运行快速动态编程算法以获得所有答案

Question

我如何运行快速动态编程算法来获得所有可能的答案。 想象我们有20个条目，它只显示1行最佳答案，我希望它一直运行并显示其他条目，直到所有条目都显示为结果，并且不允许重复。 非常感谢。真的很感激。 这是代码：

#include <iostream>
#include <set>
#include <vector>
#include <map>
#include <utility>

using namespace std;

float W ,N; //N = olcu sayisi, W = profil boyu
vector<float> numbers; //stores the set of numbers

pair<float, multiset<float>> calc(float i, float j) //returns closest sum and best subset of the first i numbers for the target value j
{
    static map<pair<float, float>, pair<float, multiset<float>>> dp; //stores results to avoid repeated calculations

    pair<float, float> p(i, j); //pair for convenience

    if(i == 0) //base case
    {
        return make_pair(0, multiset<float>(
                                      {}));
    }

    auto findResult = dp.find(p);

    if(findResult != dp.end()) //check if already calculated
    {
        return findResult->second;
    }

    auto temp1 = calc(i - 1, j); //compute result if not using number

    if(numbers[i - 1] > j) //if current number is too big
    {
        return temp1;
    }

    auto temp2 = calc(i - 1, j - numbers[i - 1]); //compute result if using number
    temp2.first += numbers[i - 1];
    temp2.second.insert(numbers[i - 1]);

    pair<float, multiset<float>> result;
    if(temp1.first != temp2.first) //compare results and choose best
    {
        result = temp1.first > temp2.first ? temp1 : temp2;
    }
    else
    {
        result = temp1.second.size() < temp2.second.size() ? temp1 : temp2;
    }
    dp[p] = result;

    return result;
}

int main()
{
    cout << "sineklik sayisi: ";
    cin >> N;
    N = 2 * N;
    cout << "Profil olcusu: ";
    cin >> W;
    numbers.reserve(N); //avoid extra reallocations

    cout << "Olculeri giriniz: ";
    for(int i = 0; i < N; i++) //input loop
    {
        float temp;
        cin >> temp;
        numbers.push_back(temp);
    }

    pair<float, multiset<float>> result = calc(N, W); //calculate

    //output below
    cout << "The best possible sum is " << result.first << " Left behind is " << W - result.first << ", obtained using the set of numbers {";
    if(result.second.size() > 0)
    {
        cout << *result.second.begin();
        for(auto i = ++result.second.begin(); i != result.second.end(); i++)
        {
            cout << ", " << *i;
        }
    }
    cout << "}.\n";
}

Answer 1

编辑：这是我以前的答案之一。当时我不是那么好，现在我知道有一个更简单，更快，更少耗费内存的解决方案来解决这个问题。如果我们在仅存储最接近的可能总和的表中自下而上计算DP，我们可以使用我们计算的表值以递归方式重建子集。

一种解决方案，输出所有数字组合，其总和等于最大可能总和不大于目标值并包含尽可能少的数字：

#include <iostream>
#include <set>
#include <vector>
#include <map>
#include <utility>

using namespace std;

int N, W; //N = number of numbers, W = target sum
vector<int> numbers; //stores the set of numbers

pair<int, set<multiset<int>>> calc(int i, int j) //returns closest sum and best subset of the first i numbers for the target value j
{
    static map<pair<int, int>, pair<int, set<multiset<int>>>> dp; //stores results to avoid repeated calculations
    pair<int, int> p(i, j); //pair for convenience
    if(i == 0) //base case
    {
        set<multiset<int>> temp;
        temp.emplace();
        return make_pair(0, temp);
    }
    auto findResult = dp.find(p);
    if(findResult != dp.end()) //check if already calculated
    {
        return findResult->second;
    }
    auto temp1 = calc(i - 1, j); //compute result if not using number
    if(numbers[i - 1] > j) //if current number is too big
    {
        return temp1;
    }
    pair<int, set<multiset<int>>> temp2 = calc(i - 1, j - numbers[i - 1]), newtemp2; //compute result if using number
    newtemp2.first = temp2.first + numbers[i - 1];
    for(const auto k : temp2.second)
    {
        multiset<int> temp = k;
        temp.insert(numbers[i - 1]);
        newtemp2.second.insert(temp);
    }
    pair<int, set<multiset<int>>> *result;
    if(temp1.first != newtemp2.first) //compare results and choose best
    {
        result = temp1.first > newtemp2.first ? &temp1 : &newtemp2;
    }
    else if(temp1.second.begin()->size() != newtemp2.second.begin()->size())
    {
        result =
                temp1.second.begin()->size() < newtemp2.second.begin()->size() ? &temp1 : &newtemp2;
    }
    else
    {
        temp1.second.insert(newtemp2.second.begin(), newtemp2.second.end());
        result = &temp1;
    }
    dp.insert(make_pair(p, *result));
    return *result;
}

