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Cheapest Palindrome Time limit:2000ms Memory limit:65536kb 64bit IO Format:%lld %lluSubmit Status DescriptionKeeping track of the cows can be a tricky tasks so farmer John has installed a system to automate it. He has installed in each cow a electronic ID tagged that the system would read as the cows pass by a scanner. Each ID tag "s contents are currently a single string with length M (1≤m≤2,000) characters drawn from a alphabet of N (1≤n≤26) differe

Cheapest Palindrome Time Limit: 2000MS Memory Limit: 65536K Total Submissions: 6013 Accepted: 2933 DescriptionKeeping track of all the cows can is a tricky task so Farmer John have installed a system to automate it. He has installed on each cow an electronic ID tag that the system would read as the cows pass by a scanner. Each ID tag ' s contents is currently a single string with length m

Test Instructions: An alphabetic sequence of length m, consisting of n letters, each with two costs: ① delete the letter; ② Add the letter. Ask the minimum cost of turning this sequence into a palindrome sequence. the f[I [j] represents the minimum cost of I-J this section into a palindrome string. when sequence s[i] = = s[J], no cost is required: f[I [j] = f[i + 1][j-1];otherwise:①-> plus or minus the I-letter, f[i + 1 [j] + w[s[I]];②-> plus or minus the J-letter, f[I [j-1] + w[s[j]];(because a

Cheapest Palindrome Topic Links:http://poj.org/problem?id=3280Test instructionsGiven a string of only lowercase letters, you can add or delete some letters (both add and delete cost and vary) to change the string to the minimum cost of a palindrome string.ExercisesSet DP[I][J] is the minimum cost of changing the interval [i,j] into a palindrome, then two cases① obvious, when S[i]==s[j], dp[i][j]=dp[i+1][j-1]② when S[i]!=s[j], Dp[i][j] has four pos

Cheapest Palindrome Time Limit: 2000MS Memory Limit: 65536K Total Submissions: 8643 Accepted: 4189 DescriptionKeeping track of all the cows can is a tricky task so Farmer John have installed a system to automate it. He has installed on each cow an electronic ID tag that the system would read as the cows pass by a scanner. Each ID tag ' s contents is currently a single string with length m

Cheapest Palindrome Time Limit: 2000MS Memory Limit: 65536K Total Submissions: 6186 Accepted: 3014 DescriptionKeeping track of all the cows can is a tricky task so Farmer John have installed a system to automate it. He has installed on each cow an electronic ID tag that the system would read as the cows pass by a scanner. Each ID tag ' s contents is currently a single string with length m

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DP[I][J] Indicates the cost of processing I to J, if s[i] = = S[j] Do not need to process, otherwise processing s[i] or s[j],For a character ch, plus ch or delete ch, the effect is the same for interval transfer, taking min.#include #include#includestring>#include#include#include#include#include#include#includeSet>#includeusing namespacestd;Const intSigma_size = -, MAXM = 2e3+5;CharS[MAXM];intAdd[sigma_size], del[sigma_size];intCost[sigma_size];intDP[MAXM][MAXM];//#define LOCALintMain () {#ifdef

() {intb//define the number of books purchased intM,n; inti; Doublesum=0; cout"Please enter the number of books you need to purchase:"Endl; CIN>>b; M=b/5; N=b%5; I=n+5; if(i!=8) { Switch(n) { Case 0: Sum=m* +*0.75; Break; Case 1: Sum=m* +*0.75+8; Break; Case 2: Sum=m* +*0.75+ -*0.95; Break; Case 3: Sum=m* +*0.75+ -*0.9; Break; Case 4: Sum=m* +*0.75+ +*0.8; Break; } } Elsesum= (M-1)* -+4*8*0.8*2; cout"Reader Purchase""The lowest price for this book is:"Endl; coutE

Poj 3280 Cheapest Palindrome (DP) Question: There is a string consisting of n lower-case letters with a length of m. You can add or delete characters to it to convert it into a return text. Adding or deleting a character has a minimum cost. Idea: DP. Dp [I] [j] indicates the str [I ~ J] The minimum cost for returning objects. Therefore, the state transition equation is: When str [I] = str [j] dp [I] [j] = dp [I + 1] [J-1];When str [I]! = Str [j] dp

POJ 3280 Cheapest Palindrome DP question When we see the Palindrome question, we should first think about the central issue, and then start from the center and think about how to solve it. DP problems are generally converted from smaller problems to larger problems, and then computed from bottom up to bottom up. It is easy to get the state transition equation. The state transition equation in this question is: If (cowID [I] = cow {j]) tbl [id] [I] = t

Topic Link: [POJ 3280]cheapest PALINDROME[DP] The analysis: There is a string of length m, all lowercase letters, given the cost of deletion and insertion of each letter in the string, ask: to constitute a palindrome string, at least how many costs. Ideas for solving problems: Set Dp[i][j] for the string s[i ~ J] to become a palindrome string the minimum cost. So there's the transfer equation: DP[I][J] = dp[i + 1][j-1] (s[i] = = S[j]); Otherwise dp[

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POJ 3280 Cheapest Palindrome (DP), pojpalindromeZookeeper Description Keeping track of all the cows can be a tricky task so Farmer John has installed a system to automatic it. he has installed on each cow an electronic ID tag that the system will read as the cows pass by a packet. each ID tag's contents are currently a single string with lengthM(1 ≤M≤ 2,000) characters drawn from an alphabetN(1 ≤N≤26) different symbols (namely, the lower-case roman al

POJ 3280 Cheapest Palindrome dynamic programming method, pojpalindrome Description Keeping track of all the cows can be a tricky task so Farmer John has installed a system to automatic it. he has installed on each cow an electronic ID tag that the system will read as the cows pass by a packet. each ID tag's contents are currently a single string with lengthM(1 ≤M≤ 2,000) characters drawn from an alphabetN(1 ≤N≤26) different symbols (namely, the lower

Test instructions: Give you a M-bit string, where the character is one of the given n lowercase letters, and then the cost of adding or removing the letter in the stringNot the same. It is the minimum cost of the string to be a palindrome after adding a delete operation.Idea: Consider D[i][j] as the minimum cost of a palindrome string I to J.such as xx.....yy, if x = = Y, obviously d[i][j] = d[i+1][j-1];if x! = Y, then there are 4 scenarios: Delete x, add an x after Y, delete y, and add y before

"The main topic"Gives a string that can delete or add some characters, each of which consumes value. Ask at least how much value, can make a string into a palindrome.IdeasIn fact the value of deleting or adding characters only needs to keep the smaller one. Assuming that you are currently converting (j,i) to a literal, it has the following three scenarios:(1) Add or remove at the end of a character that is the same as the beginning, f[j][i-1]+cost[s[i]-' a ';(2) Adding or deleting a character at

given name tag.Sample Input3 4ABCBA 1100b 700c 200 800Sample Output900HintIf We insert an "a" in the end to get "ABCBA", the cost would is 1000. If we delete the "a" on the beginning-get "BCB", the cost would is 1100. If we insert "BCB" at the begining of the string, the cost would are + + + + =, which is the minimum.SourceUsaco Open GoldA look at this problem is always felt to be a string processing problems, in fact, it is necessary to model dynamic programming method.Dynamic programming meth