Simple DP
Calculate a minimum spanning tree before DP
Status Transfer
F [I] [J] indicates the minimum cost for the first I edge to use the No. 5 material in Unit J.
F [I] [J] = min (F [I] [J], F [I-1] [J-Len] + Len * C5)
F [I] [J] = min (F [I] [J], F [I-1] [J] + Len * C6)
Then you can trace the output.
The memory is a

both strictly smaller than the corresponding base dimensions of the lower block because there has to be some spac E for the monkey to step on. this meant, for example, that blocks oriented to have equal-sized bases couldn't be stacked. your job is to write a program that determines the height of the tallest tower the monkey can build with a given set of blocks. inputThe input file will contain in one or more test cases. the first line of each test case contains an integer n, representing the nu

G. Xor-mst time limit per test 2 seconds memory limit per test megabytes input standard input output standard output
You is given a complete undirected graph with n vertices. A number AI is assigned to each vertex, and the weight of an edge between vertices I and j are equal to Ai xor aj.
Calculate the weight of the minimum spanning tree in this graph. Input
The first line contains N (

Dynamic Planning Learning Series-divide DP (2) and Plan dp
The second question of Partitioned DP, wikioi 1039, is different from the biggest idea of the product of the first question.
Question requirements: Divide a number into several parts and ask the total number of methods.Truth-solving ideas:In fact, it is a bit

Question 2: Return Character Sequence (interval DP), return dpTime Limit: 256 ms single point time limit: Ms memory limit: MB
Description
Returns the number of subsequences of input strings. The reversed Character Sequence is still the same as the original sequence. For example, in the string aba, the input sub-sequence is "a", "a", "aa", "B", and "aba". There are 5 sub-sequences in total. Subsequences

Http://www.gdutcode.sinaapp.com/problem.php?cid=1049pid=3Dp[i][state] Represents the minimum number of days to rest when the first I is processed, and then the current status is state.For example, with 1 for rest, 2 for training, and 3 for exercise.If the current allowed state of the day is only training.Then dp[i][3] = inf (indicates impossibility)Rest is possib

, which represents the balance on the card. mN=0 indicates the end of the data.Output for each set of inputs, outputs a line that contains an integer that represents the smallest possible balance on the card.Sample Input
1 50 5 10 1 2 3 2 1

Charm Bracelet
Time Limit: 1000MS
Memory Limit: 65536K
Total Submissions: 47440
Accepted: 20178
DescriptionBessie have gone to the mall ' s jewelry store and spies a charm bracelet. Of course, she ' d like-to-fill it with the best charms possible from the N (1≤ n ≤3,402) available Char Ms. Each charm I in the supplied list has a weight wi (1≤ w

description of the colors of the vertices: n integers x0, x1, ..., xn -1 (xi is either 0 or 1). If xi was equal to 1, vertex i was colored black. Otherwise, vertex i was colored white. OutputOutput a single integer-the number of ways to split the tree modulo 1000000007 (9 + 7). Sample Test (s) input30 00 1

possible path. He needs at least strength = 2 initially.Case 2:note this to start from (a) he needs at least strength = 1.Hint
Added by:
Varun Jalan
Date:
2011-12-15
Time limit:
0.336s
Source limit:
50000B
Memory limit:
1536MB
Clu

, we can find all the numbers that match the meaning of the question. This is the secret of the digital DP in this question.
Another example is 1249.
1 ~ 999
1000 ~ 1199
1200 ~ 1239
1240 ~ 1248
1249
Dp [I] [j] indicates the number of I (that is, the number of I-bits) IN THE STATUS j.
There are three statuses in this question:
①

the secret of the digital DP in this question.
Another example is 1249.
1 ~ 999
1000 ~ 1199
1200 ~ 1239
1240 ~ 1248
1249
DP [I] [J] indicates the number of I (that is, the number of I-bits) IN THE STATUS J.
There are three statuses in this question:
① DP [I] [0] indicates the number of numbers with an I length and No

Test instructions: 0~30000 There are 30,001 places, each place has one or more gold coins, the first step to D, the step size is D, the step after walk can be the last step +1,-1 or unchanged, go to a place to collect the wealth of that place, and now ask to go out (>30000) The maximum amount of wealth that can be collected before.Solution: It is easy to think of Dp

[I] [j] = max {dp [I] [j], dp [I-1] [k] + stn [I] [j]} (k indicates the k State of the I-1 row
The final answer is the maximum value in dp [n] [j.
Code:
# Include
# Include
# Include
Using namespace std; const int hpn = 18000; int state [hpn], stn [25

namespace std;Struct node{Int u;Int v;Int w;Bool friend operator Return a. w }} Edge [2, 1001];Int gcd (int n, int m) {if (n Int lcm (int n, int m) {if (n Double p [1, 1001] [51];Double dp [1, 1001] [31] [31];Int main (){Int m, t, n, I, j, k;While (scanf ("% d", m, t, n) (m | n | t )){For (I =

1 /*2 depending on the relationship, you can build a tree3 Dp[i][j] Indicates the maximum value of the first node selected J4 Dp[i][j]=max (Sigma (Dp[c[i][ki]))5 Sigma (Dp[c[i][ki]]) indicates the maximum number of J points select

selected several similar problems for you:3554 3556 3557 3558 3559Topic Analysis:Digital DP.These are state transition equations.DP[I][0]=DP[I-1][0]*10-DP[I-1][1]; The representation does not contain 49 (when the number of digits increases by one, excluding 49 increases by

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