C++ 具有最小值和最大值的多维背包

我有一个类似于背包问题的问题，更具体地说是背包问题

我有一堆对象，它们都有成本、价值和类别。我需要在最大成本下对背包进行价值优化，但每个类别中也有特定数量的物品

#include <cstdio>
#include <iostream>
#include <vector>
#include <map>
#include <algorithm>

using uint = unsigned int;

template <typename T>
struct item {
    T value;
    uint weight;
    uint category;
};

template <typename T>
std::pair<T,bool> knapSack(uint W, const std::vector<uint>& K, const std::vector< item<T> >& items) {

    std::map< std::tuple<uint, uint, std::vector<uint> >, std::pair<T,bool> > cache;

    std::function<std::pair<T, bool>(uint, uint, std::vector<uint>)> recursion;
    recursion = [&] (uint n, uint w, std::vector<uint> k) {

        auto it = cache.find(std::make_tuple(n,w,k));
        if (it != cache.end())
            return it->second;

        std::vector<uint> ccount(K.size(),0);
        for (uint c = 0; c < K.size(); ++c) {
            for (uint i = 0; i < n; ++i) {
                if (items[i].category == c)
                    ++ccount[c];
            }
        }
        for (uint c = 0; c < k.size(); ++c) {
            if (k[c] > ccount[c]) {
                auto p = std::make_pair(0,false);
                cache.insert(std::make_pair(std::make_tuple(n,w,k),p));
                return p;
            }
        }

        uint sumk = 0; for (const auto& _k : k) sumk += _k;
        if (n == 0 || sumk == 0) {
            auto p = std::make_pair(0,true);
            cache.insert(std::make_pair(std::make_tuple(n,w,k),p));
            return p;
        }

        T _v = items[n-1].value;
        uint _w = items[n-1].weight;
        uint _c = items[n-1].category;
        T nextv;
        bool nextvalid = true;
        if (_w <= w and k[_c] > 0) {
            std::vector<uint> subk = k;
            --subk[_c];
            auto take = recursion(n-1,w-_w,subk);
            auto reject = recursion(n-1,w,k);
            if (take.second and reject.second) {
                nextv = std::max(_v + take.first,reject.first);
            } else if (take.second) {
                nextv = _v + take.first;
            } else if (reject.second) {
                nextv = reject.first;
            } else {
                nextv = 0;
                nextvalid = false;
            }   
        } else {
            std::tie(nextv,nextvalid) = recursion(n-1,w,k);
        }
        std::pair<T,bool> p = std::make_pair(nextv,nextvalid);
        cache.insert(std::make_pair(std::make_tuple(n,w,k),p));
        return p;   
    };

    return recursion(items.size(),W,K);
}

int main(int argc, char *argv[]) {
    std::vector< item<int> > items = {{60,10,0}, {100,20,1}, {120,30,0}, {140,35,1}, {145,40,0}, {180,45,1}, {160,50,1}, {170,55,0}};
    int  j = 145;
    for (uint k1 = 0; k1 <= items.size(); ++k1) {
        for (uint k2 = 0; k2 <= items.size(); ++k2) {
            auto p = knapSack(j,std::vector<uint>({k1,k2}),items);
            if (p.second)
                std::cout << "K0 = " << k1 << ", K1 = " << k2 << ": " << p.first << std::endl;
        }
    }
    return 0;
}

