multiplication jigsaw puzzle

Tags: blog HTTP Io AR for SP 2014 on Log Question: The sequence of N numbers (3 Question link: http://poj.org/problem? Id = 1651 --> Status: DP [I] [J] indicates the minimum weight and value of the sequence from number I to number J. State transition equation: DP [I] [J] = min (DP [I] [J], DP [I] [k] + dp [k] [J] + A [I] * A [k] * A [J]); (enumerate the last retrieved number to update) Time Complexity: O (N ^ 3) #include Poj-1651-multiplication

Topic linksGive a string number, take out a number i the price is a[i-1]*a[i]*a[i+1], the first number and the last number can not be taken, to take away the remaining number of the minimum value of the price.#include #include#include#include#include#include#include#includeSet>#includestring>#includeusing namespacestd;#definePB (x) push_back (x)#definell Long Long#defineMK (x, y) make_pair (x, y)#defineLson L, M, rt#defineMem (a) memset (a, 0, sizeof (a))#defineRson m+1, R, rt#defineMem1 (a) mem

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 K-sections so that the maximum values of eac

Start with the question:Multiplication puzzle Time limit:1000 ms Memory limit:65536 K Total submissions:6162 Accepted:3758 DescriptionThe multiplication puzzle is played with a row of cards, each containing a single positive integer. during the move player takes one card out of the row and scores the number of points equal to

Title: http://poj.org/problem?id=1651Water problemTest instructions: A number of cards, the score of the first card is a[i] (1 n cards, in addition to the first and last card is not available, you can take one, get the score a[i–1] * a[i] * a[i + 1], and then the card removed. The title requires the final calculation of the minimum total score. Principle: Matrix multiplication.1#include 2#include 3#include 4#include string.h>5 using namespacestd;6 Lon

Multiplication puzzletime limit: 1000 msmemory limit: 65536 kbthis problem will be judged on PKU. Original ID: 1651 64-bit integer Io format: % LLD Java class name: Main the multiplication puzzle is played with a row of cards, each containing a single positive integer. during the move player takes one card out of the row and scores the number of points equal to t

Title Link: http://poj.org/problem?id=1651Idea: In addition to the top and bottom two numbers can not be taken, all the number is required to take out, each take a number to spend this number and the two adjacent number of products, the cost of the cost is minimalUse Dp[i][j] to denote the minimum cost of the interval [i,j], then there is Dp[i][j]=min (Dp[i][k]+dp[k][j]+a[i]*a[k]*a[j]) i+11#include 2#include 3#include 4#include 5#include string.h>6#include 7#include 8#include 9#include Ten#inclu

http://blog.csdn.net/libin56842/article/details/9747021Http://www.cnblogs.com/devil-91/archive/2012/06/26/2562976.html#include #includestring>#include#include#include#include#include#include#include#include#include#include#includeSet>#include#includeusing namespacestd;#defineMem (A, B) memset (A,b,sizeof (a))#definePF printf#defineSF scanf#defineSPF sprintf#definePB Push_back#defineDebug printf ("!\n")#defineINF 10000#defineMAX (A, b) a>b?a:b#defineBlank pf ("\ n")#defineLL Long Long#defineAll (

Test instructions: Give a sequence, select a number at a time (recorded as B), calculate the nearest number to his left (recorded as a), the right distance from his nearest number (recorded as C), calculate the A*b*c,sum+=a*b*c repeat operation until the end of the first two numberDifferent schemes correspond to different sums to calculate the minimum sum valueAnalysis: Typical interval dp,dp[i][j] represents the minimum value of the selected number of all from I to J.Dp[i][j]=min (dp[i][j],dp[i

Interval DP Basic problemPost Code only#include Copyright NOTICE: This article for Bo Master original article, without Bo Master permission not reproduced. POJ 1651:multiplication Puzzle

Title Link: http://poj.org/problem?id=1651Test instructions: A series of numbers, in addition to the tail can not move, each time take out a number, this number and left and right adjacent numbers of the product for its value,Finally add up all the values and ask for the minimum value.It is easy to think of greed is to first take the largest number first out so that you can meet the remaining total value as small as possible, if there are multipleThe number of time priority to take away the valu

A special interval DP is compared. The breakpoint is also enumerated when the small interval shifts the large interval. However, the breakpoint has a special meaning with the normal interval DP ratio. Indicates that the breakpoint is the last point taken out of the interval. And an interval indicates that both ends are not taken in the middle of taking away the minimum cost.#include #include#include#include#include#include#include#include#include#includeSet>#includeusing namespacestd;Const Doubl