超过23个点位的最优路径规划咨询（基于Google API）

Great question! The 23-waypoint limit in Google Directions API is definitely a pain point when dealing with large route planning tasks. Let's break down practical solutions for your scenario, addressing both your grouping confusion and greedy algorithm concerns:

1. 聚类分组+跨组连接点优化（平衡效率与最优性） #

This approach splits your points into manageable groups first, then optimizes how to link those groups with minimal extra distance:

• Step 1: Spatial Clustering
  Use a spatial clustering algorithm like K-means (or even a simpler grid-based grouping) to split all your points (including start/end) into groups, each with no more than 20 waypoints (leave a few slots for cross-group connection points to stay under the 23 limit). The goal is to keep geographically close points in the same group to reduce intra-group travel distance.
• Step 2: Optimize Intra-Group Routes
  For each group, pass the points (plus a temporary start/end for the group) to Google Directions API to get the optimized intra-group route.
• Step 3: Find Optimal Cross-Group Links
  The key problem you mentioned—how to connect groups—can be solved by:
  • Calculating pairwise distances between all points in different groups (use Google Distance Matrix API for accurate road distances, not just straight-line).
  • Identifying the pair of points (one from each adjacent group) with the shortest road distance as the connection points. Alternatively, compute the "centroid" of each group (average latitude/longitude) and find the closest point pairs between groups whose centroids are nearest.
• Step 4: Stitch Routes Together
  Replace the temporary start/end of each group with the cross-group connection points, then concatenate all intra-group routes into a single full path.

2. 改进贪心算法（避免局部最优陷阱） #

Your initial greedy idea (pick the nearest next point) is simple but often leads to local optimal paths (e.g., you might end up backtracking later). Here's how to tweak it for better results:

• Instead of only choosing the closest point to your current location, calculate a combined score for each remaining point:
  Score = Distance(current → point) + Average distance from point to all remaining points
• Select the point with the lowest score each time. This balances immediate proximity with future travel efficiency, getting you closer to a global approximate optimal path.
• Note: You'll need to use Google Distance Matrix API to fetch all pairwise distances upfront, and this method doesn't rely on Google's route optimization—you're handling the ordering yourself, then passing the ordered points to Google API in chunks of 23 or fewer.

3. 启发式TSP+API拆分（全局最优优先） #

Since your problem is essentially a Traveling Salesman Problem (TSP) with more points than Google API can handle in one go, combine heuristic TSP algorithms with API splitting:

• Step 1: Generate a Global Approximate Optimal Order
  Use a heuristic TSP algorithm like Genetic Algorithm or Simulated Annealing to compute an approximate optimal order of all your points (start → waypoints → end). These algorithms are efficient for large point sets and can give you a near-optimal path sequence.
• Step 2: Split the Sequence into API-Friendly Chunks
  Split the ordered list into segments where each segment has no more than 23 points (including the start/end of the segment). Make sure the end point of one segment is the start point of the next to ensure smooth transitions.
• Step 3: Fetch Routes for Each Chunk
  Pass each segment to Google Directions API to get the optimized route for that chunk, then stitch them together to form the full path.

Key Notes to Remember #

• Always include your original start/end points in the first/last group (or sequence segment) to avoid unnecessary detours.
• For real-time route planning, factor in live traffic data (Google API provides this) when calculating cross-group connection points or TSP scores.
• Test with small over-limit datasets first (e.g., 25 points) to validate which method works best for your use case—compare total travel distance and time across approaches.

内容的提问来源于stack exchange，提问作者Snake