cisco 2955

This problem is an introduction to the interval DP. The title asks for the longest valid parenthesis sequence length, the direct memory searches the words to be good to think, but the code is a bit long, if writes the recursive type then the code will be reduced much.State transition equation:DP[I][J] indicates the longest length from I to J can be composedIf I position and J position pair, then dp[i][j] = Dp[i + 1][j-1] + 2;The next step is to enumerate k,k from the position between i-j, dp[i][

regular brackets sequence. Given the initial sequence ([([]])] , the longest regular brackets subsequence is [([])] .InputThe input test file would contain multiple test cases. Each input test case consists of a single line containing only the characters ( , ) , [ , and ] ; St'll has length between 1 and inclusive. The End-of-file is marked by a line containing the word "end" and should not being processed.OutputFor each input case, the program should print the length of the longest possible

test case, output a line with the maximum number of millions he can expect to get while the probability of gettin G caught is less than the limit set.Notes and Constraints 0 0.0 0 0 A Bank goes bankrupt if it is robbed, and we may assume that all probabilities is INDEP Endent as the police has very low funds. Sample Input30.04 31 0.022 0.033 0.050.06 32 0.032 0.033 0.050.10 31 0.032 0.023 0.05Sample Output2461#include 2#include 3 using namespacestd;4 intMain ()5 {6 intw[ -],n,t,i,j,

To find the maximum number of matches for a brace, this question can be compared to UVA 1626.Pay attention to the same background, but do not ask for the same.Back to this question, set D (i, j) to denote the maximum number of matches in a substring of SI ~ sj, if Si and SJ can pair, D (i, j) = d (i+1, j-1)Then to enumerate the intermediate points k,d (i, j) = max{D (i, K) + D (k+1, J)}1#include 2#include 3#include 4#include 5 using namespacestd;6 7 Const intMAXN = -+Ten;8 9 intN;Ten CharS[MAXN]

For the first time, the probability is regarded as a backpack, which is increased by 100000 times. In addition, the probability of many strong banks is multiplied, not the sum. Take the amount of money you want to grab as a backpack transfer formula: DP [J] indicates the maximum probability of success when the user can grab J million DP [J] = max (DP [J], DP [J-M [I] * (1-p_bank [I]); /*************************************** * *********************************> File Name: hdu2955.cpp> aut

Poj 2955 Brackets interval dp The question is to give you a string consisting of braces and parentheses. Ask how long the longest subsequence in the string conforms to the matching rules of mathematical parentheses. At the beginning, I plan to use the legendary left-closed and right-open interval to write data. Later I found it was not suitable for me, but I still wrote back the left-closed and right-closed interval. The idea of dp is relatively simp

POJ 2955 Brackets Main Topic The number of matching parentheses in the given string, () and [] are matched. Thinking of solving problems DP[I][J] represents how many matching strings are available for i~j.Transfer equation:1, Dp[i][j]=dp[i+1][j-1]+2, when the i,j position composition can match the parentheses;2. Enumeration split point K (I≤kcode implementation #include POJ 1505 Copying Books Main Topic Divides the number of rows into successive

Poj 2955 Brackets (interval DP to find the longest matching substring) Idea: assuming that the longest matching string of the interval [I, j] is required, it must be transferred from [I, J-1, it is possible that s [j] And s [I] Have a "relationship" (matching or not matching) until s [J-1], if no "relationship" occurs ", that is, s [j] has a "relationship" with itself. Use the for loop to enumerate all possibilities and take the maximum value. Code:

POJ 2955 Brackets, poj2955bracketsBrackets Time Limit:1000 MS Memory Limit:65536 K Total Submissions:6622 Accepted:3558 Description We give the following inductive definition of a "regular brackets" sequence: The empty sequence is a regular brackets sequence, IfSIs a regular brackets sequence, then (S) And [S] Are regular brackets sequences, and IfAAndBAre regular brackets sequences, thenABIs a regular brackets

probability p without being caught. Attached code:1#include 2#include 3#include 4 using namespacestd;5 intMain ()6 {7 intt,n,m,i,j;8 Doublep;9 inta[ the];Ten Doubleb[ the],dp[10005]; Onescanf"%d",T); A while(t--) - { - intsum=0; thescanf"%lf%d",p,n); - for(i=0; i) - { -scanf"%D%LF",a[i],b[i]); +Sum+=a[i];//the bank's total money is considered a backpack capacity - } +Memset (DP,0,sizeof(DP)); Adp[0]=1;//0 dollars, the security probabilit

