topology for dummies

also different. Only 26 lowercase letters.How to solve the problem: we need to understand that there is either a unique letter behind each letter, or not. So as long as the first letter of one of my substrings does not appear in the non-first character of the other substring, then I can use this substring as a non-precursor node. such as: Abc,ab,efg,fgk. Because a does not appear as a non-first character in other substrings, a can be a non-precursor node, and e can also. So all we have to do is

Website: Http://acm.hust.edu.cn/vjudge/contest/view.action?cid=117863#problem/BThought analysis: The bare topological sort, comments in the code.Code:#include #include #include #include #include #include #include using namespace Std;const int maxn=110;Vectorint b[maxn];//Storage Dependency DegreeQueueint m,n;int main (){int x, y;while (scanf ("%d%d", n,m) (n| | m)){Memset (b,0,sizeof (b));for (int i=1;iTp[i].clear ();//Initialize!!!for (int i=0;i{scanf ("%d%d", x,y);Tp[x].push_back (y);//Record

edges in the Adjmat array, and look at the methods of Moverowup and Movecolleft below:private void moverowup (int row, int length) {for (int col = 0; col This introduces all the procedures for sorting the topology. The complete code is attached below:Package graph;/** * Topological ordering of graphs: * Topological ordering is another operation that can be simulated with diagrams, which can be used to represent a situation in which certain items or e

** quarters and laboratory markings (two pairs of points x1,y1 and X2,y2 respectively). Next m line: three integers per line, u, V, L (1≤u≤n,1≤v≤n,1≤l≤10000), there is a road between table U and V, the time required to pass this road is L. Out-of-Gegger::: A line, an integer representing the time of two people per day (that is, the length of the longest public path). An output row, an integer that represents the time (that is, the length of the longest public path) of two people per day, Sample

HDU 1285 determines the competition ranking (Topology Sorting template) It is easy to understand the meaning of the question. The key is to see the following: the matching ranking may not be unique. At this time, the team with a small number must be in the front; Idea: this is the classic application of topological sorting. You can use the queue to prioritize the queuing of small numbers. Topological sorting:Topological sorting is a sort of directed a

(￣▽￣) " //This problem is required for the serial number output, the smaller ordinal priority output, so use the priority queue//Priority queue is the highest priority pop-up value, so the last to reverse the output, is the correct outputs#include #include#include#include#include#includestring>#include#include#includeusing namespaceStd;typedefLong Longll;Const intmaxn=100005; Vectorint>G[MAXN];intdegree[maxn],l[maxn],n,m;voidToposort () {inttot=0; Priority_queueint> que;//the implementation of T

After learning about Topology Sorting today, I found that this topic is a raw question. So I tried it again as I thought. I used the queue algorithm, namely, the Kahn algorithm. The complexity o (V + E) was completed and handed in. WA... So I checked it and handed it over again, or WA... I thought it was a queue problem. I changed it to stack and WA. Then I gave up STL, knocked on the queue, or WA... I'm crazy. It's okay. I got stuck for over an hour.

each case, output the case number as shown and then print ' Yes ', if there is a ' Triangle love ' among these N PEO PLE, otherwise print "No".Take the sample output for more details.Sample Input25001001000001001111011100050111100000010000110001110Sample OutputCase #1: yescase #2: NoMain topic:T Group test data, each group of data a n represents n individuals, and then the matrix of N*n to represent the relationship between these people, input must meet if a does not like B, B must like a, and

DescriptionThere are N teams (1InputThe input has several groups, the first behavior in each group is two number n (1OutputGive a ranking that meets the requirements. There is a space between the queue numbers at the time of the output, and no space after the last.Other Notes: Qualifying rankings may not be unique, at which point the output is required to be numbered in front of the team; the input data is guaranteed to be correct, i.e. the input data ensures that there must be a qualifying rank

); - if(op[i].r[0] =='=' ) - { Asum--; + Union_set (OP[I].A, op[i].b); the } - } $ BOOLFLAG0 =false; the for(inti =0; I ) the { the if(op[i].r[0] =='=')Continue; the intA = FINDF (op[i].a), B =findf (op[i].b); - if(A = =b) in { theFLAG0 =true; the Break; About } the if(op[i].r[0] =='' ) the { the Addedge (b, a); +

