Question

我想找到近似形状的总数。

在图像中，有6个没有近似多边形。我试过以下方法

    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。像下面的数组对象

[ { “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。

Answer 1

如果你有一个有序的边列表，你肯定有一个封闭的多边形，这样每条边都会在下一个开始的顶点上结束，并且在列表中没有重复边。

我不清楚您的数据结构，但我可能因此：

定义一个标识两个顶点的对象Edge。

对于每个顶点，创建一个包含每个接触该顶点的单个边的数组。

然后，类似于Swift-ish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
}

即席等等有点像Dijkstra寻路算法的递归部分，除了潜在路径是累积而不是较短路径，在完成评估之前消除较长路径。

查找关闭多边形iOS的总数

1 个答案: