у меня есть код, который возвращает максимум, но мне нужно напечатать путь в файле

Question

этот код может найти максимальное значение ранца в наборе данных. Мне нужно опубликовать путь, хотя в файле .txt в формате, как 0,1,0,1,1,0 ......, что означает 1-й элемент = 0 (не включены) 2-й элемент = 1 (включены в решение) ... есть предложения? проверенная таблица [n] существует для того, чтобы сохранить путь, а затем сохранить в файле

#include <bits/stdc++.h> 
using namespace std; 

// Structure for Item which store weight and corresponding 
// value of Item 
struct Item 
{ 
    float weight; 
    int value; 
}; 

// Node structure to store information of decision 
// tree 
struct Node 
{ 
    // level --> Level of node in decision tree (or index 
    //           in arr[] 
    // profit --> Profit of nodes on path from root to this 
    //       node (including this node) 
    // bound ---> Upper bound of maximum profit in subtree 
    //       of this node/ 
    int level, profit, bound; 
    float weight; 
}; 

// Comparison function to sort Item according to 
// val/weight ratio 
bool cmp(Item a, Item b) 
{ 
    double r1 = (double)a.value / a.weight; 
    double r2 = (double)b.value / b.weight; 
    return r1 > r2; 
} 

// Returns bound of profit in subtree rooted with u. 
// This function mainly uses Greedy solution to find 
// an upper bound on maximum profit. 
int bound(Node u, int n, int W, Item arr[]) 
{ 
    // if weight overcomes the knapsack capacity, return 
    // 0 as expected bound 
    if (u.weight >= W) 
        return 0; 

    // initialize bound on profit by current profit 
    int profit_bound = u.profit; 

    // start including items from index 1 more to current 
    // item index 
    int j = u.level + 1; 
    int totweight = u.weight; 

    // checking index condition and knapsack capacity 
    // condition 
    while ((j < n) && (totweight + arr[j].weight <= W)) 
    { 
        totweight += arr[j].weight; 
        profit_bound += arr[j].value; 
        j++; 
    } 

    // If k is not n, include last item partially for 
    // upper bound on profit 
    if (j < n) 
        profit_bound += (W - totweight) * arr[j].value / 
                                        arr[j].weight; 

    return profit_bound; 
} 

// Returns maximum profit we can get with capacity W 
int knapsack(int W, int ** tab, int n) 
{ 
    FILE *bbdfs=fopen("pr1.txt","a");



    int checked[n];//array to store chosen values in 0 & 1
    int i;
    for(i=0;i<n;i++){
        checked[i]=0;
    }
    Item arr[n];
    for(i=0;i<n;i++){
        arr[i].value=tab[i][1];
        arr[i].weight=tab[i][0];

    }



    // sorting Item on basis of value per unit 
    // weight. 
/// sort(arr, arr + n, cmp); 

    // make a queue for traversing the node 
    queue<Node> Q; 
    Node u, v; 

    // dummy node at starting 
    u.level = -1; 
    u.profit = u.weight = 0; 
    Q.push(u); 

    // One by one extract an item from decision tree 
    // compute profit of all children of extracted item 
    // and keep saving maxProfit 
    int maxProfit = 0; 
    while (!Q.empty()) 
    { 
        // Dequeue a node 
        u = Q.front(); 
        Q.pop(); 

        // If it is starting node, assign level 0 
        if (u.level == -1) 
            v.level = 0; 

        // If there is nothing on next level 
        if (u.level == n-1) 
            continue; 

        // Else if not last node, then increment level, 
        // and compute profit of children nodes. 
        v.level = u.level + 1; 

        // Taking current level's item add current 
        // level's weight and value to node u's 
        // weight and value 
        v.weight = u.weight + arr[v.level].weight; 
        v.profit = u.profit + arr[v.level].value; 

        // If cumulated weight is less than W and 
        // profit is greater than previous profit, 
        // update maxprofit 
        if (v.weight <= W && v.profit > maxProfit) 
            maxProfit = v.profit; 



        // Get the upper bound on profit to decide 
        // whether to add v to Q or not. 
        v.bound = bound(v, n, W, arr); 

        // If bound value is greater than profit, 
        // then only push into queue for further 
        // consideration 
        if (v.bound > maxProfit) 
            Q.push(v); 

        // Do the same thing, but Without taking 
        // the item in knapsack 
        v.weight = u.weight; 
        v.profit = u.profit; 
        v.bound = bound(v, n, W, arr); 
        if (v.bound > maxProfit) 
            Q.push(v); 
    } 
    for(i=0;i<n;i++){
            fprintf(bbdfs,"%d ,",checked[i]);
        }
        fprintf(bbdfs,"\n");
    return maxProfit; 
}