Huawei computer examination 01 knapsack problem

Package com.nowcoder.huawei; 

Import java.util.Scanner; 

/ ** 
 * input line 1, two positive integers, separated by a space: N m 
 * (where N (<32000) represents the total amount of money , m (<60) to a desired number of items purchased.) 
 * 
 * 
 * from row 2 to row m + 1, j-th row gives the number j-1 is the basic data items, each row has 3 nonnegative integers vpq 
 * (where v represents the price of the item (v <10000), 
 * P indicates the degree of importance of the items (1 ~ 5), q represents the article is a primary member or attachments. 
 * If q = 0, represents the main element of the article, if q> 0, indicating that the article is attachment, q is the number of main member belongs) 
 * 
 * 
 * 
 * / 
public  class ShoppingList {
     public  static  int getMaxValue ( int [] Val, int [] weight, int [] Q, int n-,int w) {
        int[][] dp = new int[n + 1][w + 1];
        for (int i = 1; i <= n; i++) {
            for (int j = 1; j <= w; j++) {
                if (q[i-1] == 0) {  // 主件
                    if (weight[i - 1] <= j) // 用j这么多钱去买 i 件商品 可以获得最大价值
                        dp[i][j] = Math.max(dp[i - 1][j], dp[i - 1][j- weight[i - 1]]+ val[i - 1]);
                }
                else { //附件
                    if (weight[i - 1] + weight[q[i - 1]] <= j) //附件的话 加上主件一起算
                        dp[i][j] = Math.max(dp[i - 1][j], dp[i - 1][j- weight[i - 1]]+ val[i - 1]);
                }
            }
        }
        return dp[n][w];
    }

    public static void main(String[] args) {
        Scanner input = new Scanner(System.in);
        while (input.hasNextInt()) {
            int n = input.nextInt(); // 总钱数
            int m = input.nextInt(); // 商品个数
            int[] p = new int[m];
            int[] v = new int[m];
            int[] q = new int[m];
            for (int i = 0; i < m; i++) {
                p[i] = input.nextInt();        // 价格
                v[i] = input.nextInt() * p[i]; // 价值
                q[i] = input.nextInt();        // 主件还是附件
            }
            System.out.println(getMaxValue(v, p, q, m, n));
        }
    }
}

 

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Origin www.cnblogs.com/bigdata-stone/p/11241968.html