求自相交多边形最大面积的算法需求

Hey, let's work through how to solve this problem of finding the maximum area of a self-intersecting polygon. Since regular methods like the shoelace formula fall flat here (they calculate signed areas that cancel out in self-intersecting shapes), and you already know all the intersection points, we can use a combination of plane sweeping and polygon partitioning to get the right result—like your example where the total should be 40 (36 from the square plus 4 from the outer triangles). Here's a step-by-step breakdown:

The "maximum area" you're looking for is actually the sum of all enclosed regions formed by the self-intersecting polygon (including those small outer triangles in your example). With known intersection points, we can split the original polygon into non-overlapping simple polygons (no self-intersections), calculate each's area with the shoelace formula, then add them all up.

1. Merge Vertices and Intersection Points, Then Sort #

First, combine all original polygon vertices with the given intersection points. Then sort this complete set of points by their x-coordinate (and y-coordinate if x values are equal). This sorted list is the foundation for our plane sweep algorithm, and this step runs in O(m log m) time where m is the total number of points (original vertices + intersections).

2. Plane Sweep to Track Edges and Regions #

Use a sweep line that moves from left to right across the plane. As the line encounters each point, we maintain an active set of edges that intersect the sweep line:

• When we hit a starting point of an edge (either an original vertex or an intersection where a new segment begins), add the edge to the active set. Note the edge's direction (upward or downward relative to the sweep line).
• When we hit an endpoint or intersection where a segment ends, remove the corresponding edge from the active set. At the same time, calculate the area of the region formed between the previous sweep position and the current one, bounded by the active edges.

Pro tip: Since intersection points split original edges into smaller segments, make sure each split segment is treated as a separate edge in this process.

3. Partition into Simple Polygons and Sum Areas #

Alternatively, a more intuitive approach (if you have the full sequence of split segments from the original polygon's path) is to group these segments into simple polygons:

• Traverse the original polygon's path, following each split segment (from vertex to intersection to vertex, etc.).
• Whenever you form a closed loop (a simple polygon), extract it, calculate its area using the shoelace formula, and add the absolute value of that area to your total (since self-intersecting shapes can produce negative signed areas, but we want all enclosed regions regardless of orientation).

In your example, this would split the shape into the central square and four small outer triangles. Summing their areas gives 36 + 4*1 = 40, which matches your expected result.

• Sorting the points takes O(m log m). If the number of intersection points k is proportional to n (the original vertex count), this becomes O(n log n), which meets your requirement.
• The plane sweep process runs in O(m log m) because each edge insertion/removal from the active set takes O(log m) time.
• Calculating areas for all simple polygons takes O(m) time, since each point is part of a constant number of polygons.

Overall, the algorithm's time complexity is O(m log m), which is optimal for this problem.

• Ensure every intersection point is correctly inserted into the original polygon's edge sequence—splitting edges at intersections is critical for forming valid simple polygons.
• When using the shoelace formula, always take the absolute value of each polygon's area before summing. Self-intersecting paths can create polygons with clockwise and counter-clockwise orientations, which produce positive and negative signed areas.
• Since your polygon is guaranteed to be drawable in one stroke, the partitioning process will be straightforward—no disconnected segments to handle.

内容的提问来源于stack exchange，提问作者satoshi