算法设计与分析项目-有容量设施选址问题

算法设计与分析项目-有容量设施选址问题 Capacitated Facility Location Problem(CFLP)

问题描述

Suppose there are n facilities and m customers. We wish to choose:

1.which of the n facilities to open
2. the assignment of customers to facilitiesThe objective is to minimize the sum of the opening costand the assignment cost.
3.The total demand assigned to a facility must not exceed its capacity.

数据集结构

注意demand of customer和Assignment cost 都是十个数据一换行，与Facility和customer数量无关。Assignment cost第1第2个数据为：将第1第2个顾客分配到工厂1的花费，而不是将第1个顾客分配到第1第2个工厂的花费，注意顺序。
还要注意有些数据集demand of customer和Assignment cost的数据后面有一个点，这里所有数据都可以视为整型。

项目

项目使用贪心算法和SA模拟退火算法
源代码，数据集和结果上传到github

贪心算法

贪心算法思路很简单，对于每个顾客，寻找能容纳其需求的且Assignment cost最小的工厂。
主要代码：

 facility_occpy.resize(facilities.size(), 0);
    for (auto &i : facility_occpy) i = 0;
    custumer_assign.resize(demands.size());

    for (int i = 0; i < demands.size(); i++) {
        size_t mincost = std::numeric_limits<size_t>::max();
        int min_index;
        //寻找能容纳其需求的且Assignment cost最小的工厂
        for (int j = 0; j < facilities.size(); j++) {
            if (get_cost(j, i) < mincost && demands[i] <= facilities[j].capacity - facility_occpy[j]) {
                min_index = j;
                mincost = get_cost(j, i);
            }
        }
        facility_occpy[min_index] += demands[i];
        custumer_assign[i] = min_index;
    }

部分结果

数据集	花费	时间	结果
p1	9440	0.004s	8 8 1 6 3 8 2 4 4 1 9 0 3 2 0 3 4 0 9 7 3 4 6 4 2 5 1 5 0 5 2 6 0 3 9 4 4 3 0 4 1 8 1 5 7 0 4 0 4 3
p3	10126	0.003s	8 8 1 6 3 8 2 4 4 1 9 0 3 2 0 3 4 0 9 7 3 4 6 4 2 5 1 5 0 5 2 6 0 3 9 4 4 3 0 4 1 8 1 5 7 0 4 0 4 3
p5	9375	0.003s	8 8 1 6 3 8 2 4 4 1 9 8 3