原文:
给定 n 个物品的重量和值，将这些物品放入容量为 w 的背包中，得到背包中的最大总值。换句话说，给定两个整数数组，val[0..n-1]和 wt[0..n-1]分别表示与 n 个项目相关联的值和权重。同样给定一个代表背包容量的整数 w，找出项目，使得给定子集的这些项目的权重之和小于或等于 w。您不能分解一个项目，要么选择完整的项目，要么不选择它(0-1 属性)。 先决条件: 示例:
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
进场: 让 val[] = {1，4，5，7}，wt[] = {1，3，4，5} w = 7。 2d 背包表看起来像:
从 k[n][w]开始回溯。这里 k[n][w]是 9。 由于该值来自顶部(灰色箭头所示)，因此不包括该行中的项目。在桌子上垂直向上，不要把这个放在背包里。现在，这个值 k[n-1][w]是 9，它不是从顶部来的，这意味着这一行中的项目被包括在内，并垂直向上，然后按所包括项目的重量向左移动(如黑色箭头所示)。继续这个过程，在背包中包括总价值为 9 的权重 3 和 4。
c
// cpp code for dynamic programming based
// solution for 0-1 knapsack problem
#include
#include
using namespace std;
// 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];
cout<< res << endl;
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.
cout<<" "<
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 语言(一种计算机语言，尤用于创建网站)
// 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.
python 3
# 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.
服务器端编程语言（professional hypertext preprocessor 的缩写）
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
?>
java 描述语言
output:
220
30 20
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