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=st.top ();Panax Notoginseng St.pop (); -instack[v]=0; theprintf"%d", v);//outputs the number of all points that belong to this strongly connected component +} while(u!=v); Aprintf"\ n"); the } + } - intMain () $ { $ inti; - while(SCANF ("%d%d", n,m)! =EOF) - { the g.clear (); -G.resize (11000);Wuyi while(!st.empty ()) St.pop (); theNtime=1; -memset (Instack,0,sizeof(Instack)); Wumemset (DFN,0,sizeof(DFN)); -memset (Low,0,sizeof(Low)); About intu,v; $ for(i=0

The code implements the circular Unidirectional link list (the tail pointer points directly to the head pointer, there is no empty node in the middle), and the operation of successive repeating elements is removed.For example: 1 (head)->2->2->3->3->1->1 (head) after the removal of the result is 1->2->3, notice that the tail of 1 also to remove one.The time complexity is O (N)//Space complexity is O (1)//code as follows: #include Copyright NOTICE: This article for Bo Master original article, wit

Floyd#include #include #include #include #include #include using namespace STD;intn,m,edge[ -][ -],vis[ -],dist[ -]; Mapstring,int>AvoidFloyd () {intI,j,k,t; for(k=1; k for(i=1; i for(j=1; jif(edge[i][k]==-1|| edge[k][j]==-1)Continue; T=min (Edge[i][k],edge[k][j]);if(edge[i][j]==-1) edge[i][j]=t;Else if(edge[i][j]intMain () {inti,j,tot,st,ed,cas=0, U,v,w;stringS1,S2; while(~scanf("%d%d", n,m)) {memset(Edge,0,sizeof(Edge));if(n==0m==0) Break; A.clear (); tot=1; for(i=0; iCin>>s1>>s2>>w;if(

SPFA water problem. Record the maximum load capacity to this node. Because of a small error and WA a night, sad, t_t ...#include #include#include#include#include#includeusing namespacestd;Const intmaxn= ++Ten;structedge{int from, To,w;} e[1000010];intDis[maxn],vis[maxn];vectorG[MAXN];intN,m,tot;voidSPFA () {Queueint>p; memset (DIS,0,sizeof(DIS)); memset (Vis,0,sizeof(VIS)); vis[1]=1; Q.push (1); while(!Q.empty ()) { intH=q.front (); Q.pop (); vis[h]=0; for(intI=0; I) {Edgee=G[h][i]; if

To remove duplicate elements from an array: 1. Pure natural raw methods from data structures: foreach ($feerecord as $value){$fee _record_id = $value [' Feerecordid '];if (!in_array ($fee _record_id, $fee _record_id_arr)){$fee _record_id_arr[] = $fee _record_id;}Else{Continue}} return $fee _record_id_arr; Reasonable use of 2.PHP functions, code implementation needs: one line $fee _recorda_id_arr = array_values (Array_unique (Array_column ($feerecord, ' feerecordid ')); 3 Function Hints:

-dimensional array cannot be identical, removing the duplicates: The code is as follows: Output Result: before tranform the array://Original ArrayArray ([0] = = Array ([name] = James [age] = +) [1] = = Array ([name] = + Susu [age] = +) [2] =gt ; Array ([name] = James [age]=>) [new] + = Array ([name] = Kube [age] + Notoginseng) [list] =>array ([Nam E] = kube [age] = 27))After split the array://Array after splittingArray ([0] = james,30 [1] = susu,26 [2] = = james,30 [new] =>kube,37 [list] = kube

+= 1 return self.data for i in Next(3): print(i) print("-----------") n = Next(3) i = iter(n) while True: try: print(next(i)) except Exception as e: break The program running result is as follows: __next__ called 4 __next__ called 5 __next__ called 6 __next__ called ----------- __next__ called 4 __next__ called 5 __next__ called 6 __next__ called It can be seen that after the _ iter _ and _ next _ methods are implemented, the object can be iterated through the for in meth

control, S2SH integrated development; Spring integration JPA;9 Classic Java EE20, jsf21. EJB and related technologies: JNPI and RMI; session Bean and its life cycle; IOC and EJB interceptor; JMS and MDB; session bean and Web Service;22. Jpa:orm Framework and JPA Specification, JPA annotations and common API;JTA transaction and transaction management; JPQL query; EJB, JPA integration;Java Expansion, Advanced:23, Java EE Practice and Architecture: Ant+ivy or MAVEN;SVN, CVS, deep understanding of

