Clarke—Wright节约算法求解旅行商问题的伪代码
时间: 2023-08-01 19:09:40 浏览: 50
以下是使用 Clarke-Wright 节约算法求解旅行商问题的伪代码:
```
Create initial solution:
1. Create a route for each city that starts and ends at the depot
2. Sort routes by decreasing order of savings s(i,j) = d(0,i) + d(0,j) - d(i,j), where d(i,j) is the distance between cities i and j
For each pair of routes (i,j) in decreasing order of savings:
1. If i and j are not already merged and merging them doesn't violate capacity constraints:
a. Merge i and j by inserting the cities in route j into route i in the order that minimizes the increase in total distance
b. Update savings for all remaining pairs of routes that include either i or j
Output the final solution
```
初始解的构造是将每个城市看作一个单独的路线,起点和终点都是仓库。然后,按照每对城市之间的节约距离进行排序。
在每个节约距离最大的城市对 (i,j) 上,如果合并这两个城市对应的路线不会导致超过车辆容量限制,则将它们合并。合并的方法是将 j 路线上的城市按照最小化总距离增加的顺序插入到 i 路线中。然后更新所有包含 i 或 j 的城市对之间的节约距离。
重复上述步骤,直到不能再合并任何城市对为止。最终得到的解就是 TSP 的近似解。