0-1背包问题 c语言

Here you will learn about 0-1 knapsack problem in C.

在这里，您将学习C语言中的0-1背包问题。

We are given n items with some weights and corresponding values and a knapsack of capacity W. The items should be placed in the knapsack in such a way that the total value is maximum and total weight should be less than knapsack capacity.

我们给了n个具有一定权重和相应值的项目以及一个容量为W的背包。这些项目应以总值最大且总重量小于背包容量的方式放置在背包中。

In this problem 0-1 means that we can’t put the items in fraction. Either put the complete item or ignore it. Below is the solution for this problem in C using dynamic programming.

在这个问题0-1中，意味着我们不能将项目分成零。 放入完整的项目或忽略它。 下面是使用动态编程在C中解决此问题的方法。

动态编程的C背包问题程序 (

Program for Knapsack Problem in C Using Dynamic Programming)

#include
    
      int max(int a, int b) { return (a > b)? a : b; } int knapSack(int W, int wt[], int val[], int n){   int i, w;   int K[n+1][W+1];    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];       }   }    return K[n][W];} int main(){    int i, n, val[20], wt[20], W;        printf("Enter number of items:");    scanf("%d", &n);        printf("Enter value and weight of items:\n");    for(i = 0;i < n; ++i){    	scanf("%d%d", &val[i], &wt[i]);    }     printf("Enter size of knapsack:");    scanf("%d", &W);        printf("%d", knapSack(W, wt, val, n));    return 0;}
    

Output

输出量

Enter number of items:3 Enter value and weight of items: 100 20 50 10 150 30 Enter size of knapsack:50 250

输入物品数量：3 输入物品的重量和重量： 100 20 50 10 150 30 输入背包尺寸：50 250

You can watch below video to learn knapsack problem easily.

您可以观看下面的视频轻松了解背包问题。

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0-1背包问题 c语言