背包问题
来源:互联网 发布:爵迹亏了多少钱知乎 编辑:程序博客网 时间:2024/06/03 13:50
Backpack I
Problem 单次选择+最大体积
Given n items with size Ai, an integer m denotes the size of a backpack. How full you can fill this backpack?
Notice
You can not divide any item into small pieces.
Example
If we have 4 items with size [2, 3, 5, 7], the backpack size is 11, we can select [2, 3, 5], so that the max size we can fill this backpack is 10. If the backpack size is 12. we can select [2, 3, 7] so that we can fulfill the backpack.
You function should return the max size we can fill in the given backpack.
Challenge
O(n x m) time and O(m) memory.
O(n x m) memory is also acceptable if you do not know how to optimize memory.
Note
动规经典题目,用数组dp[i]表示书包空间为i的时候能装的A物品最大容量。两次循环,外部遍历数组A,内部反向遍历数组dp,若j即背包容量大于等于物品体积A[i],则取前i-1次循环求得的最大容量dp[j],和背包体积为j-A[i]时的最大容量dp[j-A[i]]与第i个物品体积A[i]之和即dp[j-A[i]]+A[i]的较大值,作为本次循环后的最大容量dp[i]。
注意dp[]的空间要给m+1,因为我们要求的是第m+1个值dp[m],否则会抛出OutOfBoundException。
Solution
public class Solution { public int backPack(int m, int[] A) { int[] dp = new int[m+1]; for (int i = 0; i < A.length; i++) { for (int j = m; j > 0; j--) { if (j >= A[i]) { dp[j] = Math.max(dp[j], dp[j-A[i]] + A[i]); } } } return dp[m]; }}
Backpack II
Problem 单次选择+最大价值
Given n items with size A[i] and value V[i], and a backpack with size m. What's the maximum value can you put into the backpack?
Notice
You cannot divide item into small pieces and the total size of items you choose should smaller or equal to m.
Example
Given 4 items with size [2, 3, 5, 7] and value [1, 5, 2, 4], and a backpack with size 10. The maximum value is 9.
Challenge
O(n x m) memory is acceptable, can you do it in O(m) memory?
Note
和BackPack I基本一致。依然是以背包空间为限制条件,所不同的是dp[j]取的是价值较大值,而非体积较大值。所以只要把dp[j-A[i]]+A[i]换成dp[j-A[i]]+V[i]就可以了。
Solution
public class Solution { public int backPackII(int m, int[] A, int V[]) { int[] dp = new int[m+1]; for (int i = 0; i < A.length; i++) { for (int j = m; j > 0; j--) { if (j >= A[i]) dp[j] = Math.max(dp[j], dp[j-A[i]]+V[i]); } } return dp[m]; }}
Backpack III
Problem 重复选择+最大价值
Given n kind of items with size Ai and value Vi( each item has an infinite number available) and a backpack with size m. What's the maximum value can you put into the backpack?
Notice
You cannot divide item into small pieces and the total size of items you choose should smaller or equal to m.
Example
Given 4 items with size [2, 3, 5, 7] and value [1, 5, 2, 4], and a backpack with size 10. The maximum value is 15.
Solution
public class Solution { public int backPackIII(int[] A, int[] V, int m) { int[] dp = new int[m+1]; for (int i = 0; i < A.length; i++) { for (int j = 1; j <= m; j++) { if (j >= A[i]) dp[j] = Math.max(dp[j], dp[j-A[i]]+V[i]); } } return dp[m]; }}
Backpack IV
Problem 重复选择+唯一排列+装满可能性总数
Given n items with size nums[i] which an integer array and all positive numbers, no duplicates. An integer target denotes the size of a backpack. Find the number of possible fill the backpack.
Each item may be chosen unlimited number of times
Example
Given candidate items [2,3,6,7] and target 7,
A solution set is:
[7][2, 2, 3]return 2
Solution
public class Solution { public int backPackIV(int[] nums, int target) { int[] dp = new int[target+1]; dp[0] = 1; for (int i = 0; i < nums.length; i++) { for (int j = 1; j <= target; j++) { if (nums[i] == j) dp[j]++; else if (nums[i] < j) dp[j] += dp[j-nums[i]]; } } return dp[target]; }}
Backpack V
Problem 单次选择+装满可能性总数
Given n items with size nums[i] which an integer array and all positive numbers. An integer target denotes the size of a backpack. Find the number of possible fill the backpack.
Each item may only be used once
Example
Given candidate items [1,2,3,3,7] and target 7,
A solution set is:
[7][1, 3, 3]return 2
Solution
public class Solution { public int backPackV(int[] nums, int target) { int[] dp = new int[target+1]; dp[0] = 1; for (int i = 0; i < nums.length; i++) { for (int j = target; j >= 0; j--) { if (j >= nums[i]) dp[j] += dp[j-nums[i]]; } } return dp[target]; }}
Backpack VI aka: Combination Sum IV
Problem 重复选择+不同排列+装满可能性总数
Given an integer array nums with all positive numbers and no duplicates, find the number of possible combinations that add up to a positive integer target.
Notice
The different sequences are counted as different combinations.
Example
Given nums = [1, 2, 4], target = 4
The possible combination ways are:
[1, 1, 1, 1][1, 1, 2][1, 2, 1][2, 1, 1][2, 2][4]return 6
Solution
public class Solution { public int backPackVI(int[] nums, int target) { int[] dp = new int[target+1]; dp[0] = 1; for (int i = 1; i <= target; i++) { for (int num: nums) { if (num <= i) dp[i] += dp[i-num]; } } return dp[target]; }}
- 【无限背包】背包问题
- 背包问题---01背包
- 背包问题--部分背包
- 背包问题
- 背包问题
- 背包问题
- 背包问题
- 背包问题
- 背包问题
- 背包问题
- 背包问题
- 背包问题
- 背包问题
- 背包问题
- 背包问题
- 背包问题
- 背包问题
- 背包问题
- 4.nginx配置文件详细说明
- 关于程序员使用的各种软件的教程总贴——这里做个汇总
- Java后端程序员1年工作经验总结
- 为AndroidStudio设置自定义类注释
- Android
- 背包问题
- Codeforces Round #452 (Div. 2) B
- 网络判断跳转设置界面
- 东忙西忙失去了重心
- 前端工程——基础篇
- android 图片工具类 (图片压缩 图片长按缓存 Bitmap转Base64 Bitmap转File File转Bitmap 打开系统相册解析URI)
- 一个优秀的CIO,应该具备如何的知识体系和管理能力?
- DCGAN、WGAN、WGAN-GP、LSGAN、BEGAN原理总结及对比
- 数据结构——序列