hfc network components

. Compute the minimal number of extensions that has to is made so this whatever school we send the new software to, it'll Reach all other schools (Subtask B). One extension means introducing one new member into the list of receivers of one school.Program Name:schlnetinput FORMATThe first line of the input file contains a integer n:the number of schools in the network (2 in the line.SAMPLE INPUT (file schlnet.in)OUTPUT FORMATYour program should write,

(Subtask B). One extension means introducing one new member into the list of receivers of one school.InputThe first line contains a integer n:the number of schools in the network (2 OutputYour program should write, lines to the standard output. The first line should contain one positive integer:the solution of subtask A. The second line should contain the solution of subtask B.Sample Input52 4 3 04 5 0001 0Sample Output12SourceIOI 1996Test instructio

[N];intN, tot, Dfs_clock, scc_cnt, top;voidAddedge (intUintV) {E[tot]. from= u; E[tot].to = v; E[tot].next = Head[u]; Head[u] = tot++;}voidInit () {memset (head,-1,sizeof(head)); tot =0;intV for(intU =1; U while(SCANF ("%d", v) v) addedge (U, v); }}voidDfsintu) {pre[u] = lowlink[u] = ++dfs_clock; Stack[++top] = u;intV for(inti = Head[u]; I! =-1; i = e[i].next) {v = e[i].to;if(!pre[v]) {DFS (v); Lowlink[u] = min (Lowlink[u], lowlink[v]); }Else if(!sccno[v]) {Lowli

Network of schools Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 14018 Accepted: 5592 DescriptionA number of schools is connected to a computer network. Agreements has been developed among those Schools:each School maintains a list of schools to which it distributes Softwa Re (the "Receiving schools"). Note that if B was in the dis

finding the number of all such critical places. Help them.InputThe input file consists of several blocks of lines. Each block describes one network. The first line of each block there is the number of places N by one space. Each block ends with a line containing just 0. The last block had only one line with N = 0;OutputThe output contains for each block except the last in the input file one line containing the number of critical places.Sample Input55

].clear (); } voidAdd_edge (int from,intTo ) {g[ from].push_back (to); Rg[to].push_back ( from); } voidDfsintv) {Used[v]=true; for(intI=0; I) if(!Used[g[v][i]]) DFS (g[v][i]); Vs.push_back (v); } voidRdfsintVintk) {Used[v]=true; CMP[V]=K; for(intI=0; I) if(!Used[rg[v][i]]) Rdfs (Rg[v][i], k); } intSCC () {memset (used,0,sizeof(used)); Vs.clear (); for(intv=1; v) if(!Used[v]) Dfs (v); memset (Used,0,sizeof(used)); intK =1; for(intI=vs.si

].push_back (x); A } + } the SCC (n); - if(scc_cnt = =1){ $printf"1\n0\n"); the Continue; the } the int inch[MAXN], out[Maxn],in_tot =0, Out_tot =0; theMemsetinch,0,sizeof(inch)); -Memset out,0,sizeof( out)); in for(inti =0; I ){ the for(intj =0; J ){ the intv =G[i][j]; About /** the iterate through each point. the traverse the point connected to the point the determine if two points are the same connec

thread is still initializing the good EventLoop object (not yet put into _loops), causing the error;Construction and destructorEventlooppool::eventlooppool (size_t loopnums): _loopnums (loopnums), _running (false), _curindex (0), _mutex (), _countdownlatch (loopnums) { assert (_loopnums > 0); _loops.reserve (loopnums); _threads.reserve (loopnums);} Eventlooppool::~eventlooppool () { if (_running) Stop ();}Start and stop IO threadsvoid Eventlooppool::start () { assert (!_r

Kernel options during boot Linux users pass the Kernel configuration options to the Boot Record, The Boot Record, and then the options to the kernel, and then fine-tune the kernel through the boot configuration parameters. During system boot, parse_args is called to parse the configuration parameters entered during boot twice. The input string parameters for parse_args function parsing are in the form of "variable name = value". Appropriate processing functions are enabled Based on the parsed ke

capabilitiesNetworkSocket.Fast.dll function1, to provide rapid construction services, without the need to focus on communication protocols and communication processes (similar to WCF);2. Provide proxy code function (similar to WCF reference service) to generate client invocation service;3. Support Async and await keywords for asynchronous programming;4, support a number of common serialization methods for data serialization (JSON, XML, Google Protobuf, etc.);5. Client functionNetworkSocket.WebS

use ACCEPT4 to get a non-blocking connection descriptor at once;InetAddressUse InetAddress to encapsulate the IP address and port of the socket, make it easy to remove the IP string name, etc.InetAddress statementClass inetaddress Final{public: inetaddress () = default; InetAddress (const char* IP, uint16_t port); uint16_t port () const; std::string ipstring () const; std::string hostnamestring () const; struct sockaddr_in address () { return _address; } Private: struct sockaddr_in

Refer to this blog:http://blog.csdn.net/ascii991/article/details/7466278#include #includestring.h>#include#include#include#includeusing namespaceStd;typedefLong LongLL;Const intN = 1e2+5;intHead[n],tot,p,h[n],n, out[N],inch[N];structedge{intU,v,next;} Edge[n*n],e[n*N];voidAddintUintv) {edge[tot].u=u; EDGE[TOT].V=v; Edge[tot].next=Head[u]; Head[u]=tot++;}voidAddedge (intUintv) {E[P].V=v; E[p].next=H[u]; H[u]=p++;}intClk,dfn[n],low[n],cnt,bel[n];BOOLInstack[n];stackint>s;voidTargin (intu) {Dfn[u]=

Topic Link: Click to open the linkTest instructionsGiven a direction graph, ask:1) Select at least a few vertices. Ability to proceed from these vertices to reach all vertices2) At least how many edges to add. Talent makes it possible to reach all vertices from whatever vertexNumber of vertices After the strong connected component is obtained, the shrinkage point is calculated, and the degree of each point's penetration is computed.The answer to the first question is the number of points with ze

Title Address: POJ 1144Cut points. There are two ways to judge whether a point is a cut point:If u is a cut point, when and only if the following 1 bars are satisfied1, if u is the root, then you must have more than 1 subtrees tree2, if you are not the root, then (u,v) for the branch edge, when Low[v]>=dfn[u].Then according to these two sentences to find a cut point on it.The code is as follows:#include POJ 1144 Network (strong connected

Title Address: POJ 1236The main idea of this question is to ask at least how many points to send the message can make any one point can receive the message and the minimum number of additional edges can make the graph a connected graph. For the first problem, you can find the number of strong connected blocks with a degree of 0, because only a strong connected block with a degree of 0 is unable to receive information from the outside world, and as long as there is an entry, then the entire conne