quip vs slack

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

A tough question First, the input is very painful. According to the online experts, the question matrix is reversed. Then I gave it back. If it is a match between n and n, there is no pressure to directly obtain the negative value of KM. However, if the points on both sides are different, it is said that there will be problems. [Cpp]# Include # Include # Include # Include # Include # Include # Define maxn505# Define MAXM 555555# Define INF 1000000000Using namespace std;Int n, m, ny, nx;Int w [

20005Using namespace std;Struct Point {Double x, y;} Ant [105], tree [105];Double path [1, 101] [2, 101];Int cnt;Double lx [101], ly [101];Int match [101];Double slack;Bool v_x [101], v_y [101];Double dist (Point p1, Point p2 ){Return sqrt (p1.x-p2.x) * (p1.x-p2.x) + (p1.y-p2.y) * (p1.y-p2.y ));}Bool dfs (int k ){V_x [k] = true;Double temp;For (int I = 0; I If (! V_y [I]) {Temp = lx [k] + ly [I]-path [k] [I];If (zero (temp )){V_y [I] = true;If (match

why calculate Setup time of the Slack need to consider adding cycles, Hold Time when you don't need it? Summary one:Because the data delay,launch edge and capture edge do not correspond to the same clock edge of the clock signal source clocks, due to the existence of the time of Setup, the addition cycle needs to be considered. When a single clock cycle is checked, the tool defaults to capture Edge-launch edge=1 cycles.Hold time because the two corre

easily solved. (It seems that a template like this, basic all similar problems can be resolved)1#include 2#include 3#include 4#include 5#include 6 using namespacestd;7 #defineMAXN 1008 #defineINF 1000000009 Ten intCases,n; One intA[MAXN][MAXN],B[MAXN][MAXN],VAL[MAXN][MAXN],SLACK[MAXN],VALX[MAXN],VALY[MAXN],LINKY[MAXN]; A BOOLVISX[MAXN],VISY[MAXN]; - - structpoint{ the intx, y; - }ans; - -Pointoperator-(point A,point b) {return(point) {a.x-b.

1 intG[n][n];2 intLx[n], ly[n];3 intSlack[n];4 intMatch[n];5 BOOLVisitx[n], visity[n];6 intN;7 8 BOOLHungary (intu)9 {TenVisitx[u] =true; One for(inti =0; I i) A { - if(Visity[i]) - Continue; the Else - { - if(Lx[u] + ly[i] = =G[u][i]) - { +Visity[i] =true; - if(Match[i] = =-1||Hungary (Match[i])) + { AMatch[i] =u; at return true; - } - } -

HDU_3488 It is worth mentioning that if we use the KM algorithm to find the perfect matching of minimum weights, we need to initialize the edge weight to the MAX-G [I] [j] and then find the perfect matching of the maximum right, and then convert the result back. If you initialize G [I] [j] to-G [I] [j] to find the perfect matching of the maximum permission, the program I write Will time out, at the moment, I have no idea whether this idea is a problem or my writing is a problem. #include#inclu

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 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);

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

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

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1. Create a raw folder under Res to place HTML files. 2. Create an HTML file in raw: mypage.html Optimis é e-Flash Player 10.2 sera pr é-install é sur certains desArdoises, mais il sera un t é chargement OTA pour la plupart desTablettes honeycomb Android disponible en «Quelques semaines». ce queCela signifie est que le dispositif de lancement de honeycomb, le xoomMotorola qui sera disponible ce jeudi, VA navire sans Flash Player. EnFait une annonce pour le xoom Verizon é

foundation for storage, reuse, and transmission of containers, and the container mirrors "sit on the wheel" and spread to every corner of the world, helping the rapid development of container technology. Unlike earlier kernel container technologies, such as Solaris Container and LXC, Docker also provides developers with a good set of tools to experience, including dockerfile for mirroring builds and a domain-specific language for writing Dockerfil. The standard method of constructing mirror ima

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