int main()
{
    cout << "Enter the number of numbers: ";
    cin >> N;
    cout << "Enter target sum: ";
    cin >> W;
    numbers.reserve(N); //avoid extra reallocations
    cout << "Enter the numbers: ";
    for(int i = 0; i < N; i++) //input loop
    {
        int temp;
        cin >> temp;
        numbers.push_back(temp);
    }
    pair<int, set<multiset<int>>> result = calc(N, W); //calculate
    //output below
    cout << "The best possible sum is " << result.first << ", which can be obtained using a set of "
            << result.second.begin()->size() << " numbers " << result.second.size()
            << " different ways:\n";
    for(const auto &i : result.second)
    {
        cout << '{';
        if(i.size() > 0)
        {
            cout << *i.begin();
            for(auto j = ++i.begin(); j != i.end(); ++j)
            {
                cout << ", " << *j;
            }
        }
        cout << "}\n";
    }
}

此解决方案不允许输出中出现的数字超过输入中出现的次数。如果您希望仅允许数字在输出中出现一次，即使它在输入中出现多次，也请将多重组numbersleft更改为一组。

#include <iostream>
#include <set>
#include <vector>
#include <map>
#include <utility>

using namespace std;

int N, W; //N = number of numbers, W = target sum
vector<int> numbers; //stores the set of numbers

pair<int, set<multiset<int>>> calc(int i, int j) //returns closest sum and best subset of the first i numbers for the target value j
{
    static map<pair<int, int>, pair<int, set<multiset<int>>>> dp; //stores results to avoid repeated calculations
    pair<int, int> p(i, j); //pair for convenience
    if(i == 0) //base case
    {
        set<multiset<int>> temp;
        temp.emplace();
        return make_pair(0, temp);
    }
    auto findResult = dp.find(p);
    if(findResult != dp.end()) //check if already calculated
    {
        return findResult->second;
    }
    auto temp1 = calc(i - 1, j); //compute result if not using number
    if(numbers[i - 1] > j) //if current number is too big
    {
        return temp1;
    }
    pair<int, set<multiset<int>>> temp2 = calc(i - 1, j - numbers[i - 1]), newtemp2; //compute result if using number
    newtemp2.first = temp2.first + numbers[i - 1];
    for(const auto k : temp2.second)
    {
        multiset<int> temp = k;
        temp.insert(numbers[i - 1]);
        newtemp2.second.insert(temp);
    }
    pair<int, set<multiset<int>>> *result;
    if(temp1.first != newtemp2.first) //compare results and choose best
    {
        result = temp1.first > newtemp2.first ? &temp1 : &newtemp2;
    }
    else if(temp1.second.begin()->size() != newtemp2.second.begin()->size())
    {
        result =
                temp1.second.begin()->size() < newtemp2.second.begin()->size() ? &temp1 : &newtemp2;
    }
    else
    {
        temp1.second.insert(newtemp2.second.begin(), newtemp2.second.end());
        result = &temp1;
    }
    dp.insert(make_pair(p, *result));
    return *result;
}

int main()
{
    cout << "Enter the number of numbers: ";
    cin >> N;
    cout << "Enter target sum: ";
    cin >> W;
    numbers.reserve(N); //avoid extra reallocations
    cout << "Enter the numbers: ";
    for(int i = 0; i < N; i++) //input loop
    {
        int temp;
        cin >> temp;
        numbers.push_back(temp);
    }
    pair<int, set<multiset<int>>> result = calc(N, W); //calculate
    //output below
    cout << "The best possible sum is " << result.first << ", which can be obtained using sets of "
            << result.second.begin()->size() << " numbers:\n";
    multiset<int> numbersleft;
    numbersleft.insert(numbers.begin(), numbers.end());
    for(const auto &i : result.second)
    {
        bool good = true;
        for(const int &j : i)
        {
            if(numbersleft.find(j) == numbersleft.end())
            {
                good = false;
                break;
            }
        }
        if(good)
        {
            for(const int &j : i)
            {
                numbersleft.erase(j);
            }
            cout << '{';
            if(i.size() > 0)
            {
                cout << *i.begin();
                for(auto j = ++i.begin(); j != i.end(); ++j)
                {
                    cout << ", " << *j;
                }
            }
            cout << "}\n";
        }
    }
}