pigtail antenna

Test Instructions: There are k points in a n*m matrix, and you must place the antennas over all the observation points. Each radar can only have two observation points on the antenna, and the two points must be adjacent. Calculates the minimum number of antennas.Practice: Connect all adjacent observation points and build a map. Run it once. The Hungarian algorithm calculates the maximum number of covers, divided by two, the number of antennas. Also ad

Antennas are rectangular, circular, and rod-shaped. Generally, the longer the communication distance is, the larger the size of the antenna. However, in actual use, when tags are attached to an item, even the RFID tags of the business card size are too large. In fact, the size and shape of the RFID tag in use need to be selected based on the communication distance. Currently, antenna materials include al

Label: style blog HTTP Io color OS AR for SP Poj 3020 antenna placement Question Link Question: Given a map, the place of '*' should be covered. Each time we can use a rectangle of 1X2 to cover the map, we can use at least a few to cover the map. Idea: bipartite graph matching to find the Maximum Independent Set, adjacent * edge, and then find the maximum independent set.Depending on the data range, you can also use the outline DP. Code: #in

"POJ 3020" antenna PlacementThe maximal independent set problem of binary graphs' O ' means that the break point requires that all * connections be connected to one or two consecutive lines per route *The maximum match can satisfy only the longest path that can be formed by only two consecutive connections, and the points that are not connected need to be circled separately.i.e. N (total * number)-m (maximum match) +m (maximum match)/2(due to the fact

Antenna package the j2s programAuthor: JeruMany people like to use ant to compile and package Java applications. They can easily generate war packages and run them in J2EE containers. I also saw an article on sun's website.How to use ant to package the j2_program: http://wireless.java.sun.com/midp/articles/ant/I tested it. It's not bad, but the disadvantage is that I need to write the manifest. mf and jad files in advance, so I am unwilling to

Program | Package Antenna Package J2ME Program Author: Jeru Many people like to use ant to compile packaged Java applications, and it's easy to generate war packages running in the Java EE container, and I also saw an article on Sun's website tutorial on using ant to package J2ME programs: http://wireless.java.sun.com/midp/articles/ant/ Test a bit, but also good, but the drawback is that the need to write their own MANIFEST.MF and jad documents, so

; - }Wuyi the intMainvoid)//POJ 3020 Antenna Placement - { Wu //freopen ("poj_3020.in", "R", stdin); - About intT scanf ("%d", t); $ while(t--) - { - intH, W; scanf ("%d%d", h, W); - for(intI=1; ii) A { +scanf ("%s", S[i] +1); the } - $Un = VN =0; the for(intI=1; ii) the { the for(intj=1; jj) the { - if(S[i][j] = ='*') in { th

Antenna Magus 5.5.0 pro\Compusoft.winner.v9.0a2_hao\EPLAN Harness ProD Studio 2.5\Ixray Ixforten 4000 version 4.9.8uu\Mintec.minesight.3d.v7.0.3\NI. Vision.v2015.SP1 1DVD NI Vision 2015\Pix4d.pix4dmapper.pro.v2.0.104.macosx\Techlog learning materials software training \Techlog2014.2\multiphase flow simulation software Schlumberger OLGA 2015.1\[CATIA. P2. V5r16 catia.cadam.drafting.v5r16sp1\338\CPFD. Barracuda.v17.0.0.rarHypermill8.4.zipMpcci v3.03 Lin

The main problem: There are N cities, to establish a radio station in these n cities, each radio station can only cover 2 neighboring cities, ask at least how many wireless power stations need to buildProblem-Solving ideas: English topic good pit, looked at half a day.This problem is similar to POJ-2446 chessboard.All cities can be divided into two point sets, then the connection between the radio station to represent the coverage of the relationship.Because all cities have to be covered, so acc

