class Solution: def change(self, amount: int, coins: List[int]) -> int: import numpy as np dp = np.zeros([len(coins) + 1, amount + 1]) for i in range(0, len(coins) + 1): dp[i, 0] = 1 for j in range(1, amount + 1): dp[0, j] = 0 for i in range(1, len(...
Knapsack problem (超大01背包) FZUOJ 2214 Knapsack problem 之前的01背包都是在枚举背包的容量,时间复杂度可以优化到O(N∗V)O(N*V)O(N∗V),但是如果背包的总的容量非常大的话,就没办法枚举背包容量来进行计算了。 题目描述(题目链接) 有一个背包,容量为BBB,同时有nnn个物品,每个物品有一个重量www和...
2. 背包问题(Knapsack Problem) 题目描述(以 0-1 背包问题为例): 给定n 种物品和一个背包,每种物品都有自己的重量和价值,在限定的最大重量内,如何选择物品使得背包内物品的总价值最大? 解题思路: 使用二维数组 dp,其中 dp[i][j] 表示前 i 件物品在总重量不超过 j 的情况下可以达到的最大价值。状态转...
// Solution 1: Sort of like a Knapsack problem. 代码1 // Code 1 495 Teemo Attacking // #495 提莫进攻 描述:给定一些进攻的时间点,和进攻的中毒持续时间,求敌人总共的中毒时间。 // #495 Description: Teemo Attacking | LeetCode OJ 解法1:这么水的题,居然是medium? // Solution 1: Too trivial ...
for i=1..N for v=V..0 f[v]=max{f[v],f[v-c[i]]+w[i]}; 其中的f[v]=max{f[v],f[v-c[i]]}一句恰就相当于我们的转移方程f[i][v]=max{f[i-1][v],f[i-1][v-c[i]]},因为现在的f[v-c[i]]就相当于原来的f[i-1][v-c[i]]。如果将v的循环顺序从上面的逆序改成顺序...
Capacity of Knapsack:Limitless Objective:Pick up at most m 0's and 1 n's vector<vector<int>> dp(M+1,vector<int>(N+1,0));for(autos : strs) {intzero = zeros(s);// a function to count zeroes in a stringintone = ones(s);// a function to count ones in a stringfor(inti=M...
// Solution 1: The problem itself is easy, except no really if you wish to attain O(N) time. The writing is a bit tricky if you wish to make it clean. First, we care only the max and min for each array. Then, we calculate the max and min for each array except itself. (Capi...
For example, if a "greedy thief" with a 10-Liter knapsack sees two types of items a 6-Liter item worth $9 (1.5 $/L) a 5-Liter item worth $5 (1.0 $/L) the thief will take 1 of the 6-Liter items instead of 2 of the 5-Liter items. Although this means the thief will only...
E.Y.-H. Lin, "A dynamic programming approach to the miltiple-choice multi-period knapsack problem and the recursive APL2 code," w: Proc. of the 17th Triennal Conference on the International Federation of Operational Research Societies, 2005....
Fractional Knapsack Problem: Items can be divided to maximize value. LRU and FIFO L1 Cache Implementation using C This is a C program to demonstrate cache mechanism by simulating a cache in C. The source code can run in any C Compiler with minor modifications if required. It can run on ...