% OUTPUT (with comments) %

K0 = 0, K1 = 0: 0
K0 = 0, K1 = 1: 180 // e.g. {} from 0, {180} from 1
K0 = 0, K1 = 2: 340 // e.g. {} from 0, {160,180} from 1
K0 = 0, K1 = 3: 480 // e.g. {} from 0, {140,160,180} from 1
K0 = 1, K1 = 0: 170 // e.g. {170} from 0, {} from 1
K0 = 1, K1 = 1: 350 // e.g. {170} from 0, {180} from 1
K0 = 1, K1 = 2: 490 // e.g. {170} from 0, {140, 180} from 1
K0 = 1, K1 = 3: 565 // e.g. {145} from 0, {100, 140, 180} from 1
K0 = 2, K1 = 0: 315 // e.g. {145,170} from 0, {} from 1
K0 = 2, K1 = 1: 495 // e.g. {145,170} from 0, {180} from 1
K0 = 2, K1 = 2: 550 // e.g. {60,170} from 0, {140,180} from 1
K0 = 2, K1 = 3: 600 // e.g. {60,120} from 0, {100,140,180} from 1
K0 = 3, K1 = 0: 435 // e.g. {120,145,170} from 0, {} from 1
K0 = 3, K1 = 1: 535 // e.g. {120,145,170} from 0, {100} from 1
K0 = 3, K1 = 2: 605 // e.g. {60,120,145} from 0, {100,180} from 1
K0 = 4, K1 = 0: 495 // e.g. {60,120,145,170} from 0, {} from 1

我已经成功地在C++中实现了原始背包算法，不需要注意类别。 当我尝试添加类别时，我发现我可以简单地将其视为多维背包问题，每个类别在新维度中的权重为0或1

我的主要问题是，我不仅有一个最大值，例如：5个类型食物对象，而且还有一个最小值，因为我需要确切地5个类型食物对象

我不知道如何在算法中加入最小值

显然，我可以使用一个一般情况，其中每个维度都有一个最大值和最小值，并对总计进行优化，因为除了一个维度之外，我的所有维度的范围都只有1，所以最终还是会对值进行优化。此外，我可以将值的最小值设置为零，以避免一维没有最小值，它仍然可以工作

我在C++中工作，但是老实说，即使伪代码也可以，我只需要算法。< /P> 显然，我也需要它的速度，如果可能的话，尽可能快

下面是一个测试用例的示例。由于这主要是一个优化问题，因此实例是巨大的，但它应该适用于任何大小的实例。可能的类别数和类别字段数是固定的

你有一个背包，最多可以装100个重量单位，还有1000个物品的列表，每个物品都有一个值、一个重量和一种类型。你特别需要带上10件食物、15件衣服和5件工具。每个对象都有一个完全任意（但大于0）的美元值和单位权重。我需要根据每种物品的最大重量和具体数量，找到价值的最佳配置

---

对象列表将始终包含至少一个有效配置，这意味着它将始终具有至少足够的每种类型的对象，这些对象最终将处于最大权重之下，因此我不必为“无答案”的情况进行计划。我只需要为（可能）大量的可用项目找到最佳答案。

知道每个类别中可以选择多少项目是一个很大的限制。考虑最简单的情况，即有一个类别。对于成本总和[w_i x_i] 普通背包问题

下面是一个简单的演示：以通过记忆解决标准0/1背包问题为起点：

#include <cstdio>
#include <iostream>
#include <vector>
#include <map>
#include <algorithm>

using uint = unsigned int;

template <typename T>
struct item {
    T value;
    uint weight;
};

template <typename T>
T knapSack(uint W, const std::vector< item<T> >& items) {

    std::map< std::pair<uint, uint>, T> cache;

    std::function<T(uint, uint)> recursion;
    recursion = [&] (uint n, uint w) {
        if (n == 0)
            return 0;
        auto it = cache.find(std::make_pair(n,w));
        if (it != cache.end())
            return it->second;
        T _v = items[n-1].value;
        uint _w = items[n-1].weight;
        T nextv;
        if (_w <= w)
            nextv = std::max(_v + recursion(n-1,w-_w),recursion(n-1,w));
        else
            nextv = recursion(n-1,w);
        cache.insert(std::make_pair(std::make_pair(n,w),nextv));
        return nextv;   
    };