Question: In a rectangle, there are n Cities '*'. Now all N cities need to cover wireless networks. If a base station is placed, it can cover at most two adjacent cities.How many base stations can be placed to ensure that all cities are wireless? Ideas: Number each city, create two-way edge, run Hungary, and then the number of cities-matching number/2 is the answer Assume that each city needs to build a base station, and then how many matches will reduce the number of base stations. Because the

#include 2#include 3#include 4 using namespacestd;5 intmatch[ the];6 structnode{7 intx, y;8}dir[4]={{1,0},{-1,0},{0,1},{0,-1}};9Vectorv;Ten intn,m,cnt; one Charmatrix[ -][ -]; a intmp[ -][ -]; - intbook[ the]; - inte[ the][ the]; the voiddeal () { - for(intx=0; x) { - for(inty=0; y) { - if(matrix[x][y]!='*')Continue; + intI=mp[x][y]; - for(intD=0, xx,yy; d4; d++) { +xx=x+dir[d].x; ayy=y+dir[d].y; at if(xx0|| X>=n | | yy0|| Yy

--){scanf"%d%d", m, n);intsum =0;///record number of ' * 'InIt (); for(i=0;i{scanf"%s", G[i]); for(j=0;j{if(G[i][j] = ='*'){///Purchase Relationship DiagramINDEX[I][J] = ++nx;///give me the number.sum++;if(J! =0 g[i][j-1]=='*'){///matches to the left.Addedge (Index[i][j], index[i][j-1]);Addedge (index[i][j-1], index[i][j]);}if(I! =0 g[i-1][J] = ='*'){///with the above can matchAddedge (Index[i][j], index[i-1][J]);Addedge (index[i-1][J], index[i][j]);;}}}}intans = Karp ();printf"%d\n", Sum-ans);}

; memset (Link,-1,sizeof(link)); for(intI=1; i) {memset (Vis,0,sizeof(VIS)); if(Dfs (i)) sum++; } returnsum;}intMain () {intans,i,j,k; while(SCANF ("%d", t)! =EOF) { while(t--) {CNT=0; memset (MM,0,sizeof(mm)); memset (Mark,0,sizeof(Mark)); scanf ("%d%d",n,m); for(i=0; i) { for(j=0; j) {scanf ("%c",temp); if(temp=='*') {Mm[i][j]=++cnt;//each point is represented by a number } } } for(i=0; i)

Reprinted please indicate the source: Thank you Http://user.qzone.qq.com/289065406/blog/1299322779 Tip: do not be misled by the circle of pictures. You can see clearly that '*' is a city, 'O' is an open space, and the antenna of an elliptical circle must cover the city '*', instead of covering the Open Space Question: In a rectangle, there are n Cities '*'. Now all N cities need to cover wireless networks. If a base station is placed, it can cover at

The basis of binary matching. You only need to find out a and B. Search for the four directions of each "*" and record their relationships; Because we read both a-> B and B-> A, we need to divide it by 2; #include#includechar str[41][11];bool mat[40

Question: In a rectangle, there are n Cities '*'. Now all N cities need to cover wireless networks. If a base station is placed, it can cover at most two adjacent cities.How many base stations can be placed to ensure that all cities are

An original Mini-game inspired by gold miners and big fish. In fact, this game has been played for two weeks and is only uploaded until now. Some bugs have not been fixed yet. No way. Busy. Recently, I am engaged in a large-scale webgame with

Black and white coloring and then finding the maximum matching of the Bipartite Graph [cpp]# Include # Include # Include Using namespace std;Const int maxn = 40 + 9;Int n, m;Char a [maxn] [maxn];Struct{Int to, next;} E [maxn * 10*4];Int head [maxn *

Question: In a rectangle, there are n Cities '*'. Now all N cities need to cover wireless networks. If a base station is placed, it can cover at most two adjacent cities.How many base stations can be placed to ensure that all cities are wireless?

Question link: http://poj.org/problem? Id = 3020 First of all, ignore the one that may mislead others (maybe I have poor understanding, or I am misled by a little bit ). An H * w graph is provided. "*" indicates point of interest, and "O" ignores it.