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試圖找到貪婪背包問題的最佳子集(python)

[英]Trying to find the optimal subset for the Greedy knapsack problem(python)

我認為這是找到最佳值的正確算法,但現在我需要找到獲得該值的最佳子集。 幫助將不勝感激!

這些是我的方向:實現一個貪心算法,按價值與重量比的降序排列項目(vi/wi for i = 1, 2, ..., n),然后 select 按此順序排列項目,直到重量下一個項目超過剩余容量(注意:在這個貪婪版本中,我們在第一個包含超過背包容量的項目之后立即停止)。

def greedy_knapsack(val, weight, W, n):
    # index = [0, 1, 2, ..., n - 1] for n items
    index = list(range(len(val)))
    # contains ratios of values to weight
    ratio = [v / w for v, w in zip(val, weight)]
    QuickSort(ratio, 0, len(ratio) - 1)
    max_value = 0
    for i in index:
        if weight[i] <= W:
            max_value += val[i]
            W -= weight[i]
        else:
            max_value += val[i] * W // weight[i]
            break
    return max_value

在許多情況下,你的貪婪方法會失敗。

一個這樣的微不足道的案例:

weight = [10, 10, 10]
value = [5, 4, 3]
W = 7

在這種情況下,您的算法將選擇 (項目 1) sum = 5,但最佳答案應該是 (項目 2 和 3),總和 = 7。

您需要一種動態編程方法來解決這個問題,並且您可以保留一個矩陣來存儲您以前的狀態,以便您可以重建解決方案並獲取項目列表。

# Prints the items which are put in a  
# knapsack of capacity W 
def printknapSack(W, wt, val, n): 
    K = [[0 for w in range(W + 1)] 
            for i in range(n + 1)] 

    # Build table K[][] in bottom 
    # up manner 
    for i in range(n + 1): 
        for w in range(W + 1): 
            if i == 0 or w == 0: 
                K[i][w] = 0
            elif wt[i - 1] <= w: 
                K[i][w] = max(val[i - 1]  
                  + K[i - 1][w - wt[i - 1]], 
                               K[i - 1][w]) 
            else: 
                K[i][w] = K[i - 1][w] 

    # stores the result of Knapsack 
    res = K[n][W] 
    print(res) 

    w = W 
    for i in range(n, 0, -1): 
        if res <= 0: 
            break
        # either the result comes from the 
        # top (K[i-1][w]) or from (val[i-1] 
        # + K[i-1] [w-wt[i-1]]) as in Knapsack 
        # table. If it comes from the latter 
        # one/ it means the item is included. 
        if res == K[i - 1][w]: 
            continue
        else: 

            # This item is included. 
            print(wt[i - 1]) 

            # Since this weight is included 
            # its value is deducted 
            res = res - val[i - 1] 
            w = w - wt[i - 1] 

# Driver code 
val = [ 60, 100, 120 ] 
wt = [ 10, 20, 30 ] 
W = 50
n = len(val) 

printknapSack(W, wt, val, n) 

參考: https://www.geeksforgeeks.org/printing-items-01-knapsack/

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