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Python implementation __python of several shortest path calculations

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

HDU 3488 Tour

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 (

Hdus 3435 A new Graph Game

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[

poj3565 Ants miles algorithm to find the minimum weight perfect match, floating-point weighted value

(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'

HDU-2255 Ben-off make a lot of money (binary chart, km algorithm, template)

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

"Turn" Bellman_ford algorithm

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

Laravel basic tutorial-SSH task

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

Two-dimensional graph with weighted matching (thinking of pushing box problem)

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

HDU2426 interesting Housing problem (km matches)

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

HDU 2255 Ben-off make money

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

"Hdoj" 2255 Ben-off make money

Maximum binary graph matching, O (n^3).1 /*2255*/2#include 3#include 4#include 5#include 6#include 7 using namespacestd;8 9 #defineMAXN 305Ten #defineINF 0XFFFFFFF One A intW[MAXN][MAXN]; - intLINK[MAXN]; - intLX[MAXN], LY[MAXN]; the intSlack; - BOOLS[MAXN], T[MAXN]; - intN; - + BOOLDfsinti) { -S[i] =true; + for(intj=1; jj) { A if(T[j]) at Continue; - intTMP = lx[i]+ly[j]-W[i][j]; - if(TMP = =0) { -T[J] =true; - if(!link[j] | |DFS (Link[j]))

Dichotomy Chart "template"

],SLACK[MAXN];intVISX[MAXN],VISY[MAXN];//MarkintMAP[MAXN][MAXN];//Storage weight value//lx[],ly[] Top label; link[] Record matching valueintFindpath (intU//backtracking to find the optimal solution{Visx[u] =1; for(inti =1; I if(Visy[i])Continue;inttemp = Lx[u] + ly[i]-map[u][i];if(temp = =0)//if (map[u][i] = = Lx[u] + ly[i])//description is an equal sub-graph{Visy[i] =1;if(Link[i] = =-1|| Findpath (Link[i])) {Link[i] = u;return 1; } }Else if(

K8s and cicd--a complete drone demo

This is a creation in Article, where the information may have evolved or changed. Objective In the previous article, we simply introduced the deployment of drone. The next step is to introduce the official documents as well as the plugins I used in the actual work. This article, is mainly a complete drone demo. Pipeline includes the Git clone pull code, the go build build, the Docker image is generated and pushed to the harbor,notify notification. On the notice, the official plug-ins are some of

Going HOME-POJ 2195 (Network Flow | Two-point matching)

The main idea: in a grid there are n small men and N houses, now want to let every little man have a house to live, but everyone to move a bit to spend ¥1, now find out the minimum total cost. PS: It can be thought that each point of the grid is a large square and accommodates all the people who can walk at the point of having a house but do not enter the house.Analysis: Man-house, very perfect with all minimum values match Ah, the cost of people to a house is their Manhattan distance between th

Give programming a chance that you love it

If a thing, you do not love him, perhaps because you fear it, dare not to touch him, so devote less time. Muddle through the living, muddle through life, will slack off, and is once the opportunity to slack, no opportunity will also create opportunities to slack off, and the first big law is "slack people will never su

The maximum weights of binary graphs match km algorithm

feasibility, that is, for the above after this step, at least one feasible edge will be added to the figure. (5) After the modification, continue to the X-side point Dfs augmentation, if it fails, continue to modify until successful;The above is the basic idea of the KM algorithm. But the simple implementation method, the time complexity of O (N4)-Need to find O (n) Secondary augmentation path, each augmentation needs to modify the O (N) sub-index, each time the top is modified to enumerate the

[SinGuLaRiTy] KM algorithm

edges, so that new edges added will always be able to add matching numbers to the sub-graph, while at the same time make the weight and the maximum increase.Implementation code/*HDU-2255*/#include#include#include#include#defineMAXN 310#defineINF 0x3f3f3f3fusing namespacestd;intNx,ny;intG[MAXN][MAXN];intLINKER[MAXN],LINKX[MAXN],LINKY[MAXN];intSLACK[MAXN];BOOLVISX[MAXN],VISY[MAXN];BOOLDfsintk) {Visx[k]=true; for(inty=0; y) { if(Visy[y])Continue; inttmp=linkx[k]+linky[y]-G[k][y]; if(!tm

HDU 2818 One fihgt one

HDU_2813 To avoid time-out when reading data, hash can be used for processing, and then non-Repeated names can be taken out for a vertex, then use the KM algorithm to find the optimal matching for the bipartite graph. #include#include#include#define MAXD 210#define MAXM 40010#define INF 1000000000#define HASH 10007int headL[HASH], nextL[MAXM], headC[HASH], nextC[MAXM];int N, M, K, G[MAXD][MAXD], yM[MAXD], MAX;int A[MAXD], B[MAXD], slack[MAXD];int vis

HDU 1533 Going Home

HDU_1533 The shortest distance between people and houses is the distance between Manhattan. After creating the image, apply the KM algorithm to obtain the optimal matching of the Bipartite Graph. #include#include#include#define MAXD 110#define MAX 10010#define INF 1000000000int N, hx[MAXD], hy[MAXD], px[MAXD], py[MAXD];int a[MAXD][MAXD], G[MAXD][MAXD], yM[MAXD];int A[MAXD], B[MAXD], slack, visx[MAXD], visy[MAXD];char b[MAXD];int init(){int i, j, k, n

HDU 2583 Assignment

HDU_2853 We can think that if we start matching from scratch and if the final part of matching changes but does not affect the final result, we do not need to change it into a new matching method. Therefore, we consider adding a "preference" to the original edge to ensure that it matches the original edge first without affecting the result. When processing data, we can expand each edge to a certain value (multiply by 10 in my program), and then add the original edge weight to 1, in this way, w

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