replay backpack

that each calculation is using the upper-left data, so that is, if the inverse is calculated:For I:1..NFor J:m. Need (i)F (j) = Max{f (J), F (j-need (i)) +value (i)}It will not overwrite the "top left" data, and can successfully complete the calculation of the entire state transfer amount.Code is actually the state transfer, so the problem of dynamic planning, as long as the correct state transfer can be listed, from the AC is not far away.Impl:1#include 2 using namespacestd;3 4 intMain ()5 {6

Http://codeforces.com/problemset/problem/283/CAt the beginning, look at the sample do not understand, why 5 * A1 + A3 do not? It's 17.The original is, the topic requires the number of coins A3 > A4 > A2, then, if not selected, is not legal. is 0, 0, 0 This is not legal, because a3 = A4.Then you know, A3 at least choose two.So how do you maintain this relationship? , the idea is to rely on the backpack, the number of A3 need more than A4, you can turn

Big backpack Question: There are n a weight and price value respectively W[i] and V[i] project. The total weight of these products does not exceed the W project. Finds the maximum sum value for all selected scenario price values.Among them, 1≤n≤40, 1≤w[i], v[i]≤10^15, 1≤w≤10^15.The first feeling of this problem is the normal 01 backpack.Just after reading the data range will be found. This time the value and weight can be very large values, compared t

 Fateproblem DescriptionRecently Xhd is playing a game called fate, in order to get the best equipment, XHD in the constant killing blame do tasks. Over time XHD began to hate the killing monsters, but also had to kill the blame to rise to the last level. Now the problem is, XHD up the last level also need n experience value, XHD still have M endurance, each kill a strange xhd will get corresponding experience, and reduce the corresponding endurance. When endurance drops below 0 or 0, XHD wil

HDU2602 (0-1 backpack problem), hdu26020-1 backpack ProblemN-01 backpackTime Limit:1000 MSMemory Limit:32768KB64bit IO Format:% I64d % I64u DescriptionMany years ago, in Teddy's hometown there was a man who was called "Bone Collector ". this man like to collect varies of bones, such as dog's, cow's, also he went to the grave... The bone collector had a big bag with a volume of V, and along his trip of coll

or the 01 backpack time limit:10000Ms | Memory Limit:228000KB Difficulty:5 Describe There are n items with a weight and value of WI and VI, from which the total weight of items not exceeding W is selected, and the maximum value of the sum of the values in all selection schemes is obtained. Input Multiple sets of test data. The first row of each set of test data entered N

positive integers n and m, indicating the number of prizes, and the number of lotteries in small Ho's hands.The next n lines describe each line describing a prize, where I behaves two integers need (i) and value (i), meaning as described earlier.Test Data AssuranceFor 100% of data, the value of n does not exceed the value of 500,m no more than 10^5For 100% of data, need (i) does not exceed 2*10^5, value (i) does not exceed 10^3OutputFor each set of test data, output an integer ans that represen

ZOJ3662Math Magic (group backpack + full backpack), zoj3662mathmagicI-Math MagicTime Limit:3000 MSMemory Limit:32768KB64bit IO Format:% Lld % lluSubmit Status Description Yesterday, my teacher taught us about math: +,-, *,/, GCD, LCM... as you know, LCM (Least common multiple) of two positive numbers can be solved easily because of a * B = GCD (a, B) * LCM (a, B ). In class, I raised a new idea: "how to ca

N, m, and k indicate that there are n pairs of k kinds of shoes, and you have m capacity; Each pair of shoes has capacity p and value v. If you want to buy all k types of shoes, the maximum value of the shoes can be bought if the capacity is m; Each pair of shoes can only be bought once (01 backpack). Each type of shoes must have at least one (group backpack: each group can have only one. Note initializa

The main effect of the topic: There are n pieces of furniture, each of which weighs (1≤wi≤100) and is transported to the destination with two handlebar vehicles. The two vehicles have a load limit of C1, C2 (1≤ci≤100), asking at least a few trips to get all the furniture to the destination. The two cars must go with each other, even if a car is not carrying anything. Ideas: A First on your own way: Enumerating all the combinations of furniture that the first vehicle may carry, then a binar

Explanation of backpack 01, full, and multiple PS: If you think you can write well, just help me pin my blog. Thank you. First, let's talk about dynamic planning. Dynamic Planning is just like recursion. We can only find local relationships. If we want to list them all, it is very difficult, such as the tower of Hanoi. You can say that it is an intuitive relationship to move all the layers except the last layer to 2, move the last layer to 3, and

Two questions: 1) Dynamic Planning solution for knapsack problems2) Another Implementation Method of Dynamic Planning, memory recursion (mentioned in the first article of DP, also mentioned in professional). I will explain it here and add all the articles in detail, this can be regarded as a dynamic rule. The last one is not detailed. Certificate --------------------------------------------------------------------------------------------------------------------------------------------------- Gi

[Problem description]Select some items from the items in the T group and put them in the backpack to find the minimum value of the remaining space.Restriction: You must select consecutive items from each group. That is to say, if there are n items in this group: item 1, Item 2, Item 3 ,... , Item n, you can only select item I, item I + 1 ,... , Item j, where 1 [Input]The first two positive integers v and T separated by spaces. The number of spaces and

Analysis: First, this question cannot be considered a normal 01 backpack problem, because w Question link http://acm.nyist.net/JudgeOnline/problem.php? PID = 1, 860 Code: # Include

1. Multiple backpacks void Zoreonepack (int val,int cost) {for (int i=v;i>=cost;i--) { if (dp[i-cost]+val>dp[i)) { dp[i]=dp[i-cost]+val; } }} void Completepack (int val,int cost) {for (int i=cost;i 2.O (n^2) TSP #include 3. Packet backpack Memset (Dp,0,sizeof (DP)); for (int i=1;i 4. Multiply LCA Online void Dfs (int u,int from) {for (int i=head[u];i!=-1;i=e[i].next) {

2214 knapsack problemAccept:6 Submit:9Time limit:3000 mSec Memory limit:32768 KB problem DescriptionGiven a set of n items, each with a weight w[i] and a value v[i], determine a-to-choose the items into a knapsack so T Hat The total weight are less than or equal to a given limit B and the total value is as large as possible. Find the maximum total value. (Note that each item can is only chosen once).InputThe first line contains the integer T indicating to the number of test cases.For each test c

(This old man's explanation won my mind, Special Recommendation: http://blog.csdn.net/insistgogo/article/details/8579597) There are n items and a backpack with a capacity of v. The cost of the I-th item is C [I], and the value is W [I]. Solving which items are loaded into a backpack can make the total cost of these items not exceed the capacity of the backpack, a

P01: 01 question about backpack There are n items and a backpack with a capacity of v. The cost of the I-th item is C [I], and the value is W [I]. Solving which items are loaded into a backpack can maximize the total value.Basic Ideas This is the most basic problem with a backpack. It features that each item has only o

Document directory Problem description Basic Ideas Optimize space complexity Initialization details Problem description Conversion to 0-1 backpack Further Optimization-O (CN) Solution This article mainly comes from the nine lectures on backpacks. I have selected a simple 0-1 backpack problem and a summary of the full backpack problem. I added the Python

"Backpack Ninth Lecture"p01:01 knapsack problemTopicThere are n items and a backpack with a capacity of V. The cost of article I is c[i], the value is w[i]. The solution of which items are loaded into the backpack allows the sum of the costs of these items to be no more than the backpack capacity and the maximum value.