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HDU 3718 Similarity

HDU_3718 First, we need to convert the characters in the string into the number of [0, k-1] with the same meaning, and then scan the array sequentially, we can get the maximum value of various matching between the two sets of numbers, this completes the graph creation. Then use the KM algorithm to find the optimal match. #include#include#define MAXD 30#define MAXN 10010#define INF 1000000000char a[MAXN][5], b[MAXN][5];int visa[MAXN], visb[MAXN], x[MAXN], y[MAXN];int G[MAXD][MAXD], yM[MAXD], N,

HDU 3395 Special Fish

HDU_3395 We can split a fish into two points, which respectively represent attack and attack, and then set the Edge Weight of G [I] [j] = 1 to value [I].^ Value [j], and then use the KM algorithm to find the optimal matching of the Bipartite Graph. #include#include#define MAXD 110#define INF 1000000000int yM[MAXD], G[MAXD][MAXD], N, value[MAXD];int A[MAXD], B[MAXD];int visx[MAXD], visy[MAXD], slack;char b[MAXD];int init(){int i, j; scanf("%d", N);

hdu2255 Ben-off make money, Max right match, km algorithm

tree, The Edge (I,J) may be added to the equal sub-graphIn order for the A[i]+b[j]>=w (I,J) to always be true, and at least one edge is added to the equal sub-graph, d=min{a[i]+b[j]-w (i,j)},i in the interlaced tree, J is not in the interlaced treetime complexity: need to find an O (n) time augmentation path. Each augmentation requires an O (n) sub-index change. Each time you change the top, enumerate the edges to find the D value, the complexity is O (N2), and the total complexity is O (N4). S

Bellman-ford algorithm--Solving the shortest path problem of single source point

distance s. When 1 times slack is made on each side, the branches that start from S and have a maximum level of 1 are generated. That is, the shortest path to the vertices with a maximum of 1 edges associated with S is found, and the 2nd-pass relaxation of each edge creates a 2nd-level branch, which means that the shortest path of those vertices connected by 2 edges is found .... Because the shortest path contains up to |v|-1 edges only, you only nee

"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

HDU 3772 Card Game

HDU_3772 We can split a string into two vertices to represent the inbound and outbound degrees, and then use the KM algorithm to perform Optimal Matching for the bipartite graph. #include#include#define MAXD 210#define INF 1000000000char b[MAXD][1010];int G[MAXD][MAXD] , yM[MAXD], N;int A[MAXD], B[MAXD], slack;int visx[MAXD], visy[MAXD];int check(char *str1, char *str2){int i, j, num = 0; i =strlen(str1) - 1;for(j = 0; str1[i] == str2[j] i >= 0

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