telecharger slack

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]))

],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(

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

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

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

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

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_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 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_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 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

Uva_000046 This topic can be done with the minimum cost and the maximum flow. In order to review the optimal matching of the bipartite graph, we wrote a KM program. At first, because I had not processed the double KM before, I was going to multiply the decimal number by a certain power of 10 into an integer, but I found that WA was always there, and then I simply wrote it using double, and then the AC. #include#include#include#define MAXD 30#define INF 1000000000int N, M, xM[MAXD], yM[MAXD] ,vis

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

intmaxn= -+Ten;Ten One intn,m; A intLX[MAXN],LY[MAXN],VISX[MAXN],VISY[MAXN]; - intLINK[MAXN],SLACK[MAXN],W[MAXN][MAXN]; - the intDfsintx) - { -visx[x]=1; - for(inty=1; yif(w[x][y]!=-1) + { - if(Visy[y])Continue; + intt=lx[x]+ly[y]-W[x][y]; A if(t==0) at { -visy[y]=1; - if(link[y]==-1||DFS (Link[y])) - { -link[y]=x; - return 1; in } - } to El