jira slack

This post was last edited by Jimila on 2012-10-5 12:55 Transferred from: http://www.12306ng.org/thread-911-1-1.html Original: http://www.ltesting.net/ceshi/open/kyrjcsxw/2012/0713/205272.html How did the open source community form? How does open source project manage? In this article, I'd like to share some of the management tools and collaborative processes I've used in the AS7 development process, as well as some understanding of the open source community. The AS7 development process involves

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

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