officejet 4635

Hdu 4635 strongly connected component + contraction point Problem Description Give a simple directed graph with N nodes and M edges. please tell me the maximum number of the edges you can add that the graph is still a simple directed graph. also, after you add these edges, this graph must NOT be strongly connected. A simple directed graph is a directed graph having no multiple edges or graph loops.A stronugly connected digraph is a directed graph in

http://acm.hdu.edu.cn/showproblem.php?pid=4635Problem descriptiongive A simple directed graph with N nodes and M edges. Please tell me the maximum number of the edges you can add that the graph was still a simple directed graph. Also, after you add these edges, the this graph must isn't be strongly connected.a simple directed graph was a directed graph having no multiple edges or graph loops.A strongly connected digraph is a directed graph in which it's possible to reach all node starting from a

Question link: Http://acm.hdu.edu.cn/showproblem.php? PID = 1, 4635 Strongly CTED Time Limit: 2000/1000 MS (Java/others) memory limit: 32768/32768 K (Java/Others)Total submission (s): 1568 accepted submission (s): 654 Problem descriptiongive a simple directed graph with n nodes and m edges. please tell me the maximum number of the edges you can add that the graph is still a simple directed graph. also, after you add these edges, this graph must not b

(head));}voidAddedge (intUintv) {edge[tot].to=v; Edge[tot].next=Head[u]; Head[u]=tot++;}voidTarjan (intu) { intv; Low[u]=dfn[u]=++Index; Stack[top++]=u; Instack[u]=true; for(inti=head[u];i!=-1; i=Edge[i].next) {v=edge[i].to; if(!Dfn[v]) {Tarjan (v); if(Low[u]>low[v]) low[u]=Low[v]; } Else if(instack[v]low[u]>Dfn[v]) Low[u]=Dfn[v]; } if(low[u]==Dfn[u]) {SCC++; Do{v=stack[--top]; INSTACK[V]=false; BELONG[V]=SCC; NUM[SCC]++; } while(v!=T); }}voidSolveintN) {memset

/* HDU 4635 stronugly connected: A simple (no duplicate edge, no self-ring: You just direct yourself to your side) Directed Graph. If it is strongly connected, output-1. Otherwise, the output can add up to a maximum number of edges but not strongly connected. The status of the added edge is that there are two strongly connected blocks (including the number of points n, m ), any two points inside each of them have two sides in different directions, and

; SCC++; while(!St.empty ()) {num=St.top (); St.pop (); Id[num]=SCC; SUM[SCC]++; if(num = = u) Break; } }}intMain () {intT,n,m,a,b,ca =0 ; scanf ("%d",t); while(t--) {scanf ("%d%d",n,m); Tot=0; memset (Head,-1,sizeof(head)); for(inti =0; I ) {scanf ("%d%d",a,b); Edge[i].to=b; EDGE[I].NXT=Head[a]; Head[a]=i; } init (); for(inti =1; I ) { if(!Dfn[i]) Tarjan (i); } printf ("Case %d:",++CA); if(SCC = =1) {puts ("-1"); Continue; } for(intU =1; u )

() {intt,cas=0; scanf ("%d",T); while(t--) {printf ("Case %d:",++CAs); scanf ("%d%d",n,m); for(intI=1; ii) {Head[i]=-1;d fn[i]=0; instack[i]=false; Bcc[i]=0; } CLK=cnt=tot=0; for(intI=1; ii) { intu,v; scanf ("%d%d",u,v); Add (U,V); } for(intI=1; ii)if(!Dfn[i]) targin (i); if(cnt==1) {printf ("-1\n");Continue; } for(intI=1; iinch[i]= out[i]=0; for(intI=0; ii) { intu=bel[edge[i].u],v=BEL[EDGE[I].V]; if(U==V)Continue; ++ out[u];++inch[v]; } intans=INF; for

answer. Because if you do not reduce the degree to 0 or the out of 0 related edges, then the point itself contains the in and out edges, plus the edge is always a strong connected graph. So can only be removed with the degree of 0 or out of the relative edge of 0 points, only one direction of the edge, or all (N-minnum) points to the minnum points in the edge, then is minnum points to (N-minnum) points out of the edge.#include Copyright NOTICE: This article for Bo Master original article, with

