slack markup

the KM algorithm can be O (N3. Each y vertex is given a "relaxation amount" function slack, Which is initialized to an infinite number every time the augmented path is searched. When you check the edge (I, j) in the process of searching for the augmented path, if itNot equal subgraph, The slack [J] is changed to a smaller value of the original value and a [I] + B [J]-W [I, j. In this way, allY vertex not i

data contains multiple sets of test cases, the first line of each set of data entered N, indicating the number of houses (also the number of people in the home), followed by n rows, the number of n per row represents the first village name of the room price (nOutputPlease output the maximum revenue value for each set of data, one row for each group of outputs.Sample Input2 100 10) 15 23Sample Output123 template questions, organize the template/************************************************aut

the village leader assigns).Test Instructions Description: Because this is titled Chinese title, this is not translated here. algorithm Analysis: maximum weight matching problem,km algorithm for the entry question. Baidu km algorithm has a lot of relevant introduction, here no longer described. Explanation Point, km optimization: The slack array saves the value min (Lx[x]-ly[y]-w[x][y]) of each y in the Y-set (binary graph: X-Set and Y-set) that is n

KM algorithm1#include 2 #defineN 15003 #defineINF 9999999994 using namespacestd;5 inta[n],bs[n],nx=0, ny=0, K;6 intLinky[n],lx[n],ly[n],slack[n];7 intVisx[n],visy[n],w[n][n];8 intMinintAintb) {return(aa:b;}9 intFindintx) {Tenvisx[x]=1; One for(inty=1; y){ A if(Visy[y])Continue; - intt=lx[x]+ly[y]-W[x][y]; - if(t==0) {visy[y]=1; the if(linky[y]==-1||find (Linky[y])) { -Linky[y]=x;return 1; - } -

metVis_boy[boy] =true; if(Match[boy] = =-1|| DFS (Match[boy])) {//find a guy who doesn't have a match, or the boy's sister can find someone else .Match[boy] =girl; return true; } } Else{Slack[boy]= Min (Slack[boy], GAP);//slack can be understood as the boy who wants to get a girl's heart. How much will it take to have a minimum spare tire appearance "cove

intMAXN = the;Const intINF =0x3f3f3f3f;intLOVE[MAXN][MAXN];//record the goodwill of every sister and every boyintEX_GIRL[MAXN];//the expectations of every sisterintEX_BOY[MAXN];//the expectations of every boyBOOLVIS_GIRL[MAXN];//record every match of the match GirlBOOLVIS_BOY[MAXN];//keep track of each match, match the boys.intMATCH[MAXN];//record each boy's match to the sister, if not, 1.intSLACK[MAXN];//keep track of the number of expectations that every man needs at least if he can be attrac

],slack[n];intN,ans,nx,ny;CharE[n][m];voidGetmap ()//Building Map{ inti,j,x,y,count,l; for(x =1; x ) { for(y =1; Y ) { if(x = =y) w[x][y]=0; Else{L= strlen (e[x]+1); Count=0; for(i = L,j =1; i >0e[y][j]!=' /'; I--, J + +) { if(E[x][i] = =E[y][j]) Count++; Else Break; } W[x][y]=count; } } } return; }intDfsintX//find an augmented path{ inty,tmp; V

One or Unvisited[neighbour] > newdistance: #如果两个点之间的距离之前是无穷大或者新距离小于原来的距离 unvisited[neighbour] = newdistance# Update distance Visited[current] = currentdistance# This point has been slack, record del unvisited[current] #从未访问过的字典中将这个点删除 if not unvisited:b reak# if all the points are slack, jump out of this cycle Candidates = [node for node in Unvisited.items () if NODE[1]] #找出目前还有拿些点未松弛过 Current, currentd

Contest (6)--host by BIT problem solving: Just like the cyclic tour in front of you.1#include 2 using namespacestd;3 Const intMAXN =310;4 Const intINF =0x3f3f3f3f;5 intW[MAXN][MAXN],LX[MAXN],LY[MAXN],SLACK[MAXN];6 intN,LINK[MAXN];7 BOOLS[MAXN],T[MAXN];8 BOOLMatchintu) {9S[u] =true;Ten for(intv =1; V v) { One if(T[v])Continue; A intD = lx[u] + ly[v]-W[u][v]; - if(!d) { -T[V] =true; the if(Link[v] = =-1||match (