Test instructions: Roy wants to rob a bank, every bank has a certain amount and the probability of being caught, know that Roy was caught the maximum probability p, ask Roy in the case of arrest, rob the most.Analysis:1, the amount of money can be added to each bank to get the total amount of money, and then as the capacity of the backpack.2, each bank's money as weight.3, the probability of not being caught as a value, pay attention here to multiply instead of Add.4, therefore has the state tra

Pj.Outputfor each test case, output a line with the maximum number of millions he can expect to get while the probability of Getting caught is less than the limit set.Notes and Constraints0 0.0 0 0 0.0 A Bank goes bankrupt if it is robbed, and your may assume that all probabilities be independent as the police have very lo W funds.Sample Input30.04 0.022 0.033 0.050.06 + 0.032 0.033 0.050.10 to 0.032 0.023 0.05 sample Output2461#include 2#include 3#include 4 using namespacestd;5 6 Const intmaxn

{8 if(x>y)returnx;9 Else returny;Ten } One intMain () A { - intt,m,i,j; - DoubleN; the while(~SCANF ("%d",t)) - { - while(t--) - { +scanf"%lf%d",n,m); - intsum=0; + for(i=1; i) A { atscanf"%d%lf",a[i],b[i]); -sum+=A[i]; - } -Memset (DP,0,sizeof(DP)); -dp[0]=1; - for(i=1; i) in { - for(j=sum;j>=a[i];j--) toDp[j]=max (Dp[j], (1-b[i]) *dp[j-A[i]]); + } -

Title Link: http://poj.org/problem?id=2955Idea: Bracket matching problem, find the longest matching length of the given sequence (can be mismatched in the middle) For example [(])] have [()] meet the conditions, length of 4DP[I][J] represents the maximum number of brackets that match from interval I to interval J, first, assuming mismatch, then dp[i][j]=dp[i+1][j]; and then looking for i+1~~j with a wood with the first bracket.If there is, Dp[i][j]=max (dp[i][j],dp[i+1][k-1]+dp[k][j]+2) .....1#i

Outputs the number of parentheses within a string that can be matchedState transition equation:if (s[i]== ' (' s[j]== ') ' | | s[i]== ' [' s[j]== '] ')dp[i][j]=dp[i+1][j-1]+2;And then merge the intervals.1 //#pragma COMMENT (linker, "/stack:167772160")//Manual expansion of the ~~~~HDU with C + + intersection2#include 3#include 4#include 5#include 6#include 7#include 8#include 9#include Set>Ten#include One#include A //#include - using namespacestd; - #defineCLC (A, B) memset (A,b,sizeof (a)) t

#include 　　HDU 2955 Robberies (01 backpack)

01 backpack + card precision HDU 2955

1#include 2#include 3#include 4#include 5 #defineSC (x) scanf ("%d", (x))6 #definePF (x) printf ("%d\n", X)7 #defineCL (x, y) memset (x, y, sizeof (x))8 #defineMax (A, B) (a > B? a:b)9 using namespacestd;Ten Const intMAX = the; One structNode//Set as knapsack problem, structure body A { - intp; - DoubleW; the }bag[max]; - DoubleDp[max*max], P;//the probability of being caught - intm, T, I, J, N; - intMain () + { - SC (T); + while(t--) A { atm =0; -scanf"%lf%d", p, N); -

Test instructionsNow we define an R string that must meet the following conditions:1) when its string is empty, then it is the R string.2) when it is R string, then (s) or [s] is also the R string.3) when both A and B are r strings, then AB is also the R string.I did not fully understand the meaning of the topic here, so I found that the recursion is not over. In fact, its essence is the parentheses match.This means that the legal sequence here is that the parentheses can match 22.if ((a[s]== '

backpack capacity C, and then the probability of not being caught 1-PJ as a value of W to beg. Then his biggest not caught the probability of robbing the most of the money is the product, and finally enumerate it on the line, why to use the product here, think of the mathematics involved in the problem of probability, rather than the sum of simple probabilities,. This calculates the probability that the robbery to J money will not be caught, multiplied to save it.Dp[j] Indicates the probability