[英]Knapsack using Dyanmic Programming Fails to give correct answer
我用的是我在這個環節中發現的算法實現背包問題的片段背包問題
我之前在Stack Overflow上問過這個問題,我在從程序的return語句獲取輸出時遇到了問題。 您可以在此處檢查上一個問題
我也在這里也附上了算法的代碼片段。
我為該算法編寫了以下python代碼段。 這里是:
def knapsack(v,w,n,W):
V = [[None for x in range(W+1)] for x in range(len(v)+1)]
keep = [[0 for x in range(W+1)] for x in range(len(v)+1)]
# print keep
for wy in range(W+1):
V[0][wy] = 0
for i in range(1,n+1):
for wx in range(W+1):
# print i,wx
if w[i-1] <= wx:
V[i][wx] = max(V[i-1][wx], v[i-1]+V[i-1][wx-w[i-1]])
keep[i][wx] = 1
else:
V[i][wx] = V[i-1][wx]
keep[i][wx] = 0
K = W
# print keep
for i in range(n,0,-1):
if keep[i][K] == 1:
print i
K = K - w[i-1]
return V[n][W]
print knapsack(v = [10,40,30,50], w=[5,4,6,3],n=4,W=10)
我應該得到4,2作為我的值,但是得到4,3 。 請更正我要去的地方。
如果存在問題:
if w[i-1] <= wx:
V[i][wx] = max(V[i-1][wx], v[i-1]+V[i-1][wx-w[i-1]])
keep[i][wx] = 1
else:
V[i][wx] = V[i-1][wx]
keep[i][wx] = 0
如果(w[i-1] <= wx) and (v[i-1]+V[i-1][wx-w[i-1]] <= V[i-1][wx])
是傻瓜
V[i][wx] = max(V[i-1][wx], v[i-1]+V[i-1][wx-w[i-1]])
keep[i][wx] = 1
但是你應該
V[i][wx] = V[i-1][wx]
keep[i][wx] = 0
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