你把配送中心的坐标设置为（0,0）试试，其他配送点的坐标围着配送中心展开

好的，我将配送中心的坐标设置为(0, 0)，并将其他配送点的坐标围绕配送中心展开。同时，我还会相应地更新距离矩阵。以下是修改后的代码示例：

import gurobipy as gp
from gurobipy import GRB

model = gp.Model()

# 导入数据
customer = 5
vehicle = 2
capacity = 10

# 设置配送点坐标
coordinates = [
    (0, 0),
    (2, 1),
    (-1, 3),
    (3, -2),
    (-2, -1)
]

# 计算距离矩阵
distance = []
for i in range(customer):
    distance_row = []
    for j in range(customer):
        x1, y1 = coordinates[i]
        x2, y2 = coordinates[j]
        dist = ((x2 - x1) ** 2 + (y2 - y1) ** 2) ** 0.5
        distance_row.append(dist)
    distance.append(distance_row)

# 添加配送中心
customer += 1
coordinates.append((0, 0))
distance.append([0] * customer)
for i in range(customer):
    distance[i].append(0)

# 创建变量
x = {}  # x={k,j}表示配送点j是否由无人车k进行配送
for k in range(vehicle):
    for j in range(customer):
        x[k, j] = model.addVar(vtype=GRB.BINARY)

# 建立约束
for j in range(customer):
    model.addConstr(gp.quicksum(x[k, j] for k in range(vehicle)) == 1)

for k in range(vehicle):
    model.addConstr(gp.quicksum(x[k, j] for j in range(customer)) <= capacity)

# 添加配送中心约束
center = customer - 1
for k in range(vehicle):
    model.addConstr(gp.quicksum(x[k, center] for j in range(customer)) == 1)
    model.addConstr(gp.quicksum(x[k, j] for j in range(customer)) == gp.quicksum(x[k, j] for j in range(customer + 1)))

# 目标函数
obj = gp.quicksum(distance[i][j] * x[k, i] * x[k, j] for k in range(vehicle) for i in range(customer) for j in range(customer))
model.setObjective(obj, GRB.MINIMIZE)

model.optimize()

if model.status == GRB.OPTIMAL:
    print('最优解：')
    for k in range(vehicle):
        print(f'车辆{k + 1}的路径')
        print('配送中心 ->', end=' ')
        for j in range(customer):
            if x[k, j].x > 0.5:
                print(f'客户{j + 1} ->', end=' ')
        print('配送中心')
else:
    print('未找到最优解')

修改后的代码将根据配送点的坐标计算距离矩阵，并在最优解时输出每辆车的路径，包括从配送中心出发和返回配送中心。请注意，这只是一个示例，具体问题可能需要根据实际情况进行调整和修改。