#include #include #include #include using namespace STD;intmp[ -+5][ -+5], in[ -+5],u[ -+5];intMain () {int_, I, J, K, n, cas =0, t;Chars[ -+5];BOOLFlagscanf("%d", _); while(_--) {memset(MP,0,sizeof(MP));memset(In,0,sizeof(in));memset(U,0,sizeof(u));scanf("%d", n); flag=0; for(i =0; I scanf('%s ', s); for(j =0; J if(S[j] = =' 1 ') {Mp[i][j] =1; in[j]++; } } } Stackint>S while(!s.empty ()) S.pop (); for(i=0; iif(in[i]==0) S.push (i); } while(!s.empty ())

if(!inch[v]) {Bayi if(flag)return false; the S.push (v); theFlag =true; - } - } the } the return true; the } the - the intMain () { the intT; thescanf"%d",T);94 while(t--){ thescanf"%d%d",n,m); the init (); the for(inti =1; I ){98 intu,v; Aboutscanf"%d%d",u,v); - addedges (u,v);101 }102 FIND_SCC ();103 if(SCNT = =1) {104Puts"Yes"); the Continue;106 }107 for(inti =

Simple and simple is true. Those high-end foreign copy topologies are purely self-abuse.There are about four principles for implementing replication:① A master database can have multiple slave Databases② One slave database can only have one master database③ Each slave Database Server ID is globally unique④ Log_slave_updates has the utility of pass-throughThe following is a brief introduction to the Topology Design of several types of replication, thei

the main thing is that the search for zero points is especially This topic input storage is a difficult point I have been using the map is actually a method only a zero in the points can produce CHAMPIONS!!!! Start blogging again!!1#include 2 using namespacestd;3mapstring,int>inch;4mapstring,mapstring,int> >MP;5mapint,string>dis;6mapstring,int>PiS;7queuestring>Q;8 intMain ()9 {Ten intN; One intCoun; A while(SCANF ("%d", n)! =EOF) - { - if(n==0) the Break; -

predicate method takes two geometries as parameters and returns a Boolean value that indicates whether the geometry has a specified spatial relationship. The spatial relationships it supports are: equal (equals), detach (disjoint), intersect (intersect), phase (touches), cross (crosses), inclusive (within), inclusive (contains), Overwrite/ Covered in (overlaps). The General relational (relate) operator is also supported. Relate can be used to determine the dimension extension of the nine-inters

Common topology of MySQL ReplicationThe replication architecture has the following basic principles:(1) Each slave can have only one master;(2) Each slave can have only one unique server ID;(3) Each master can have many slave instances;(4) If you set log_slave_updates, slave can be the master of other slave, thus spreading master updates. MySQL does not support multi-master Replication-that is, one slave can have multiple masters. However, through som

point, if there are two points of the dis is the same, then the relationship between the two points is not clear, or there is a certain contraction point it does not have the edge and the edge, and its rank is less than the total number of N, indicating that the set is isolated, the set of points Of course, you first have to deal with the contraction point offline, or in case the input order changes, the addition of the edge is not rightTo two sets of data:3 11 = 0Uncertain2 11 = 0OkCode:#inclu

;i3; i++){ inchI [A+n]++;g[i][b].push_back (A +N); inchI [B+n]++;g[i][a].push_back (b +N); } } Else if(ch[0]=='X'){ inch[0][b]++;g[0][a+N].push_back (b); } Else if(ch[0]=='Y'){ inch[1][b]++;g[1][a+N].push_back (b); } Else{ inch[2][b]++;g[2][a+N].push_back (b); }} printf ("Case %d:",++CAs); if(Topo ()) {printf ("possible\n"); for(intI=1; i) printf ("%d %d%d%d%d%d\n", val[0][i],val[1][i],va

]--; for(intk=2; k){ if(dp[it->id][k]>dp[j][k-1]+it->W) {Dp[it->id][k]=dp[j][k-1]+it->W; Fromx[it->id][k]=J; } } } Break; } } for(inti=n;i>=2; i--){ if(dp[n][i]t) {printf ("%d\n", i); Stackint>SS; intx=n,y=i; Ss.push (n); while(fromx[x][y]!=1) {Ss.push (fromx[x][y]); X=Fromx[x][y]; Y--; } Ss.push (1); while(!Ss.empty ()) { intpp=Ss.top (); Ss.pop (); printf ("%d", pp); }

Initial error:Dynamicbrokersreader [ERROR] node/brokers/ids/0 does not existSet up:Brokerhosts brokerhosts = new Zkhosts (Zks, "/kafka/brokers");Then there is the following error:2016-04-13t15:07:10.657+0800b.s.util[error]asyncloopdied! java.lang.runtimeexception:java.lang.runtimeexception:org.apache.zookeeper.keeperexception$ nonodeexception:keepererrorcode=nonodefor/kafka/brokers/topics/testtopic/ Partitionsatstorm.kafka.dynamicbrokersreader.getbrokerinfo ( dynamicbrokersreader.java:81) ~[stor