HDU_3435 This is a perfect match for minimum weights. When I use the KM algorithm to create an adjacent table, I need to initialize the edge weight to MAX-w [e]. then, find the perfect match of the maximum right and convert the result back. Because it is an undirected graph, we need to create two directed edges for each undirected edge. #include#include#define MAXD 1010#define MAXM 20010#define MAX 10001#define INF 1000000000int first[MAXD], next[MAXM], v[MAXM], w[MAXM], N, M;int yM[MAXD], wM[

(Fabs (GAP) //If you meet the requirements, pay attention to floating point ah. Vis_boy[boy] =true; if(Match[boy] = =-1|| DFS (Match[boy])) {//find a guy who doesn't have a match, or the boy's sister can find someone else .Match[boy] =girl; return true; } } Else{Slack[boy]= Min (Slack[boy], GAP);//slack can be understood as the boy who wants to get a girl'

of their income. (Villagers who have the money to buy a house but not necessarily can buy it depends on what the village leader assigns).Input data contains multiple sets of test cases, the first row of each set of data input n, indicating the number of houses (also the number of people home), followed by n rows, the number of n per row represents the price of the room of the second village name (nOutput make the maximum revenue value for each set of data, one row for each set of outputs.Sample

Embedded RealOne Player controlsusing markup: Note:Only oneTag on the page requiresSRCParameter. In contrast, eachTag requires an identicalSRCParameter. For bothAnd, The optionalAUTOSTARTParameter is needed in only one tag on the page.All markup: Note:This full control panel drops some controls when youReduce its width or height below certain pixel values.Controlpanel

Original link: http://www.cnblogs.com/Jason-Damon/archive/2012/04/21/2460850.htmlExcerpt from Baidu EncyclopediaThe Bellman-ford algorithm is a single-source shortest path algorithm with negative weights, which is very inefficient, but the code is easy to write. That is, the continuous relaxation (relaxation), each slack to update each edge, if the n-1 can be updated after the relaxation, then the picture has a negative ring (that is, the negative pow

after the task is executed. This command receives an API token, the name of the room, and the user name of the sender displayed in the message: @servers(['web' => '192.168.1.1'])@task('foo', ['on' => 'web']) ls -al@endtask@after @hipchat('token', 'room', 'Envoy')@endafter If necessary, you can also send custom messages to the HipChat room. When building a message, the available variables of the task are also available in the message: @after @hipchat('token', 'room', 'Envoy', "$task ran in

staggered road X4, Y2, X3, Y0 X1, not found? there is a X4---------Y2, X3, X0--Y1-X2 The inverse of the path attribute becomes the upper (right) graph. At this point, all the vertices in set X already have corresponding matching, that is, complete matching! That is, the maximum weight of this binary graph match! X0-Y1X1-Y0X2-Y4X3-Y3X4-Y2Maximum power value is 30What about the minimum weight matching requirement? Very simple, before solving the ownership value of the opposite number, the results

Test instructions: N students arranged to M dormitory, each student to the dormitory has a rating, positive, 0, negative, now the evaluation is negative, can not let this student go to this room, ask how to arrange to let all the students live in the dormitory and the most evaluation.Idea: When building the weight of a graph, filter out negative edges.#include #include #include #include #include #include #include #include #include #include #include #include #define Lson (rt#define Rson (rt#defin

KM algorithm binary graph maximum weighted value matching#include #include#include#includeConst intMAXN =356;Const intINF = (1 to)-1;intW[MAXN][MAXN];intLX[MAXN],LY[MAXN];intLINKY[MAXN];intVISX[MAXN],VISY[MAXN];intSLACK[MAXN];intNx,ny;BOOLFindintx) {Visx[x]=true; for(inty =0; Y ) { if(Visy[y])Continue; intt = lx[x] + ly[y]-W[x][y]; if(t==0) {Visy[y]=true; if(linky[y]==-1||find (Linky[y])) {Linky[y]=x; return true; } } Else if(Slack