moschino backpack

3269 Hybrid Backpacktime limit: 1 s space limit: 256000 KB title level: Diamonds DiamondTitle DescriptionDescriptionBackpack volume is V, give n items, each item occupies a volume of VI, the value of WI, each item can take up to 1 pieces, or the maximum number of MI pieces (Mi > 1), or the amount of unlimited, in the total volume of the items installed under the premise of the value of the item is the largest value of the sum of what?Enter a descriptionInput DescriptionThe first line two number

http://acm.hdu.edu.cn/showproblem.php?pid=2191Chinese topic, do not say nonsense ...Analysis: The problem is a multi-pack, but can be used 01 backpack to do, each kind of rice has a few bags a separate storage, so it becomes 01 backpack ...#include #include#includestring.h>#includestring>#include#include#include#include#include#includeusing namespacestd;#defineMet (A, b) memset (A, B, sizeof (a))#defineMAXN

), which is the short board principle. Then consider this node, if bought, directly on the short board to add damage, provided that this node only allowed to put 1 bombs.How to transfer? Start with a backpack that costs 0 of the subtree, and if there's a better substitute for him. This step temporarily represents dp[t][j] as a subtree with T as the root, if you get J, you can at least attack the enemy's HP. State equation: Dp[t][j]=max (dp[t][j], min

Piggy-bank are 100.This is impossible.Source Central Europe 1999 test instructions: N Types of money have corresponding weight and value in a piggy bank that accommodates f-e quality money. Question the minimum amount of money in the piggy bank: Full backpack (01 backpack reverse)1#include 2#include 3#include 4#include 5#include 6#include 7#include 8 #definell __int649 #defineINF 0XFFFFFFFTen #definePi ACO

comments:1#include 2 3 using namespacestd;4 Const intn=1010;5 intA[n];6 intB[n][n];7 intMain ()8 {9 intn,m,m;Ten while(SCANF ("%d%d%d", n,m,m) = =3) One { A for(inti =0; I ) -scanf"%d", a[i]);//number of pieces per product - the for(inti =0; I ) -b[i][0] =1; - - for(inti =0; I //n Kinds of goods + for(intj =1; J //pick out M-pieces - { + if(J-1-A[i] >=0)//is greater than the number of items in Part I Ab[i+1][J] =

requirements. You can use a backpack to handle it. DP[I][J] Indicates if there is a J person after the first set. DP[K][P1] = 1 means existence and uniqueness, dp[k][p1] = 0 means no, dp[k][p1]>1 represents more than one. Then the key is how to output all the satisfied people. At this time, you can record the path in the backpack, the status of the current satisfaction is the state of the above push over.#

#1043: Full Backpack Time limit: 20000ms single point time limit: 1000ms memory limit: 256MB descriptionSay then before the story, small hi and small ho struggled to finally get the vast number of lottery tickets! And now, it's time for Little Ho to pick up the reward!Wait, why is this story familiar? This is going to start with the parallel universe theory ... All in all, in another universe, Little Ho's problems have changed slightly!Small Ho now h

to be divided. The lines consist of six non-negative integers n1, n2, ..., N6, where NI is the number of marbles of value I. So, the example from above would is described by the Input-line ' 1 0 1 2 0 0 '. The maximum total number of marbles would be 20000.The last line of the input file would be ' 0 0 0 0 0 0 '; Do not process this line.Outputfor each colletcion, output "Collection #k:", where k is the number of the "the test case" and then either "Can be div ided. ' or ' Can ' t be divided.Ou

in the Piggy-bank is 100. This is impossible. The main idea: for a piggy, you don't know how much money is in it, you can only estimate the weight of the pig and the weight of the corresponding coin, so that you output the minimum amount of money in the pig. Thinking analysis: A little thinking about this is a knapsack problem, while the topic does not limit the number of each coin, that is, all coins are infinite, so should apply the full backpack t

Test instructionsGives the n set of data with one type for each set of data.0 represents at least one, 1 represents at most one, and 2 represents any selection.Give the backpack capacity.If the backpack does not meet the most basic requirements output-1.Ideas:Backpack Problem disguised Inspection ~When 0 is initialized to-inf, then it is guaranteed to select at least one.When 1 or 2 initializes the value of

