給定n個元素的重量和其對應的價值,將這些物品放在一個容量為W的背包中,並使得總價值最大。數組val [0 . . n - 1]和wt [0 . . n - 1],它們分別代表價值和重量。 總重量W代表背包容量,
之前也寫過0-1背包問題:https://www.cnblogs.com/xiximayou/p/12004082.html
今天看到了個遞歸的方法,挺簡潔的,記錄一下:
def knapSack(W,wt,val,n): if n==0 or W==0: return 0 if wt[n-1]>W: return knapSack(W,wt,val,n-1) else: return max(val[n-1]+knapSack(W-wt[n-1],wt,val,n-1),knapSack(W,wt,val,n-1)) val = [60, 100, 120] wt = [10, 20, 30] W = 50 """ val=[5,4,6,2] wt=[2,4,5,3] W=8 """ n = len(val) print(knapSack(W , wt , val , n) )
輸出:220
遞歸方法會出現子問題重復計算問題,可用以下方法解決:
def knapSack(W, wt, val, n): K = [[0 for x in range(W+1)] for x 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] return K[n][W]
參考:https://www.geeksforgeeks.org/0-1-knapsack-problem-dp-10/