运行快速动态规划算法以获得所有答案

Question

我如何运行快速动态规划算法来获得所有可能的答案。 假设我们有 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出来，并且不允许重复。 太感谢了。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：这是我较早的答案之一。那时我还没有那么好，现在我知道有一个更简单、更快、内存消耗更少的解决方案来解决这个问题。如果我们在只存储最接近的可能和的表中自下而上计算 DP，我们可以稍后使用我们计算的表值递归地重建子集。

一种解决方案，输出总和等于最大可能总和但不大于目标值且包含尽可能少的数字的所有数字集：

#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";
    }
}

此解决方案不允许数字在输出中出现的次数超过它在输入中出现的次数。如果您只想允许一个数字在输出中出现一次，即使它在输入中出现多次，请将 multiset numbersleft 更改为一个集合。

#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";
        }
    }
}