Enter n inequalities to find the maximum number of inequalities at the same time, data range nAnalysis: This problem has to change the idea to do, the known data not more than 1000, so enumeration 0~1000 within the number, in order to calculate how many inequalities to meet, and then update the answer, that is, a double cycle, 1000*50However, this question does n

)-the prices of pork in the villages.
Output
Output N numbers, the J-th number is the number of pig to sell in the J-th Village. The pigs are numbered from 1 in the order they are listed in the input file.
Sample Input
3 1
50 70 60
Sample Output
3 2 1
Source
Northeastern Europe 2007, Northern subregion
Train of thought: first, according to the distance from a village and the cost of a unit journey, as well as the price of local pork, we can arrive at each village to sell pig unit weig

http://acm.hdu.edu.cn/showproblem.php?pid=3480Given a set of numbers S, size n, which requires that the set be divided into m subsets, each sub-subset of the cost is a subgroup (Max-min) ^2, to find the minimum cost.The beginning of DP transfer is easy to imagine.First, the collection is ordered from small to large, and dp[i][j] represents the minimum cost of the first I element being divided into J subsets. Then enumerate the last subset.DP[I][J] = Min{dp[k-1][j-1] + cost (k, i)};This transfer

#1223: Inequality time Limitation: 10000ms single point time limit: 1000ms memory limit: 256MB descriptionGiven n an inequality about X, the maximum number of questions is set.Each inequality is one of the following forms:X X X = CX > CX >= CInputThe first line is an integer n.The following n rows, one inequality per line.Data range:1OutputAn integer line that represents the maximum number of inequalities that can be set at the same time.

Title Description DescriptionThere are n heap of stones in a row, each pile of stones have a weight of w[i], each merge can merge adjacent two piles of stones, the cost of a merger is the weight of two piles of stone and w[i]+w[i+1]. Ask what sort of merger sequence it takes to minimize the total merger cost.
Enter a description input Description The first line is an integer n (
n≤
）Second row n integers
w 1 , w 2 ... w n ( w i ≤ the

Set $$\bex k\geq 2,\quad f\in c^k (\bbr), \quad m_j=\sup_{x\in\bbr}|f^{(j)} (x) |\ (j=0,1,\cdots,k). \eex$$ $$\bex M_j\leq 2^\frac{j (k-j)}{2}m_0^{1-\frac{j}{k}}m_k^\frac{j}{k}\ (j=0,1,\cdots,k). \eex$$Prove:(1). Only the conclusion of the $0(2). Toward the $k $ as a mathematical induction method. When $k =2$, to $j =1$, by $$\bex F (x+h) =f (x) +f ' (x) h+\frac{f ' (\xi)}{2}h^2, \eex$$ $$\bex f (x-h) =f (x)-F ' (x) h+\frac{f "(\eta)} {2}h^2. \eex$$ subtract and have $$\bex |f ' (x) |\cdot 2h \l

Set D (I, j) to the minimum length of the tree connecting the point I to the J point, there is a state transition equation:D (i, j) = min{D (i, K) + D (k + 1, j) + P[k].y-p[j].y + p[k+1].x-p[i].x}Then we use the quadrilateral inequality to optimize it.1#include 2#include 3#include 4#include 5#include 6 #defineMP Make_pair7 #defineX First8 #defineY Second9 using namespacestd;Ten Onetypedef pairint,int>PII; A - Const intMAXN = ++Ten; - Const intINF =0x3f3f3f3f; the intN; - - PII P[MAXN]; - + i

Time limit:10000msSingle Point time limit:1000msMemory Limit:256MB
Describe
Given n an inequality about X, the maximum number of questions is set.
Each inequality is one of the following forms:
X
X
X = C
X > C
X >= CInput
The first line is an integer n.
The following n rows, one inequality per line.
Data range:
1Output
An integer line that represents the maximum number of inequalities that can be set at the same time.

