Create knapsack

knapsack

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+// C++ program to solve knapsack problem using
+// branch and bound
+#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, Item arr[], int n)
+{
+    // 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);
+    }
+
+    return maxProfit;
+}
+
+// driver program to test above function
+int main()
+{
+    int W = 10; // Weight of knapsack
+    Item arr[] = {{2, 40}, {3.14, 50}, {1.98, 100},
+                {5, 95}, {3, 30}};
+    int n = sizeof(arr) / sizeof(arr[0]);
+
+    cout << "Maximum possible profit = "
+        << knapsack(W, arr, n);
+
+    return 0;
+}