smartnet coverage

Test instructions: Selects all the points in the minimum path overlay, and the path can be crossed, that is, allow the path to have duplicate points.Analysis: The difficulty of this problem is how to solve the problem of repeating points ~ method is to use Floyd to find the closure, that is, the indirect connection point directly connected to the top, and then is to find the minimum path coverage. I'm going to explain why it's right, first we have to

The main topic: to find the minimum point of two-dimensional coverage and maximum independent set.Title Analysis: If a point is selected, then all edges connected to this point are overwritten, so that the minimum point set that all edges are overwritten is called the minimum point overlay, which equals the maximum match, and the largest point set with no edges connected between any two points is called the maximum independent set, which equals the to

function dependency, then, after removing it from F, it can be exported according to the inference rules of the Armstrong axiom system.Because Cg→d (known)So CGA→AD,CGA→ACD (augmented law)Because Acd→b (known)So Cga→b (Transfer law)Because C→a (known)So Cg→b (pseudo-transitive law)Therefore, the cg→b is redundant.② the same: Ce→a is superfluous.③ again because of c→a, the function depends on the acd→b attribute A is superfluous, remove A to cd→b.So the minimum function dependency set is: F={ab→

Vertex Cover problem ' s Link Mean:Give you an out-of-the-box graph that allows you to dye the nodes in the graph so that at least one vertex of each of the two vertices of each edge is dyed. The minimum number of staining vertices to be obtained. Analyse:Bare minimum point coverage problem, the maximum matching of the binary graph, directly set template can be.Time complexity:o (n^2) View CodeBinary graph matching + minimum

Alpha to coverageIn the game, a polygonal model with translucent information textures is often used to simulate complex objects, such as grass, leaves, barbed wire, etc. If a real model is used, a grass with uneven edges may consume hundreds of polygons, but with a transparent texture, it can be solved with just two or three polygons.However, when using such a texture with translucent information, unsightly jagged edges often appear on the boundary lines of its opaque and transparent parts. This

Hihocoder-weekly223-interval CoverageTopic 1:interval Coverage time limit: 10000ms single point time limit: 1000ms memory limit: 256MB descriptionYou are given N intervals [S1, T1], [S2, T2], [S3, T3], ... [SN, TN] and a range [X, Y]. Select minimum number of intervals to cover range [X, Y].InputThe first line contains 3 integers n, x and Y. (1 The following N lines each contain 2 integers Si, Ti denoting an interval. (1 OutputOutput the minimum numbe

Click to open linkTest instructions: Give a graph, and then n the power supply range of the city, each now requires each city's D day can have electricity, for city a power, then all the cities adjacent to it will have electricity, but the problem is that each city can only be powered once a day, otherwise it will be broken, And each city has a range of days of power supply and each city can only open the switch once, and can not be used, that is, city A's power supply time must be continuous, t

1052: [HAOI2007] Coverage problem time limit: ten Sec Memory Limit: 162 MB Submit: 1095 Solved: 500 [Submit] [Status] DescriptionSomeone planted n small saplings on the mountain. Winter came, the temperature dropped rapidly, the small saplings fragile vulnerable, so the tree owners want to use some plastic film to cover up these small trees, after a long thought, he decided to use 3 l*l square plastic film to cover up the sm

The error of this problem is less consideration of a situation, the route coverage of the diameter is less than the time of the road width, the direct error!!! Without calculating the distance between routesThe idea is to calculate the route spacing for the square root of the 2 (r^2-(W/2) ^2), which has a data type conversion problem, W/2 This value is not accurate, and sqrt to negative numbers is also unrecognized, so to manually add 2*r > W.And the

, divide the 2^k * 2^k checkerboard into 4 2^ (k-1) * 2^ (k-1) sub-chessboard as shown.Special squares must be located in one of the 4 smaller sub-chessboard, with no special squares in the remaining 3 sub-chessboard. In order to convert the 3 sub-chessboard without special squares into a special chessboard, we can cover the 3 smaller chessboard with an L-shaped domino, as shown above, the squares covered by the L-shaped dominoes on the 3 sub-chessboard become special squares on the chessboard,

