Bilibili website-dynamic programming basic backpack problem (1) 01 backpack

Video URL-Bilibili website-Dynamic programming basic backpack problem (1) 01 backpack

Scrolling array

Knapsack problem-optimization algorithm-step jump point

// #include "everything.h"
#include <bits/stdc++.h> // 错误代码

using namespace std;

int N, V, dp[1001][100001], c[100001], w[100001]; // c重量、w价值

int main()
{
    int i, j;
    cin >> N >> V;
    for (i = 1; i <= N; i++)
        cin >> c[i] >> w[i]; // 读入每个物品的重量与收益
    for (i = 1; i <= N; i++)
    {
        for (j = V; j >= 0; j--)
        {
            if (j < c[i])
                break;
            dp[j] = max(dp[j], dp[j - c[i]] + w[i]);
        }
    }
    cout << dp[V] << endl;
    system("pause");
    return 0;
}

Bilibili Website-[Dynamic Programming] Backpack Problem

Video URL-Bilibili website- [Dynamic Programming] Backpack problem

Backtracking: Nothing is the same as 10, so item 4 must be packed into the backpack. The backpack space is 8, and the volume of object 4 is 5. The remaining space "3" in the backpack is used to hold other items. Consider the best combination of items when the backpack capacity is 3. --> Consider whether the item No. 3 has been loaded into the backpack --> If the item No. 3 has not been loaded into the backpack, consider the first two items loaded into the backpack.

Java code implementation: https://www.cnblogs.com/jiyongjia/p/13475026.html

Code-original link

package Z;

import java.util.ArrayList;
import java.util.LinkedHashMap;
import java.util.List;
import java.util.Map;

public class Beibao {

	public static void main(String[] args) {
		Map<Integer, Integer> map = new LinkedHashMap<>();
		map.put(2, 3);
		map.put(3, 4);
		map.put(4, 5);
		map.put(5, 6);
		System.out.print("背包可容纳最大价值：" + getMaxValue(map, 8));
	}

	private static Integer getMaxValue(Map<Integer, Integer> gems, int capacity) {
		int[][] maxValue = new int[gems.size() + 1][capacity + 1];
		List<Integer> gemList = new ArrayList<>();
		int choose, notChoose;
		for (int i = 0; i < gems.size() + 1; i++) {
			maxValue[i][0] = 0;
		}
		for (int i = 0; i < capacity + 1; i++) {
			maxValue[0][i] = 0;
		}
		gemList.add(0);
		for (Integer gemKey : gems.keySet()) {
			gemList.add(gemKey);
		}

		for (int i = 1; i < gems.size() + 1; i++) {
			for (int j = 1; j < capacity + 1; j++) {
				if (gemList.get(i) > j) {
					maxValue[i][j] = maxValue[i - 1][j];
				} else {
					choose = gems.get(gemList.get(i)) + maxValue[i - 1][j - gemList.get(i)];
					notChoose = maxValue[i - 1][j];
					maxValue[i][j] = Math.max(choose, notChoose);
				}
			}
		}

		for (int i = 0; i < gems.size() + 1; i++) {
			for (int j = 0; j < capacity + 1; j++) {
				System.out.print(maxValue[i][j] + " ");
			}
			System.out.println();
		}
		getDetails(maxValue, gems, gemList, gems.size() + 1, capacity + 1);
		return maxValue[gems.size()][capacity];
	}

	private static void getDetails(int[][] maxValue, Map<Integer, Integer> gems, 
			List<Integer> gemList, int rows, int cols) {
		List<Integer> details = new ArrayList<>();
		while (rows > 1 && cols > 1) {
			if (maxValue[rows - 1][cols - 1] != maxValue[rows - 2][cols - 1]) {
				details.add(rows - 1);
				rows--;
				cols = cols - 1 - gemList.get(rows - 1);
			} else {
				rows--;
			}
		}
		System.out.println("装入背包的有：");
		for (int i = 0; i < details.size(); i++) {
			System.out.println(
					"体积为" + gemList.get(details.get(i)) + ",价值为" + gems.get(gemList.get(details.get(i))) + "的石头");
		}
	}

}

Thanks for watching,

The basis of dynamic programming [knapsack problem]

Bilibili website-dynamic programming basic backpack problem (1) 01 backpack

Bilibili Website-[Dynamic Programming] Backpack Problem

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