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] = + +K; S.push (RT); for(inti = Pre[rt]; I! =-1; i =Edge[i].next) { intt =edge[i].to; if(!Dfn[t]) {DFS (t); LOW[RT]=min (low[rt],low[t]); } Else if(Vis[t]) {LOW[RT]=min (low[rt],dfn[t]); } } if(Low[rt] = =Dfn[rt]) { ++num; while(!S.empty ()) { intTP =S.top (); S.pop (); VIS[TP]=0; F[TP]=num; if(TP = = RT) Break; } }}intX[maxn],y[maxn],ans,inch[MAXN];intok () {Queueint>Q; for(inti =1; I ){ if(inch[I] = =0) {Q.push (i); }

; A if(G[u][i].second) +VIS[V] =true; the Dfs (v, u); - } $ } the the intMain () the { the intN, M, U, V, art, S, E, Heheda =1; -scanf"%d%d", n, m); in init (); the for(inti =0; I ) { thescanf" %d%d%d", u, v, art); About Add (U, V, art); the Add (V, U, art); the if(Art = =1) theHeheda =0; + } -scanf"%d%d", s, e); the if(Heheda) {Bayiprintf"no\n"); the return 0; the } -Tarjan (1, -1); -End_point =Block[e]; the for(intU =1; U ) { the

Town Field Poem:——— Dream who feel, the water month Build blog. Baiqian tribulation, only know the vicissitudes of the world.——— today holds the Buddhist language, the technology is boundless willing to learn. Willing to do what you learn, cast a conscience blog.——————————————————————————————————————————Step 1Step 2Step 3Press the TAB keyShow——————————————————————————————————————————The essence of the blog, in the technical part, more in the town yard a poem.Python is a good language and deserve

make the Dag strong, the answer to question 2 is how muchWays to add Edges:To add an in edge for each point with a degree of 0, add an edge for each point with a degree of 0Suppose there are n points with an entry degree of 0, and a point with an M-out of 0, how do I add edges?Number of points with all degrees 0 0,1,2,3,4 .... N-1Each time for a number I in the degree of 0 points can reach the out of 0 points, add an out edge, connected to the number of (i+1)%N of the 0 points,This needs to add

similar tasks, and has already started proving some implications. Now I wonder, what many more implications does I have to prove? Can you help me determine this?Inputon the first line one positive number:the number of testcases in most 100. After that per TestCase:* One line containing integers n (1≤n≤20000) and M (0≤m≤50000): The number of statements and the number of IMP Lications that has already been proved.* m lines with-integers s1 and S2 (1≤S1, S2≤n and S1≠S2) each, indicating that it ha

graph strongly. Solving: The degree and the degree of each SCC, and then the number of in and out of the degree of 0 in and out, take in and out of the larger one; because the degree or the degree of 0 proves that the SCC is not connected to another SCC and needs to be connected by an edge, which is the edge to join, Also because an SCC may be connected to more than one SCC, that is, only considering the degree of or only considering the degree of inaccuracy#include 　　Hdoj 3836 equivalent Sets

a line.Sample INPUT4 41 21 31 42 30 0Sample Output0 Test instructions: N points m Edge, asked to create a new edge, you can make the number of bridges to a minimum, the output of the bridge is the least: the number of the bridge to find out all the numbers, the bridge connected to the various graphs, according to the diameter of the tree to find the longest path, the number of bridges minus the diameter of the treeSCC on behalf of the double-unicom component writing habit of the right to write

];BOOLINSTACK[MAXN]; - voidTarjan (intu) { +dfn[u]=low[u]=++DN; -Stack[++top]=u; instack[u]=1; + for(intI=head[u]; i!=-1; I=Edge[i].next) { A intv=edge[i].v; at if(dfn[v]==0){ - Tarjan (v); -low[u]=min (low[u],low[v]); -}Else if(Instack[v]) { -low[u]=min (low[u],dfn[v]); - } in } - if(dfn[u]==Low[u]) { to intV ++Bn; + Do{ -v=stack[top--]; theinstack[v]=0; *Belong[v]=bn; ++Size[bn]; $} while(u!=v);Panax Notoginseng } - } the intD[MAXN]; +

(inti =0; I ) { inscanf"%d%d", u, v); the Add (U, v); the } About for(inti =1; I ) the if(!Dfn[i]) Tarjan (i); the for(intU =1; U ) { the for(inti = Head[u]; ~i; i =edge[i].nxt) { + intv =edge[i].v; - if(Belong[u]! =Belong[v]) { the Add (Belong[u], belong[v]);Bayi } the } the } - intAns =0; - for(inti =1; I ) theAns =Max (ans, DFS (i)); theprintf"%d\n", ans); th

Holy warHoly war, a yarn.Android good or IOS good, Windows good or Mac good, editor good or IDE good, underline good or underline good ...Write code for so many years, you will encounter a lot of meaningless nonsense of the so-called brain residue controversyYesterday @willerce classmate just got into the job to ask a question:缩进用 Tab 还是空格的规范貌似没有啊?Everyone suddenly realized that there was this problem ... At least prove it's not a big problem, or a little problem that hasn't been exposed.Even...

