查找背包中携带的物品
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【中文标题】查找背包中携带的物品【英文标题】:Find which items are taken in knapsack 【发布时间】:2015-12-25 10:33:51 【问题描述】:递归解决背包问题,我想知道袋子里拿了哪些物品(物品的重量)给出的最大值。 到目前为止,我有这个:
int MAX(int a, int b) return (a > b) ? a : b ;
int thief(int W, int weight[], int value[], int n)
int a,b,c;
//basecase:
if(n == 0 || weight <= 0) return 0;
// each item's weight can't be more than W:
if(weight[n-1] > W)
return thief(W, weight, value, n-1);
a=value[n-1] + thief(W-weight[n-1], weight, value, n-1);// a: nth item included
b=thief(W, weight, value, n-1);// b:nth item not included
c= MAX(a,b);//answer is the maximum of situation a and b
if (c==a) //if situation a occurs then nth item is included
cout<<weight[n]<<endl;
return c;
考虑 n=4 和最大重量 (W) = 30 让权重为:30 10 20 5 和值:100 50 60 10 但此代码输出:20 5 20 10 5 我只想输出 10 和 20。 我还尝试定义一个默认值为 false 的 bool 数组,如果发生 c==a,它的第 n 个元素将更改为 true,但这也不会给出正确的结果。 我应该递归地这样做。
【问题讨论】:
【参考方案1】:您的基本算法不起作用。当您测试不同的组合时,您无法进行打印。
但是,首先你必须修复一个错误:
cout<<weight[n-1]<<endl; // n-1 instead of n
你的算法是这样做的:
a = value[3] + thief(30-weight[3], weight, value, 3); // Use item 3
b = thief(30, weight, value, 3); // Don't use item 3
第二行会导致
a = value[2] + thief(30-weight[2], weight, value, 2); // Use item 2
b = thief(30, weight, value, 2); // Don't use item 2
第二行会导致
a = value[1] + thief(30-weight[1], weight, value, 1); // Use item 1
b = thief(30, weight, value, 1); // Don't use item 1
第二行会导致
a = value[0] + thief(30-weight[0], weight, value, 0); // Use item 0
b = thief(30, weight, value, 0); // Don't use item 0
这导致
a = 30
b = 0
所以您的代码将选择 item 0
并打印 30
但这是一个错误!
所以正如我在开头所说的:你不能在测试不同的组合时进行打印。
相反,您需要跟踪您在不同组合中使用了哪些元素,并且只保留“最佳”。
我没有测试下面的代码,但我认为你可以这样做(假设你的代码正确计算了最佳组合):
#include <vector>
// The vector v is used for holding the index of the items selected.
// The caller must supply a vector containing the items included so far.
// This function will test whether item "n-1" shall be included or
// excluded. If item "n-1" is included the index is added to the vector.
int thief(int W, int weight[], int value[], int n, vector<int>& v) // Added vector
vector<int> v1, v2; // Vector to hold elements of the two combinations
int a,b,c;
//basecase:
if(n == 0 || weight <= 0) return 0;
// each item's weight can't be more than W:
if(weight[n-1] > W)
return thief(W, weight, value, n-1, v2);
v1.push_back(n-1); // Put n-1 in vector v1 and pass the vector v1
a=value[n-1] + thief(W-weight[n-1], weight, value, n-1, v1);// a: nth item included
// Don't put anything in v2 but pass the vector v2
b=thief(W, weight, value, n-1, v2);// b:nth item not included
c= MAX(a,b);//answer is the maximum of situation a and b
if (c==a) //if situation a occurs then nth item is included
// cout<<weight[n-1]<<endl;
// Copy elements from v1 to v
for (auto e : v1)
v.push_back(e);
else
// Copy elements from v2 to v
for (auto e : v2)
v.push_back(e);
return c;
int main()
vector<int> v;
int weight[4] = 30, 10, 20, 5;
int value[4] = 100, 50, 60, 10;
cout << "result=" << thief(30, weight, value, 4, v) << endl;
// Print the elements used
for (auto e : v)
cout << "elem=" << e << endl;
return 0;
最后请注意 - 您的蛮力方法在执行时间方面非常昂贵,因为 n 的起始值很高。有很多更好的方法可以解决这个问题。
【讨论】:
【参考方案2】:你能用c写代码吗? 我写了这个,但没有工作。 (我认为区别在于粗体)
int knapSack(int W, int wt[], int val[], int n, int arr[])
int x, y, c, j, arr1[50], arr2[50];
// Base Case
if (n == 0 || W == 0)
return 0;
// If weight of the nth item is more than Knapsack capacity W, then
// this item cannot be included in the optimal solution
if (wt[n - 1] > W)
return knapSack(W, wt, val, n - 1, arr2);
// Return the maximum of two cases:
// x nth item included
// y not included
**arr1[n - 1] = val[n - 1];**
x = val[n - 1] + knapSack(W - wt[n - 1], wt, val, n - 1, arr1);
//copyArr(arr, out, n);
y = knapSack(W, wt, val, n - 1, arr2);
if (x > y)
c = x;
else
c = y;
if (c == x)
for (j = 0; j < 50; j++)
arr[j] = arr1[j];
else
for (j = 0; j < 50; j++)
arr[j] = arr2[j];
return c;
【讨论】:
欢迎来到 SO。哪个位是粗体?以上是关于查找背包中携带的物品的主要内容,如果未能解决你的问题,请参考以下文章