给定n个物料的权重和值，将这些物料放在容量为W的背包中，以在背包中获得最大的总价值。换句话说，给定两个整数数组val [0..n-1]和wt [0..n-1]，它们分别表示与n个项目相关联的值和权重。还要给定代表背包容量的整数W，找出物品，使给定子集的那些物品的权重之和小于或等于W。选择它(0-1属性)。
先决条件： 0/1背包
例子 ：
Input : val[] = {60, 100, 120};
wt[] = {10, 20, 30};
W = 50;
Output : 220 //maximum value that can be obtained
30 20 //weights 20 and 30 are included.
Input : val[] = {40, 100, 50, 60};
wt[] = {20, 10, 40, 30};
W = 60;
Output : 200
30 20 10
方法 ：
令val [] = {1，4，5，7}，wt [] = {1，3，4，5}
W = 7。
2d背包表如下所示： 
从K [n] [W]开始回溯。此处K [n] [W]为9。
由于此值来自顶部(由灰色箭头显示)，因此不包括此行中的项目。在桌子上垂直向上走，但不包括在背包中。现在，该值K [n-1] [W]为9并不是从顶部开始的，这意味着该行中的项目已被包含并垂直向上移动，然后由所包含项目的权重向左移动(如黑色所示)箭)。继续进行此过程，将背包中的权重3和权重4总计为9。
C++
// CPP code for Dynamic Programming based
// solution for 0-1 Knapsack problem
#include
// A utility function that returns maximum of two integers
int max(int a, int b) { return (a > b) ? a : b; }
// Prints the items which are put in a knapsack of capacity W
void printknapSack(int W, int wt[], int val[], int n)
{
int i, w;
int K[n + 1][W + 1];
// Build table K[][] in bottom up manner
for (i = 0; i <= n; i++) {
for (w = 0; w <= W; w++) {
if (i == 0 || w == 0)
K[i][w] = 0;
else if (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
int res = K[n][W];
printf("%d\n", res);
w = W;
for (i = n; i > 0 && res > 0; i--) {
// 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.
printf("%d ", wt[i - 1]);
// Since this weight is included its
// value is deducted
res = res - val[i - 1];
w = w - wt[i - 1];
}
}
}
// Driver code
int main()
{
int val[] = { 60, 100, 120 };
int wt[] = { 10, 20, 30 };
int W = 50;
int n = sizeof(val) / sizeof(val[0]);
printknapSack(W, wt, val, n);
return 0;
} Java
// Java code for Dynamic Programming based
// solution for 0-1 Knapsack problem
class GFG {
// A utility function that returns
// maximum of two integers
static int max(int a, int b)
{
return (a > b) ? a : b;
}
// Prints the items which are put
// in a knapsack of capacity W
static void printknapSack(int W, int wt[],
int val[], int n)
{
int i, w;
int K[][] = new int[n + 1][W + 1];
// Build table K[][] in bottom up manner
for (i = 0; i <= n; i++) {
for (w = 0; w <= W; w++) {
if (i == 0 || w == 0)
K[i][w] = 0;
else if (wt[i - 1] <= w)
K[i][w] = Math.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
int res = K[n][W];
System.out.println(res);
w = W;
for (i = n; i > 0 && res > 0; i--) {
// 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.
System.out.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
public static void main(String arg[])
{
int val[] = { 60, 100, 120 };
int wt[] = { 10, 20, 30 };
int W = 50;
int n = val.length;
printknapSack(W, wt, val, n);
}
}
// This code is contributed by Anant Agarwal.Python3
# Python3 code for Dynamic Programming
# based solution for 0-1 Knapsack problem
# 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)
# This code is contributed by Aryan Garg.C#
// C# code for Dynamic Programming based
// solution for 0-1 Knapsack problem
using System ;
class GFG {
// A utility function that returns
// maximum of two integers
static int max(int a, int b)
{
return (a > b) ? a : b;
}
// Prints the items which are put
// in a knapsack of capacity W
static void printknapSack(int W, int []wt,
int []val, int n)
{
int i, w;
int [,]K = new int[n + 1,W + 1];
// Build table K[][] in bottom up manner
for (i = 0; i <= n; i++) {
for (w = 0; w <= W; w++) {
if (i == 0 || w == 0)
K[i,w] = 0;
else if (wt[i - 1] <= w)
K[i,w] = Math.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
int res = K[n,W];
Console.WriteLine(res);
w = W;
for (i = n; i > 0 && res > 0; i--) {
// 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.
Console.Write(wt[i - 1] + " ");
// Since this weight is included its
// value is deducted
res = res - val[i - 1];
w = w - wt[i - 1];
}
}
}
// Driver code
public static void Main()
{
int []val = { 60, 100, 120 };
int []wt = { 10, 20, 30 };
int W = 50;
int n = val.Length;
printknapSack(W, wt, val, n);
}
}
// This code is contributed by Ryuga.PHP
0 && $res > 0; $i--)
{
// either the result comes from the top
// (K[i-1][w]) or from (val[i-1] + K[i-1]
// [w-wt[i-1]]) as in Knapsack table. If
// it comes from the latter one/ it means
// the item is included.
if ($res == $K[$i - 1][$w])
continue;
else
{
// This item is included.
echo $wt[$i - 1] . " ";
// Since this weight is included
// its value is deducted
$res = $res - $val[$i - 1];
$w = $w - $wt[$i - 1];
}
}
}
// Driver code
$val = array(60, 100, 120);
$wt = array(10, 20, 30);
$W = 50;
$n = sizeof($val);
printknapSack($W, $wt, $val, $n);
// This code is contributed by ita_c
?>输出：
220
30 20