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have such data ...... /* Pku2942 knights of the Round Table */ # Include # Include # Define CLR (a) memset (A, 0, sizeof ()) # Define min (A, B) (a)> (B )? (B) ()) # Define n 1005 # Define M 4000005 Typedef struct NS {Int J; struct NS * Next;} node; Node mem [m]; int memp; Void addedge (node * E [], int I, Int J ){ Node * P = mem [memp ++]; P-> J = J; P-> next = E [I]; E [I] = P; } Int ANC [N], Mark [N], deep [N], stack

Kruskal to reconstruct the tree.#include #include#include#include#defineMAXV 200050#defineMaxe 2000050#defineINF 2147483647using namespacestd;structedges{intU,v,w;} Es[maxe];structedge{intV,NXT;} E[maxe];intn,m,q,h[maxv],hash[maxv],x,y,z,lastans=0, val[maxv],g[maxv],nume=0, FATHER[MAXV];intdis[maxv],l[maxv],r[maxv],fdfn[maxv],anc[maxv][ +],times=0;introot[maxv],tot=0, tots=0, ls[maxv* at],rs[maxv* at],sum[maxv* at],len;BOOLcmp (edges x,edges y) {retur

(indgr[v]==0) Stk.push_back (v); * } $ }Panax Notoginseng } - the intLcaintUintv) { + if(depth[u]Depth[v]) swap (U,V); A intt = depth[u]-Depth[v]; the for(intI=0; T t>>=1, i++ ) + if(t1) u=Anc[u][i]; - if(U==V)returnu; $ for(intP=MAXP; p>=0anc[u][0]!=anc[v][0]; p-- ) $ if(Anc[u][p]!=

[u]; Head[u]= tot++;}/************* LCA calls DFS (1,0,0) and Getanc () ***************/intAnc[n][deg],dep[n];voidDfsintUintFaintd) {Dep[u]=D; anc[u][0]=FA; for(inti=head[u];i!=-1; i=edge[i].next)if(EDGE[I].TO!=FA) DFS (edge[i].to,u,d+1);}voidGetanc () { for(intI=1; i) for(intj=1; j) Anc[j][i]=anc[anc[j][i-1]]

(intx) at { - return(x (-x)); - } - - voidModifyintXintYintDelta) - { in if(xy) Swap (x, y); -X + +; to while(Xlowbit (x); + while(Ylowbit (y); - } the * intQueryintx) $ {Panax Notoginseng intret=0; - while(x) ret^=e[x],x-=lowbit (x); the returnret; + } A the /*initialization of DFS sequence parts and LCA*/ + voidDfsintUintFaintd) - { $dep[u]=D; $anc[u][0]=FA; -start[u]=++CNT; - for(intI=0; I) the if(E[

Social Net ZOJ, socialnetzoj Social Net ZOJ-3649 Question: I cannot understand the original question... Reference: http://www.cnblogs.com/names-yc/p/4922867.html A graph is provided to find the maximum Spanning Tree and the sum of output edge weights, and query the tree: Two node numbers x and y are given, and the path from x to y is obtained, the range in a sequence consisting of the weights of each node-requires that the subtrahend be followed by the subtrahend, that is, the {a1, a2... Aj... A

Ah good to bother with this problem ....Basic ideas are available online.One thing to note is that if the source of the army is not matched at the time of the match, then the source is matched first. (for no cost).But the data good water ....#include #include#include#include#defineMAXV 50050#defineMaxe 100500using namespacestd;intN,x,y,z,g[maxv],nume=0, m,c[maxv],anc[maxv][ -],dis[maxv][ -],l=1, r=0, disr[maxv],kr=0, FATH[MAXV];inttop=0, cnt=0;BOOLVIS

up and down the mismatch edge when the length of the node If violence is good for random data, it will be a special data card. So we can do it with multiplication, so the time complexity of this part is O (q*size*logsize).The total time complexity is O (n^2/size+q*size*logsize), the measured SIZE takes [150,200].#include #include#include#include#include#include#include#defineRep (i,s,t) for (int i=s;i#defineren for (int i=first[x];i;i=next[i])using namespaceStd;inlineintRead () {intx=0, f=1;Cha

1 ,. it is a wildcard that represents any character, for example,. c can match anc, abc, acc; 2. []. You can specify the matching characters in []. For example, a [nbc] c can match anc, abc, and acc; however, ancc cannot be matched. a to z can be written as [a-z], 0 to 9 can be written as [0-9]; 3. Number... syntaxHighl 1, ". "is a wildcard that represents any character, for example,". c can match "

back to D (subtree B is dyed black), the traversal of the query B is associated with a black node 4.In summary, it should be possible to understand the reason why Tarjan offline. It is to store all of the queries first, and then in a DFS process, utilize and check the set (Maintenance gray) and DFS sequence (maintain black and white) "by the way" to find out the LCA.The code is as follows :/* * ID:J.SURE.1 * PROG: * lang:c++ * *#include #include #include #include #include #include #include #inc

matrix (v^3)) In fact, it seems that the time complexity of the flower tree is the same as the time complexity of the Hungarian algorithm (obviously the constant with the flower tree is much larger).Template:#defineMAXN 250#defineSET (A, B) memset (A,b,sizeof (a))dequeint>Q;//g[i][j] Storage diagram: I,j whether there is an edge, Match[i] store I match points//map start initializing G//The final match scheme is match//complexity O (n^3)//the point is from 1 to nBOOLG[MAXN][MAXN],INQUE[MAXN],INB

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