] = ++index;Stack[++top] = u;Instack[u] =true; for(intJ=head[u]; j!=-1; J=e[j].next){v = e[j].v;if(!dfn[v]){Tarjan (v);Low[u] = min (Low[u], low[v]);}Elseif(Instack[v] = =true)Low[u] = min (Low[u], dfn[v]);}if(Low[u] = = Dfn[u]){++BNT; Do{v = stack[top--];INSTACK[V] =false;BLG[V] = BNT;nblg[bnt]++;} while(U! = v);}}intMain (){intT, t=1;scanf"%d", t); while(t--){intI, J, u, V, N, M;scanf"%d%d", n, m);InIt (N); for(i=0;i{scanf"%d%d", u, v);Addedge (U, v);} for(i=1;i{if(!dfn[i])Tarjan (i);} for(i=

To meet the business needs of an organization, a server is built for data storage. Because the data stream is small, the access database is used as the database. It was okay at first, but later I found a problem with the access database, that is,

Find and contract all SCC, and then select the block with the smallest number of points in the block with the outbound degree 0 or the inbound degree 0 as the independent block, this makes it hard to connect to other blocks, and the rest of the

Tarjan Algorithm After adding an edge, the graph can be divided into two parts X and Y. Only the side from X to Y has no edge from Y to X, so we need to make the number of sides as much as possible, the X part must be a complete graph, and the y

After reading the report, the idea is the same. I just need to paste it. After adding an edge, the graph can be divided into two parts X and Y. Only the side from X to Y has no edge from Y to X, so we need to make the number of sides as much as

The life cycle of an Android application is more clear than the explanation.Create a demo with the following code:public class Mainactivity extends Activity {private static final String TAG = "Android_life"; @Overrideprotected void Oncre Ate (Bundle savedinstancestate) {super.oncreate (savedinstancestate); LOG.I (TAG, "onctreate ()"); Setcontentview (R.layout.activity_main);} protected void OnStart () {Super.onstart (); LOG.I (TAG, "OnStart ()");} protected void Onresume () {super.onresume (); L

Almost all fax machines accept a smaller number of faxes than the original, because the proportion of the reduction is very small, so the impact is very small.This phenomenon is not a hardware failure, but a fax machine to receive faxes, in order to save paper and open the "Auto zoom" function caused by the problem. The paper introduces the reason and solution of the automatic reduction of the fax which is printed when the fax is received by the HP Inkjet All-in-one machine.Cause of failureOne m

Test environment: ubuntuyuntop7.10/HPofficejetProL7580 has a new HPofficejetProL7580 printer. This printer is very good, it is a fax, scanning, copy print all-in-one machine; with automatic paper feed and dual-sided printing unit, built-in Nic, WEB server. The scan function can be implemented through WEB pages. Haha, it's a bit like advertising for HP. It's really a good thing, and it's very fast. First, follow the instructions to start and school Test environment: Ubuntu DeskTop 7.10/HP

Ubuntu 14.04Page http://www.linuxidc.com/Linux/2014-04/100370.htm Install Ubuntu 14.04 permanent update address on a hard disk in Windows 7: Http://www.linuxidc.com/Linux/2014-04/100369.htm In addition to adding support for new printers and systems, some prominent problems have also been solved in this version. Fixes include the HP Officejet 6100 feature to locate problems, prevent interactions in the second Fax queue, and the HP

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Severity: IOException while loading persisted sessions:java.io.EOFExceptionJava.io.EOFExceptionAt Java.io.objectinputstream$peekinputstream.readfully (objectinputstream.java:2303)At Java.io.objectinputstream$blockdatainputstream.readshort (objectinputstream.java:2772)At Java.io.ObjectInputStream.readStreamHeader (objectinputstream.java:778)At Java.io.objectinputstream.At Org.apache.catalina.util.customobjectinputstream.At Org.apache.catalina.session.StandardManager.doLoad (standardmanager.java:3

cmake_install_prefix=/usr/local. Found CMake version is too old, so download from the official website: http://www.cmake.org/download/ I am lazy, download the binary:cmake-3.0.2-linux-i386.tar.gz And then the extracted copy into/usr inside And then you can build it. CMake Error at 3rdparty/ippicv/downloader.cmake:75 (message):icv:failed to download ICV package:ippicv_linux_20140513.tgz.status=6; " Couldn ' t resolve host name ' I can't afford to be here, there's a reply: http://answers.open