    return recursion(items.size(),W);
}

这个的输出是

K0 = 0, K1 = 0: 0; contents are {}
K0 = 0, K1 = 1: 180; contents are {5, }
K0 = 0, K1 = 2: 340; contents are {5, 6, }
K0 = 0, K1 = 3: 480; contents are {3, 5, 6, }
K0 = 1, K1 = 0: 170; contents are {7, }
K0 = 1, K1 = 1: 350; contents are {5, 7, }
K0 = 1, K1 = 2: 490; contents are {3, 5, 7, }
K0 = 1, K1 = 3: 565; contents are {1, 3, 4, 5, }
K0 = 2, K1 = 0: 315; contents are {4, 7, }
K0 = 2, K1 = 1: 495; contents are {4, 5, 7, }
K0 = 2, K1 = 2: 550; contents are {0, 3, 5, 7, }
K0 = 2, K1 = 3: 600; contents are {0, 1, 2, 3, 5, }
K0 = 3, K1 = 0: 435; contents are {2, 4, 7, }
K0 = 3, K1 = 1: 535; contents are {1, 2, 4, 7, }
K0 = 3, K1 = 2: 605; contents are {0, 1, 2, 4, 5, }
K0 = 4, K1 = 0: 495; contents are {0, 2, 4, 7, }

算法复杂性

---

这不是我的强项，但我相信运行时的复杂性是psuedo多项式，因为该算法与标准背包算法非常相似

对于你的问题，我实际上没有一个直接的答案，无论是伪代码还是特定语言中算法的实际实现，但我在这里能做的是给你一个参考列表，我认为这些参考文献与主题相关，可以帮助你开发一个工作算法：

虽然其中许多可能并不完全是背包算法的问题；我认为这些主题可能在某种程度上有助于你的算法的整体成就。我认为这些将是有用的，因为首先也是最重要的是背包问题本身是分区算法的一个变体，它有许多实现和方案。此外，并行编程和多线程编程的使用也可能对大型数据集有所帮助。我发现了这几本书和白皮书，我认为它们会很值得一读

基本上，您有一个背包

K

，它有一个卷

KV

，需要细分为更小的

{KV1，KV2，…KVn}

，其中包含不同的数据类型，每种类型都有一个

值，
重量
和
类别或分类
，项目的
重量
表示其消耗的体积部分。您还有一个约束条件，即存在一个
[min，max]
边界，但限制条件是每个
类别
或
分类
必须至少有一个。然后，将这些参数作为基本方案，然后您希望最大化
KV
以包含尽可能多的
元素
，但希望尽可能高效地使用最少的时间，希望
线性到多项式-时间和空间复杂度
避免
二次和/或指数-时间和空间的复杂性

查看其他独特不同的算法，如分区算法、人口密度和增长、图像压缩等，可以让您深入了解您的具体问题，因为这些算法的总体基础和注意事项本质上是相似的
 你能粗略估计一下投入的规模吗？多少个类别，每个类别的条目，等等。我的原始解决方案有一个bug（if-else的顺序应该颠倒），我已经重新发布了一个（更）正确的解决方案。@Mann输入可能很大，因为这大部分是错误的
#include <cstdio>
#include <iostream>
#include <vector>
#include <map>
#include <algorithm>

using uint = unsigned int;

template <typename T>
struct item {
    T value;
    uint weight;
    uint category;
};

template <typename T>
std::pair<T,bool> knapSack(uint W, const std::vector<uint>& K, const std::vector< item<T> >& items) {

    std::map< std::tuple<uint, uint, std::vector<uint> >, std::pair<T,bool> > cache;

    std::function<std::pair<T, bool>(uint, uint, std::vector<uint>)> recursion;
    recursion = [&] (uint n, uint w, std::vector<uint> k) {

        auto it = cache.find(std::make_tuple(n,w,k));
        if (it != cache.end())
            return it->second;

        std::vector<uint> ccount(K.size(),0);
        for (uint c = 0; c < K.size(); ++c) {
            for (uint i = 0; i < n; ++i) {
                if (items[i].category == c)
                    ++ccount[c];
            }
        }
        for (uint c = 0; c < k.size(); ++c) {
            if (k[c] > ccount[c]) {
                auto p = std::make_pair(0,false);
                cache.insert(std::make_pair(std::make_tuple(n,w,k),p));
                return p;
            }
        }

        uint sumk = 0; for (const auto& _k : k) sumk += _k;
        if (n == 0 || sumk == 0) {
            auto p = std::make_pair(0,true);
            cache.insert(std::make_pair(std::make_tuple(n,w,k),p));
            return p;
        }

