" : 100, "line_end_point_y" : 10, }, ... ]
请帮帮我。 提前致谢。
最佳答案
如果你有一个有序的边列表,那么你肯定有一个封闭的多边形,这样每条边都在下一条开始的顶点上结束,并且列表中没有重复的边。
我不清楚你的数据结构,但我可能因此:
定义一个对象,Edge
,它标识两个顶点。
对于每个顶点,创建一个数组,其中包含接触该顶点的每条边。
然后,类似 Swift 风格的伪代码:
var successfulPaths: [Edge] = []
for edge in edges
{
let usedEdges = [edge]
attemptTraversalFrom(edge.vertex1, edge.vertex2, usedEdges, successfulPaths)
attemptTraversalFrom(edge.vertex2, edge.vertex1, usedEdges, successfulPaths)
}
print("There were \(successfulPaths.count) successful paths")
[...]
func attemptTraversalFrom(startingVertex, endingVertex, usedEdges, successfulPaths) {
let vertexEdges = endingVertex.edges
for edge in (edges not in usedEdges) {
let newEndingVertex =
(edge.vertex1 == endingVertex) ? edge.vertex2 : edge.vertex1
if newEndingVertex == startingVertex {
successfulPaths.add(usedEdges)
return
} else {
let newUsedEdges = userEdges.addItem(edge)
attemptTraversalFrom(startingVertex, newEndingVertex, newUsedEdges, successfulPaths)
}
}
// Note: will automatically fall through to here and return
// without adding anything to `successfulPaths` if there are
// no further traversable edges
}
临时等。有点像 Dijkstra 寻路算法的递归部分,除了潜在路径是累积的,而不是较短的路径在完成评估之前消除较长的路径。
关于ios - 查找关闭多边形 iOS 的总数,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/38074652/