1 PackageCom.java7.mysnow.main;2 Importjava.awt.*;3 ImportJavax.swing.JFrame;4 ImportJavax.swing.JPanel;5 6 Public classMysnow {7 Public Static voidMain (string[] args) {8Frame W =NewJFrame ();9W.setsize (1024, 768);Ten OneMypanel MP =NewMypanel (); A W.add (MP); - -Thread T =NewThread (MP); the T.start (); - -W.setvisible (true); - } + } - + classMypanelextendsJPanelImplementsRunnable { A intX[] =New int[300]; at intY[] =New int[300]; - intFx[] =New int[300]; - in

Title: There is an unlimited load of trucks to transport goods, in the city each road has a maximum capacity to withstand the weight,Now use trucks to transport goods from one city to another, asking for the maximum shipping weight.Analysis: graph theory, shortest path, minimum spanning tree. Find a path from the starting point to the end point, making the narrowest section the widest.It spreads from the beginning to the periphery, just like the Dijkstra algorithm and the prime algorithm, only m

The weather was fine(When I got up early in the morning, I found that a heavy snow had been quietly cleared last night, and the weather was fine, As a result, Shi xingfa enjoys his own entertainment, and he is logged on to the crowded bus on the congested road !) The sky is gray "Every day", the magician will show his magic again. He said the performance may take a long night. We recommend that you take a rest first. When I wake up

The shortest path or the maximum Spanning Tree deformation. Ask about the small weight that can be passed between two locations. Use the shortest path to assign the initial value to INF, And the other value to 0. Then find dis [v] = min (DIS [u], d ); The spanning tree is to find the maximum spanning tree until the start and end points can communicate with each other, and the smallest side is. Kruskal: # Include Poj 2263 heavy cargo (zoj

1, Bubble sorting method: five-star focus2, go to weight: also important, principleCause: At the same time output 123456 more output undefined---undefined means to occupy a bit but the missing value. So x does not need to be the last cycle, can be xExam Deduction Reason:The de-weight is not compared with the neighboring numbers; the first number is more than the rest of the number, and the second number is compared to every other!!!!!3. Perpetual calendarRefer to the dynamic clock written in 201

One. Bubble Sort methodTwo. Go to the heavyThree. Perpetual calendarBubble sort, go heavy, perpetual calendar

1. Do not take type wording: 1 List listWithoutDup = newArrayList(newHashSet(listWithDup)); 2. Style with type (for example, String type):1) Java 7 The following notation: 1 ListnewArrayListnewHashSet 2) Java 7 and above: 1 ListnewArrayListnewHashSet Examples: 1234567891011121314151617 importjava.util.ArrayList;importjava.util.HashSet;importjava.util.List;publicclassTest {publicstatic

I haven't looked at object-oriented development for a long time. I have returned many conceptual things to books. Today I have posted some posts on some forums on google, follow the examples of high people to familiarize themselves with the differences between heavy load and coverage:When you inherit an object-oriented class, you need to reuse methods in the class. You can reload or reuse these methods. During the reload, the names of methods are the

1. Method overloading-[polymorphism during compilation] and heavy-load-compilation Polymorphism Let's talk about method overloading and conceptual things. Please refer to MSDN. Let's take a look at the example below: Example 1: 1. Create a console program: Add a class [Overload ]: Using System; using System. collections. generic; using System. linq; using System. text; using System. threading. tasks; namespace OOP1 {/// In the main function: Using Sy

, because it was previously reversed) is also 0, then skip, otherwise read as "0"Break}Newary = zero + newary;zero = ";}else {//If not "0"Newary = This.ary0[parseint (Ary[i]) + this.ary1[i% 4] + newary; The expected character is converted into a numeric type, and as an index number of the array ary0, in order to get the corresponding Chinese reading method, followed by its first class unit (empty, ten, hundred or thousands) and finally added to the previous deposit reading content.}}if (Newary.i

The road to wudao-one day during the refining period (from the high ground in wanshu) and from the high ground in a heavy City List some simple basic methods: Example: A list Usernames = ['Li si', 'wang 2', 'zhang san', 'machin'] Index: usernames [0] # Starts from 0 and ends from-1 (left to right) Slice: usernames [] = ['wang 2'], that is, list [header: tail: Step Size], Gu header regardless of the end (left closed and right open), usernames [2:], fro

Heavy-duty operators and conversions-conversions and class types [Part 1] Introduction: As mentioned above: A non-explicit constructor that can be called with a real parameter can define an implicit conversion.When an object of the real parameter type needs a class object, the compiler will use this conversion.Therefore, this constructor defines the conversion to the class type. In addition to class type conversion, you can also define the conversion