replay backpack

There is a backpack can be put into the item weight of S, existing n items, the weight is w1,w2,w3,...wn. Ask if you can select several items from the N-item into the backpack, so that the sum of the weights placed is just s. If there is a choice to meet the conditions, then the backpack has a solution, or the knapsack problem is not solved.DescriptionFirst act i

Title: Uva-10280old Wine into New bottles (full backpack + pruning)The main topic: Now to put the old wine into the new bottle, each bottle has the smallest maximum capacity requirements, and then give you l wine, in the give you n bottles, each bottle of the specification also give you, each kind of bottle supply is unlimited, ask how to arrange these wine to make the remaining wine the least.Problem-solving ideas: The problem is a complete

This is a simple 01 backpack, but the knapsack problem is fairly simple, is the front of the details of the time to pay attention to, test instructions roughly said three restrictions bar1. Only A, B, c three types of invoices can be reimbursed, the other will not be reimbursed2. No more than 600 reimbursement for item items3. The total amount of each invoice does not exceed 1000With these three, there is one to be careful is the amount of reimburseme

-negative integerV P q(wherevindicates the price of the item (v),Pindicates the importance of the item ( 1~5 ),QIndicates whether the item is a main piece or an attachment. Ifq=0, indicating that the item is the main piece, ifq>0, indicating that the item is an attachment,Qis the number of the owning main part)outputs description output Description there is only one positive integer, which is the maximum value of the sum of the product of the price and the importance of the item not exceeding th

, respectively the smartness and funness for each cow.Output* Line 1:one integer:the optimal sum of TS and TF such that both TS and TF is non-negative. If no subset of the cows has non-negative TS and non-negative TF, print 0.Sample Input5-5 78-66-32 1-8-5Sample Output8HintOUTPUT DETAILS:Bessie chooses cows 1, 3, and 4, giving values of TS = -5+6+2 = 3 and TF= 7-3+1 = 5, so 3+5 = 8. Note that adding cow 2 would improve the valueof TS+TF to ten, but the new value of TF would was negative, so it's

View code /* Give you n items, each of which has a certain volume, give you a certain total volume of a certain backpack, ask you to use this backpack to install these items, there are several different combinations of items in a backpack. One combination is legal: The remaining backpack capacity must be the mini

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?NoteYou cannot divide item in small pieces and the total size of items you choose should smaller or equal to M.ExampleGiven 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. Soluti

Test Instructions: The value of 1,2,3,4,5,6 items respectively a[1],a[2],a[3],a[4],a[5],a[6], ask can be divided into two piles of equal value. Solution: Convert to multiple backpacks1#include 2#include string.h>3#include 4#include 5 using namespacestd;6 7 intdp[120010];8 inta[Ten];9 Ten intNvalue; One A //0-1 backpack, at cost, to get the value of weight - voidZeroonepack (intCostintweight) - { the for(inti=nvalue;i>=cost;i--) -Dp[i]=max (dp[i]

/* Poj 1276 the meaning of multiple knapsack questions is probably given a currency m to be raised, given the number of currencies owned N [I] and the value V [I], q: How much can you raise is a multi-backpack problem DP [J] indicates the maximum number of valuable items that a backpack with a capacity of J can hold. Generally, the solution to multiple backpacks is to convert them. complete

Poj (2184) -- Cow Exhibition (01 backpack deformation) In fact, I don't think I have a deep understanding of this question. However, after doing this today, we will first record our current ideas. The general meaning of the question is: There is a nheaded ox, each of which has a s value representing the IQ value, and a f value representing its humorous value. Then, ask you what the total and maximum values of the IQ and humor values are, and ensure th

contains four numbers N (1 N M (1 W (1 W X (1 X lt;= N )-The starting point ofNitori.Then followed N by lines, each line contains the integers which means the shop on tourist attraction i sells the c2/> TVi and Things (1 TWi TVi Next, there is M lines, each line contains three numbers, Xi , Yi and Li , which means there is a one-way road From tourist attraction Xi to Yi , and the length is Li (1 Xi , Yi Li OutputFor each test case, output the answer as the description required.Sample Input4 4

There are n items and a backpack with a capacity of M, with unlimited pieces available for each item. The volume of the article I is weight, and the value is values. The solution of which items are loaded into the backpack allows the volume of these items to be no more than the backpack capacity, and the value of the largest sum.1 //"n" is the number of kinds.2 /

Description There are n items and a backpack with a capacity of V. The cost of item I is c[i], value is w[i]. These items are divided into groups, and items in each group conflict with each other, with a maximum of one selected. To find out what items to pack into the backpack can make the sum of these items does not exceed the backpack capacity, and the total va

Problem: There are n items and a backpack with a capacity of V. The cost of item I is c[i], value is w[i]. These items are divided into groups, and items in each group conflict with each other, with a maximum of one selected. To find out what items to pack into the backpack can make the sum of these items does not exceed the backpack capacity, and the total value

Title Link: https://vjudge.net/problem/POJ-1276Test instructions: There are now cash, and n kinds of coins, each coin has NI, the value of Di, for all kinds of coins composed of not more than cash of the largest amount of money ....Idea: binary split into 01 backpack, or converted to a full backpack are all possibleFull backpack://This problem can let w,v start f

Appendix: backpack problems in usaco Usaco, short for USA computing Olympus, organizes many global computer competitions. Usaco trainng is a very suitable Question Bank for beginners. I think it is characterized by a high quality and gradual process of questions. It also has good texts and question analysis. Text knapsack problems is also worth reading. In addition, usaco contest is a global competition series organized by usaco all the year roun

Tags: algorithm Dynamic Planning 0-1 backpack problems: There are n items and a backpack with a capacity of v. The weight of the I-th item is W [I], and the value is V [I]. Solving which items are loaded into a backpack can make the total weight of these items not exceed the size of the backpack, and the tota

subset of the cows has non-negative TS and non-negative TF, print 0.Sample Input5-5 78-66-32 1-8-5Sample Output8HintOUTPUT DETAILS:Bessie chooses cows 1, 3, and 4, giving values of TS = -5+6+2 = 3 and TF= 7-3+1 = 5, so 3+5 = 8. Note that adding cow 2 would improve the valueof TS+TF to ten, but the new value of TF would was negative, so it's notAllowed. The main idea: there are two sets of data, the two sets of data to be added after the maximum value (must satisfy both sets of data and are posi

With the previousVernacular backpack 01 backpackTo see what a backpack is. I hope I can see it clearly in the future. It is best to help the children's shoes I just got in touch. I. About full backpacks There are n items. Each item (with an infinite number of items) has a weight of W [I] and a value of VA [I]. There is a backpack that can carry M-weight items. No

I really want to talk about it. Every topic has to work hard to rank data for two days. It is really hard for me to be a stupid and hard-working person, there are a lot of things that can only be understood-to cheer up-otherwise it will be despised Backpack-generally speaking, it is the question of filling a backpack. Its requirement is generally to maximize its value. 01