1, Cgrectinsetcgrect cgrectinset (cgrect rect, cgfloat dx, cgfloat dy), the application of this structure is centered on the original Rect, then refer to dx,dy, zoom or enlarge. 2, Cgrectoffsetcgrect cgrectoffset (cgrect rect, cgfloat dx, cgfloat dy); Offsets relative to the source rectangle Origin rect (the point at the upper-left corner) along the X-and Y-axes, and then on a rect basis along the X-and Y -axes3. Specific examples #import"ChangeLocationTextField.h"@implementation Changelocationt

cases. And then followed T cases.The first line for each case contains integers n, m (0 Outputthe output should contain T lines. For each test case, you should just output a integer which is the least number of states the king has to divide into.Sample Input13 21 21 3 Sample Output2Source2011 multi-university Training Contest 3-host by BITRecommendlcyTest instructions: N Cities M-bar has an edge, divides these cities into several states, the principle is (1) you and V can reach each other if bo

=0; for(intI=1;i{Head1[i] = head2[i] =-1;Dfn[i] =0;Fa[i] =0;}}voidTarjan (intU{intVLow[u] = dfn[u] = ++index;Stack[++top] = u; for(intJ=head1[u]; j!=-1; J=e1[j].next){v = e1[j].v;if(E1[j].vis = =false){E1[j].vis = e1[j^1].vis =true; if(!dfn[v]){Tarjan (v);Low[u] = min (Low[u], low[v]);}ElseLow[u] = min (Low[u], dfn[v]);}}if(Low[u] = = Dfn[u]){++BNT; Do{v = stack[top--];BELONG[V] = BNT;} while(U! = v);}}intMain (){intN, M; while(SCANF ("%d%d", n, m), n+m){intI, J, u, V;InIt (N); while(m--){scanf

can reach each other (direct introduction can be), the cost is 0, that is, the cost of a point connection in an SCC is 0. So first the SCC shrinks a new composition and forms a DAG graph (directed acyclic graph). Note: It is not necessary to connect points within an SCC, but it is a cost to connect two SCC, so we need to find the least cost of the entire SCC in each SCC so that we can spend the least amount of time connecting all the SCC. Also note that the 0-point SCC costs 0.#include Copyrig

= Head[i]; J! =-1; j = edge[j].next) {intU = belong[i];intv = belong[edge[j].v];if(U = = v)Continue; Edge Newedge = {u, V, EDGE[J].W,0} ;if(EDGE[J].W returnTMP;}intDfsintU,intPre) {intMi_f = inf, mi_s = INF; for(inti =0; i intv = vec[u][i].v;if(v = = Pre)Continue;intsum = min (vec[u][i].w, DFS (V, u));if(Sum Elsemi_s = min (mi_s, sum); } ans = min (ans, mi_s);returnMi_f;}voidDebug () { for(inti =1; I for(intj =0; J printf("%d%d%d\n", I, VEC[I][J].V, VEC[I][J].W);}intMain () {//freopen ("In.txt

coordinates of the arrival of the minerals owned by the edge of the weight value. The question then becomes: the longest path the minecart can reach from node 0. But in addition to the right and down the edge, considering the transmission point and the target coordinates are composed of edges, the original image will be much more back to the edge, forming a lot of the forward ring. The emergence of a forward ring, so that the mine can be picked up a part of the mineral, as long as it can go int

]=min (low[u],low[v]); }Else if(!Belong[v]) Low[u]=min (low[u],dfn[v]); } if(low[u]==Dfn[u]) {SCC++; while(true){ intx=st[top--]; BELONG[X]=SCC; SIZE[SCC]++; if(X==u) Break; } }}voidSCC () {DFC=scc=0; C (DFN); C (Low); C (belong); C (size); Top=0; for(intI=1; iif(!Dfn[i]) DFS (i);} Edge Es[n];inths[n],cs=0; inlinevoidInssintUintv) {CS++; ES[CS].V=v;es[cs].ne=hs[u];hs[u]=CS;}voidbuildgraph () {CS=0; C (HS); for(intu=1; u){ intA=Belong[u]; for(intI=h[u];i;i=e[i].n

Implement in css The code is as follows:Copy code . Viewbox. content p {Text-indent: 24px;}

; I ) { - if(inch[I] = =0!Dfn[i]) { the Tarjan (i); the } the }94 for(inti =1; I ) { the if(!Dfn[i]) { the Tarjan (i); the }98 } About intsum =0; - for(inti =1; I ) {101 for(intj = Head[i]; ~j; j =Ed[j].next) {102 inty =ed[j].v;103 if(C[i] = = C[y])Continue;104 Addedge1 (C[i], c[y]); thein1[c[y]]++;106 } 107 }108s =C[s];109 for(inti =1; I ) { the if(I! = s in1[i] = =0) {111sum++; the