        T _v = items[n-1].value;
        uint _w = items[n-1].weight;
        uint _c = items[n-1].category;
        T nextv;
        bool nextvalid = true;
        if (_w <= w and k[_c] > 0) {
            std::vector<uint> subk = k;
            --subk[_c];
            auto take = recursion(n-1,w-_w,subk);
            auto reject = recursion(n-1,w,k);
            if (take.second and reject.second) {
                nextv = std::max(_v + take.first,reject.first);
            } else if (take.second) {
                nextv = _v + take.first;
            } else if (reject.second) {
                nextv = reject.first;
            } else {
                nextv = 0;
                nextvalid = false;
            }   
        } else {
            std::tie(nextv,nextvalid) = recursion(n-1,w,k);
        }
        std::pair<T,bool> p = std::make_pair(nextv,nextvalid);
        cache.insert(std::make_pair(std::make_tuple(n,w,k),p));
        return p;   
    };

    return recursion(items.size(),W,K);
}

int main(int argc, char *argv[]) {
    std::vector< item<int> > items = {{60,10,0}, {100,20,1}, {120,30,0}, {140,35,1}, {145,40,0}, {180,45,1}, {160,50,1}, {170,55,0}};
    int  j = 145;
    for (uint k1 = 0; k1 <= items.size(); ++k1) {
        for (uint k2 = 0; k2 <= items.size(); ++k2) {
            auto p = knapSack(j,std::vector<uint>({k1,k2}),items);
            if (p.second)
                std::cout << "K0 = " << k1 << ", K1 = " << k2 << ": " << p.first << std::endl;
        }
    }
    return 0;
}

% OUTPUT (with comments) %

K0 = 0, K1 = 0: 0
K0 = 0, K1 = 1: 180 // e.g. {} from 0, {180} from 1
K0 = 0, K1 = 2: 340 // e.g. {} from 0, {160,180} from 1
K0 = 0, K1 = 3: 480 // e.g. {} from 0, {140,160,180} from 1
K0 = 1, K1 = 0: 170 // e.g. {170} from 0, {} from 1
K0 = 1, K1 = 1: 350 // e.g. {170} from 0, {180} from 1
K0 = 1, K1 = 2: 490 // e.g. {170} from 0, {140, 180} from 1
K0 = 1, K1 = 3: 565 // e.g. {145} from 0, {100, 140, 180} from 1
K0 = 2, K1 = 0: 315 // e.g. {145,170} from 0, {} from 1
K0 = 2, K1 = 1: 495 // e.g. {145,170} from 0, {180} from 1
K0 = 2, K1 = 2: 550 // e.g. {60,170} from 0, {140,180} from 1
K0 = 2, K1 = 3: 600 // e.g. {60,120} from 0, {100,140,180} from 1
K0 = 3, K1 = 0: 435 // e.g. {120,145,170} from 0, {} from 1
K0 = 3, K1 = 1: 535 // e.g. {120,145,170} from 0, {100} from 1
K0 = 3, K1 = 2: 605 // e.g. {60,120,145} from 0, {100,180} from 1
K0 = 4, K1 = 0: 495 // e.g. {60,120,145,170} from 0, {} from 1

#include <cstdio>
#include <iostream>
#include <vector>
#include <map>
#include <set>
#include <algorithm>

using uint = unsigned int;

template <typename T>
struct item {
    T value;
    uint weight;
    uint category;
};

template <typename T>
std::tuple<T,bool,std::set<size_t> > knapSack(uint W, std::vector<uint> K, const std::vector< item<T> >& items) {

    std::map< std::tuple<uint, uint, std::vector<uint> >, std::tuple<T,bool,std::set<std::size_t> > > cache;

    std::function<std::tuple<T,bool,std::set<std::size_t> >(uint, uint, std::vector<uint>&)> recursion;

    recursion = [&] (uint n, uint w, std::vector<uint>& k) {

        auto it = cache.find(std::make_tuple(n,w,k));
        if (it != cache.end())
            return it->second;

        std::vector<uint> ccount(K.size(),0);
        for (uint i = 0; i < n; ++i) {
            ++ccount[items[i].category];
        }

