算法-动态规划-完全背包

0. 动态规划五部曲：

1. 确定dp数组（dp table）以及下标的含义
2. 确定递推公式
3. dp数组如何初始化
4. 确定遍历顺序
5. 举例推导dp数组

1. 完全背包问题

完全背包问题中，每个物品都有无数个，可以重复选择。

• 二维dp数组

  int[][] dp = new int[n][totalWeight+1];
  for(int i = 0; i<n; ++i) {for(int j = 0; j<= totalWeight; ++j) {if(j < weight[i])dp[i][j] = dp[i-1][j];else // 01背包是dp[i-1][j-weight[i]]+value[i]，一个物品不能重复选dp[i][j] = Math.max(dp[i-1][j], dp[i][j-weight[i]]+value[i]);}
  }
  

• 一维dp数组：顺序遍历，允许同种物品被重复选择。

      int[] dp = new int[totalWeight+1];for(int i = 0; i<n; ++i) {for(int j = weight[i]; j<=totalWeight; ++j) {dp[j] = Math.max(dp[j], dp[j-weight[i]] + value[i]);}}
  

import java.util.Scanner;public class Main{public static void main(String[] args) {Scanner scanner = new Scanner(System.in);int n = scanner.nextInt();int totalWeight = scanner.nextInt();int[] weight = new int[n];int[] value = new int[n];for(int i = 0; i<n; ++i) {weight[i] = scanner.nextInt();value[i] = scanner.nextInt();}int[] dp = new int[totalWeight+1];for(int i = 0; i<n; ++i) {// 顺序遍历，同一个物品可以重复选for(int j = weight[i]; j<=totalWeight; ++j) {dp[j] = Math.max(dp[j], dp[j-weight[i]]+value[i]);}}System.out.println(dp[totalWeight]);}
}