Topic Links:poj2125Test instructionsA forward graph of the N-vertex m-edge is given.For each vertex x, there are two actions:1, delete all the edges that enter X, and spend A;2. Delete all the edges that go out from X, and spend a B.Ask for the minimum cost required to delete all the edges in the diagram. and output the corresponding operation.Problem Solving Ideas:By the topic conditions (delete into the side, delete out side) The first thought should be the break-point. The problem of this top

Test instructions: N squares, side length is s[i], oblique 45 degrees in order to lay on the axis, as far as possible to the left, but not with any previous intersection, ask from the top down, which squares are visible.Solution: We first expand the edge length to sqrt (2) times the edge length, so that you can directly perform integer operations.And then in two steps:1. Find out all the b[i]2. Then the interval coverage decisionThe first step: when w

Test instructions: In a n*n matrix, there are m points, each elimination of a row or a column of points, the minimum number of operations;Idea: The x-axis and the y-axis as two different sets, each row or column as a point, each given point as an edge, so that the two-part map is built;Eliminate all points, the matrix is completely covered, so as to the minimum point coverage problem;Minimum cover number = maximum number of matches, so the maximum num

The main idea of the topic: not seen. It's all about finding the smallest round overlay anyway.Idea: A magical algorithm--random increment method. It can be proved that the algorithm can find the minimum circle coverage in the time complexity of O (n). Although it seems to be able to get out of the way, but add a random_shuffle on the card will not fall.The specific process is this:Records a circle globally, representing the current minimum circle ove

The main idea: in an n * n grid, there are m obstacles, and now you have a weapon, this weapon can eliminate any line or a row of obstacles, now requires all obstacles to eliminate, ask at least how many times the use of this weaponProblem-solving ideas: Think of long time how to solve the problems of row and column ... How to represent two point setsFinally suddenly thought, since do not know how to deal with the ranks, the ranks are divided into two points, the point on behalf of the relations

Test instructions: There are some points in a matrix that cover these points with a small matrix of 1*2, and the minimum number of small matrices required;Reference: http://blog.csdn.net/lyy289065406/article/details/6647040Idea: Non-graph minimum edge coverage = number of vertices-maximum number of matches/2;Each of the points to be matched is represented by a unique number, which is equivalent to discretization and easy to build;Each point to be matc

Original title AddressIt mainly introduces two theorems:1. Two the maximum number of matches = Two The minimum point coverage number of the graph2. Two graph minimum point cover number = Two figure vertex number-two fractal minimum point cover numberLook, it's a two-part picture.Code: (Hungarian algorithm)1#include 2#include 3 4 using namespacestd;5 6 #defineMax_n 10247 #defineMax_m 163848 9 intN, M;Ten intF[max_n]; One intN[max_m]; A intU[max_m]; - i

Source: Light OJ 1406 Assassin's Creed Directed Graph sends the least people to all cities and no one can walk anywhere else. Ideas: the minimum number of people who can finish the whole graph is obviously the problem of least path coverage. Here there may be loops, so we need to shrink the point. However, we can see that a strongly connected component may have to be split into N at most 15, so the state is compressed. Divide the full graph into sub-S

Using the dancing link to solve Sudoku For details, refer to the lrj training guide andDancing LinkS application in search Dancing link is used to solve the problem of data independence, which is precisely covered by the solution. Precise coverage is to give a 01 matrix, requiring us to select some rows so that each column has only one For the sudoku problem, the row is our choice, that is, to put the number k in column J of row I, so we can select

What is board coverage? Board coverage covers all squares except special squares on a given special board with four different forms of L-shaped bone cards, and no two L-type dominoes can overlap. Yes, that's easy! This is an example application that I learned when I used credit-based algorithms in university algorithms. I learned this algorithm and practiced Java swing (mainly because of my graduation desig