        for (uint c = 0; c < k.size(); ++c) {
            if (k[c] > ccount[c]) {
                auto p = std::make_tuple(0,false,std::set<std::size_t>{});
                cache.insert(std::make_pair(std::make_tuple(n,w,k),p));
                return p;
            }
        }
        uint sumk = 0; for (const auto& _k : k) sumk += _k;
        if (n == 0 || sumk == 0) {
            auto p = std::make_tuple(0,true,std::set<std::size_t>{});
            cache.insert(std::make_pair(std::make_tuple(n,w,k),p));
            return p;
        }

        T _v = items[n-1].value;
        uint _w = items[n-1].weight;
        uint _c = items[n-1].category;
        T nextv;
        bool nextvalid = true;
        std::set<std::size_t> nextset;
        if (_w <= w and k[_c] > 0) {
            --k[_c];
            auto take = recursion(n-1,w-_w,k);
            ++k[_c];
            auto reject = recursion(n-1,w,k);
            T a = _v + std::get<0>(take);
            T b = std::get<0>(reject);
            if (std::get<1>(take) and std::get<1>(reject)) {
                nextv = std::max(a,b);
                if (a > b) {
                    nextset = std::get<2>(take);
                    nextset.insert(n-1);
                } else {
                    nextset = std::get<2>(reject);
                }
            } else if (std::get<1>(take)) {
                nextv = a;
                nextset = std::get<2>(take);
                nextset.insert(n-1);
            } else if (std::get<1>(reject)) {
                nextv = b;
                nextset = std::get<2>(reject);
            } else {
                nextv = 0;
                nextvalid = false;
                nextset = {};
            }   
        } else {
            std::tie(nextv,nextvalid,nextset) = recursion(n-1,w,k);
        }
        auto p = std::make_tuple(nextv,nextvalid,nextset);
        cache.insert(std::make_pair(std::make_tuple(n,w,k),p));
        return p;   
    };

    return recursion(items.size(),W,K);
}

int main(int argc, char *argv[]) {
    std::vector< item<int> > items = {{60,10,0}, {100,20,1}, {120,30,0}, {140,35,1}, {145,40,0}, {180,45,1}, {160,50,1}, {170,55,0}};
    int  j = 145;
    for (uint k1 = 0; k1 <= items.size(); ++k1) {
        for (uint k2 = 0; k2 <= items.size(); ++k2) {
            auto p = knapSack(j,std::vector<uint>({k1,k2}),items);
            if (std::get<1>(p)) {
                std::cout << "K0 = " << k1 << ", K1 = " << k2 << ": " << std::get<0>(p);
                std::cout << "; contents are {";
                for (const auto& index : std::get<2>(p))
                    std::cout << index << ", ";
                std::cout << "}" << std::endl;
            }
        }
    }
    return 0;
}

K0 = 0, K1 = 0: 0; contents are {}
K0 = 0, K1 = 1: 180; contents are {5, }
K0 = 0, K1 = 2: 340; contents are {5, 6, }
K0 = 0, K1 = 3: 480; contents are {3, 5, 6, }
K0 = 1, K1 = 0: 170; contents are {7, }
K0 = 1, K1 = 1: 350; contents are {5, 7, }
K0 = 1, K1 = 2: 490; contents are {3, 5, 7, }
K0 = 1, K1 = 3: 565; contents are {1, 3, 4, 5, }
K0 = 2, K1 = 0: 315; contents are {4, 7, }
K0 = 2, K1 = 1: 495; contents are {4, 5, 7, }
K0 = 2, K1 = 2: 550; contents are {0, 3, 5, 7, }
K0 = 2, K1 = 3: 600; contents are {0, 1, 2, 3, 5, }
K0 = 3, K1 = 0: 435; contents are {2, 4, 7, }
K0 = 3, K1 = 1: 535; contents are {1, 2, 4, 7, }
K0 = 3, K1 = 2: 605; contents are {0, 1, 2, 4, 5, }
K0 = 4, K1 = 0: 495; contents are {0, 2, 4, 7, }