算法 - 动态规划(0-1背包问题)

雨点打透心脏的1/2处 2023-07-03 10:35 55阅读 0赞

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# 推出公式： #

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第三个就是：让上一次的和（新加入商品容量+(总空间-新加入的商品容量）也就是剩余空间的最大值，剩余空间的最大值去上一层找）

package Algorithm.dac.knapsack;
    
    public class KnapsackProblem { 
        public static void main(String []args){ 
            int [] weight = { 1, 4, 3};//物品的重量
            int [] value = { 1500, 3000, 2000};//物品的价值
            int m = 4; //背包容量
            int n = value.length; //物品的个数
    
            //创建二维数组
            //v[i][j]表示在前i个物品中能够装入容量为j的背包中的最大价值
            int [] [] v = new int[n+1][m+1];
            int [] [] path = new int[n+1][m+1];
            //初始化第一行和第一列，这里可以不处理，默认就是0
            for (int i = 0; i < v.length; i++) { 
                v[i][0] = 0; //第一列设置为 0
            }
    
            for (int i = 0; i < v[0].length; i++) { 
                v[0][i] = 0;//将第一行设置为 0
            }
    
            //根据公式来动态规划处理
            for (int i = 1; i < v.length; i++) { 
                for (int j = 1; j < v[0].length; j++) { 
                    if (weight[i -1] > j){ 
                        v[i][j] = v[i-1][j];
                    }else { 
                        //v[i][j] = Math.max(v[i-1][j], value[i-1] + v[i-1][j-weight[i-1]]);
                        if (v[i-1][j] < value[i-1] + v[i-1][j-weight[i-1]]){ 
                            v[i][j] = value[i-1] + v[i-1][j-weight[i-1]];
                            path[i][j] = 1;
                        }else { 
                            v[i][j] = v[i-1][j];
                        }
                    }
                }
            }
            //输出一下v
            for (int i = 0; i < v.length; i++) { 
                for (int j = 0; j < v[0].length; j++) { 
                    System.out.print(v[i][j]+" ");
                }
                System.out.println();
            }
    
            System.out.println("===========");
    
            int i = path.length - 1;
            int j = path[0].length - 1;
            //v[i][j] = value[i-1] + v[i-1][j-weight[i-1]];
            //如果是1那么，v[i][j] = value[i-1] + v[i-1][j-weight[i-1]]; 就是 物品value[i-1] + v[i-1][j-weight[i-1]]，在判断v[i-1][j-weight[i-1]]是1还是0,是1继续循环，是0就退出。
            while (i > 0 && j > 0){ 
    
                if (path[i][j] == 1) { 
                    System.out.println("物品:" + i + "加入背包");
                    j = j - weight[i - 1];
                }
                i = i - 1;
            }
        }
    }

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