The main problem: there are n enemy arsenals in a straight line, each arsenal has a value VI, and any adjacent two libraries are connected to each other. For any one linked Arsenal chain, its threat to us can be represented by the function W (i,j) as: W (i,j) =vi*sum (i+1,j) +w (i+1,j) iW (i,j) =0 i=j;Now, you have a M bomb, each of which can blow up the channel between adjacent two libraries, seeking the ultimate total minimum threat value.Title Analysis: the definition State DP (I,J) means tha

Topic 1: Inequality time limit:10000msSingle Point time limit:1000msMemory Limit:256MBDescribeGiven n an inequality about X, the maximum number of questions is set.Each inequality is one of the following forms:X X X = CX > CX >= CInputThe first line is an integer n.The following n rows, one inequality per line.Data range:1OutputAn integer line that represents the maximum number of inequalities that can be set at the same time.
Sample inp

Use W[i][j] to show that there is no cost between I and J for destruction.There is a good proof that w[i][j] satisfies the conditions of quadrilateral inequalities. if (W[i+1][j]-w[i][j]) is about J's subtraction function, is satisfies the condition. It can be proved that this w[i][j] is not a condition of hiding.#include Copyright NOTICE: This article for Bo Master original article, without Bo Master permission not reproduced. Optimization of HDU 28

sample--learning algorithm A cannot be freely screened in h---Existence of some H makes e-in (h) Far from E-out (h)According to the above table, it can be seen that training data sets such as D-1126 are relatively high-quality.Given any d, it is the probability that some H's bad sample is:The less the number of alternative functions in H, the larger the sample data n, the smaller the probability that the sample becomes a bad sample. At an acceptable probability level, learning algorithm a only

, and F) if the corresponding edge or its extension line is intersecting, the three intersections are collocated.59. Disha lattice Theorem 2: two triangles △abc and △def exist on the different planes, and their corresponding vertices (A and D, B, E, C, and F) are set) if the corresponding edge or its extension line is intersecting, the three intersections are collocated.60. bryansong's theorem: links the vertex A, D, B, E, C, and F relative to the hexagonal ABCDEF of the circle, then the three-l

1. Upper-grade boys + lower-grade girls = practiceWhen new students enter school every year, senior boys take the initiative to pick up new students, more active than what they do. After a few days, the human resources of the freshman girl were

(This article assumes that the reader already has the following knowledge: The basic nature of the shortest path, the Bellman-ford algorithm.) ）For example, there is such a set of inequalities:
X1-X2 X1-X5 X2-X5 X3-X1 X4-X1 X4-x3 X5-X3 X5-x4
Inequalities Group (1)
It's all two unknowns. The difference is less than or equal to a constant (greater than or equal to, since left and right multiplied by-1 can be

Here is a detailed description of the differential constraint system, and the solution ~ excerpt from Xuezhongfenfei (he seems to be rotating ...). ）Differential constraint systemX1-X2 X1-X5 X2-X5 X3-X1 X4-X1 X4-x3 X5-X3 X5-x4 Inequalities Group (1) It's all two unknowns. The difference is less than or equal to a constant (greater than or equal to, since left and right multiplied by-1 can be converted to less than or equal). Such an inequality gro

(This article assumes that the reader has the following knowledge: the basic nature of the shortest path and the Bellman-Ford algorithm .)For example, there is a set of inequalities:
X1-X2 X1-X5 X2-X5 X3-X1 X4-X1 X4-X3 X5-X3 X5-X4
Inequality group (1)
All are
TwoUnknown
Less than or equalA constant (or greater than or equal to, because it can be converted to less than or equal by multiplying the left and right by-1 ). Such an inequalit

stones is counted as the cost of the merger. Calculates The minimum cost of merging N heap stones into a heap. ?For example: 1 2 3 4 , there are a number of consolidation methods1 2 3 4 = 3 3 4 (3) = 6 4 (9) = 10 (19)1 2 3 4 = 1 5 4 (5) = 1 9 (+) + 10 (24)1 2 3 4 = 1 2 7 (7) = 3 7 (Ten) + 10 (20)?The total cost in parentheses shows that the first method has the lowest cost and now gives N Count the number of stones, calculate the minimum combined cost. InputSection 1 Line: N ( 2?Section 2?-? N

Main features of this algorithm
The problem of two-dimensional cropping is converted into a secondary one-dimensional cropping problem, and the problem of cropping is transformed into solving a set of inequalities;
The improvement of the Cohen-Sutherland encoding algorithm is only applicable to the shortcomings of the line segments on the same side (or left, or right, or top or bottom) of the window.
The algorithm is divided into one and

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