hp 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

; 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=

(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

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