Test instructions: Enter n ordered numbers, preceded by a number of + or-, to see if there is such a and, so that and can be divisible by the number kThere are only positive or negative two cases in front of each digit, so it fits the 0,1 backpack, and the weight of the backpack is the remainder of the K (all positive)172 KB297 msc++785 b#includePOJ 1745 divisibility (0,1

For each course, the learning time is different, the harvest is different, in a course spent two different time to learn is mutually exclusive, that is, they belong to the same group of items, directly do group backpack can.Be aware that the order of the triple loops is immutable!1#include 2#include 3#include 4 using namespacestd;5 6 Const intINF =-99999999;7 Const intN =101;8 intA[n][n];9 intDp[n];Ten intN, M; One A intMain () - { - while(SCANF

#1038:01 Backpack time Limit:20000mscase Time Limit:1000msmemory LIMIT:256MBDescriptionSay then last week's story, Small hi and small ho struggled to finally get the vast number of lottery tickets! And now, it's time for Little Ho to pick up the reward!Small Ho now has M-ticket, and the prize area has n prizes, respectively labeled 1 to N, of which the first prize of the need (i) to redeem the lottery, but also can only redeem once, in order to make t

Title DescriptionState F[i, J] indicates that the selection of the first I species has been decided, the total need does not exceed J;State transition Equation F[i, j] = Max{f[i, j–need[i]] + val[i], f[i–1, j]};The status of the result is expressed as f[n, M].Note the difference between the state equation and the 01 backpack, which reflects whether the item can take one or the other.The difference that is reflected in the code is that the order of the

Test instructions: N points m-edge, two endpoints of each edge are known, to make up the maximum number of edges that a complete binary graph can add;Reference: http://blog.csdn.net/acmhonor/article/details/47072399Idea: Not a two-figure problem ...N points constitute a complete two-dimensional graph is, the difference between the two sides of the minimum number of edges (X+y=n, Max (x*y));That is, given some edges, the requirement is that the number of points in the two sets of the smallest dif

fractional fraction (with%), indicating the maximum success rate.Sample input10 34 0.14 0.25 0.3Sample output44%Title Link: http://acm.njupt.edu.cn/acmhome/problemdetail.do?method=showdetailid=1860Title Analysis: 0/1 knapsack problem + probability of the problem, the probability of pre-processing, get the probability of not passing, because the situation is more difficult to calculate, and then the bare 0/1 backpack.DP[J] Indicates the minimum failure rate when the

problems for you:1203 2159 2955 1171 2191Main topic:That is, give you a n and V and n items of v[i] and w[i], let you find out, and each time you put an item in the backpack, let you find out the maximum number of items (the volume of the object Problem Solving Ideas:Go straight to the 0-1 backpack model:Definition status: dp[i+1][j] The maximum value that can be brought by an item that is not more than J

marbles of value I. So, the example from above would is described by the Input-line "1 0 1 2 0 0". The maximum total number of marbles would be 20000.The last line of the input file would be "0 0 0 0 0 0"; Do not process this line.OutputFor each collection, Output ' collection #k: ', where k is the number of the ' Test case ', and then either ' Can be divided. ' OR "Can ' t be divided."Output a blank line after each test case.Sample Input1 0 1 2 0 01 0 0 0 1 10 0 0 0 0 0Sample OutputCollection

Links: http://hihocoder.com/problemset/problem/1038?sid=469496Accept Code:1#include 2#include 3 using namespacestd;4 5 intdp[501][100001];6 intneed[501];7 intvalue[501];8 9 intMain ()Ten { One intNum,amount; ACin>>num>>amount; - for(intI=1; i) - { theCin>>need[i]>>Value[i]; - } - - for(intj=1; j) + { - for(intI=0; i) + { A if(iNeed[j]) at { -dp[j][i]=dp[j-1][i]; - } - Else - { -Dp[j][i]=max (d

doesn't matter. = = Stick directly on itThe topic is very simple, almost backpack entry problem, in fact, strictly speaking is not a backpack, but the choice of principle is the sameThe interpretation of the reference, the two-dimensional equation f[i,j]:=f[i-1,j]+f[i-1,j-a[i]];The array is represented in the first I course, the total price is J.You can not order the dish (f[i-1,j]), or Point (F[i-1,j-a[i]