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) {Vis[v]=1; sz[v]=0;intmx=0; for(intI=head[v];i;i=E[i].next) { if(!Vis[e[i].to]) {FINDW (E[i].to,all); MX=_max (mx,sz[e[i].to]); sz[v]+=Sz[e[i].to]; }} Sz[v]++;mx=_max (mx,all-Sz[v]); if(weight[0]GT;MX) weight[0]=mx,weight[1]=v; VIS[V]=0;}intanc[maxn][ -],LEN[MAXN],TO[MAXN],TDIS[MAXN]; LL dis[maxn][ -];voidComp (intVintAncintd) {Vis[v]=1; sz[v]=0; anc[v][++len[v]]=anc;dis[v][len[v]]=D;

Scoi the second part of the history of tears.Multiply + combine linear radicals.Time is at the bottom ...#include #include#include#include#include#defineMAXV 20050#defineMaxe 40050using namespacestd;Long LongN,q,w[maxv],nume=0, g[maxv],x,y,anc[maxv][ -],DIS[MAXV],FATH[MAXV];Long Longregis[ $],table[ $];Long Longk1[ $],k2[ $];structedge{Long LongV,NXT;} E[maxe];struct Base{ Long Longins[ $];} r[maxv][ -];voidgettable () {table[0]=1; for(intI=1; i -

] = true; if (match [u] =-1) break; u = pre [match [u];} while (1) {v = base [v]; if (inpath [v]) return v; v = pre [match [v] ;}} void reset (int u, int anc) {while (u! = Anc) {int v = match [u]; inblossom [base [u] = 1; inblossom [base [v] = 1; v = pre [v]; if (base [v]! = Anc) pre [v] = match [u]; u = v ;}} void contract (int u, int v, int n) {int

Test instructions: Each operation new plus two leaf nodes, each time after the operation to ask the diameter of the tree.Maintain the two endpoints of the tree's diameter u,v, each time calculating the distance from the newly added leaf node to the u,v two point, if there is a longer update.Because the diameter of the tree according to the method, if the new diameter, must be to u or to V distance farthest.1#include 2#include 3#include 4#include 5 6 Const intMAXN =1000000+ +;7 Const intLOGMAXN =

In fact, the whole chain is folded in half to get a tree. Also using a multiplier table, consider maintaining four values: Max,min,premax,submax, which can be solved.You can also use the line segment tree: The principal update and multiplication method are the same, and the same four values are recorded in each interval. Tip: Return a struct directly on return.#include #include #include #include #define MAXN 200005using namespace Std;Long Long n,ma[maxn][30],mi[maxn][30],m,x,y,price[maxn],submax

cancelling headphones sell well, specifically how to realize the process of canceling the noise. To generate sound waves to counteract noise, the first thing to do is to get the noise information. Active Noise cancelling headphones feature a feedback microphone for capturing noise signals in the environment. There is a sequential problem: the noise is collected first, but the noise can be reduced by genera

Copy the code classajaxpage {var $ resuvar $ anchors; var $ total ;... The code is as follows: Class ajaxpage {Var $ result;Var $ anchors;Var $ total;Var $ db;Var $ webpage effect p/jsp.html target = _ blank> jsp tutorial agination;Function _ construct ($ qry, $ starting, $ recpage, $ jspagination){$ This-> db = $ globals ['DB'];$ This-> jspagination = $ jspagination;$ Rst = $ this-> db-> query ($ qry );$ Numrows = $ this-> db-> num_rows;$ Qry. = "limit $ starting, $

The code is as follows Copy Code Class ajaxpage{var $result;var $anchors;var $total;var $db;var $ web Page special effects p/jsp.html target=_blank >jsp tutorial agination;function __construct ($qry, $starting, $recpage, $jspagination){$this->db = $globals [' DB '];$this->jspagination = $jspagination;$rst = $this->db->query ($qry);$numrows = $this->db->num_rows;$qry. = "Limit $starting, $recpage";$this->result = $this->db->get_results ($qry, array_a);$next = $starting + $recpage

I-touristsLCA Nude Questions#include using namespacestd;#definell Long Long#defineLS rt#defineRS Rt#definePB Push_backConst intinf=0x3f3f3f3f;Const intn=2e5+Ten; Vectorint>G[n];vectorEdge[n];intanc[ -][n];intDeep[n];ll H[n];BOOLvis[n]= {false};ints;voidDfsintOintu,ll W) { if(o!=u) deep[u]=deep[o]+1, h[u]=h[o]+W; for(intj=0; J) { if(g[u][j]!=o) {anc[0][g[u][j]]=u; for(intI=1; i -; i++) anc[i][

]=1; for(i=0; iRoot; while(!S.empty ()) { Now=s.top (); S.pop (); if(now!=root) for(i=1; i) {Anc[now][i]=anc[anc[now][i-1]][i-1]; G[now][i]=min (g[now][i-1],g[anc[now][i-1]][i-1]); } for(i=last[now];i;i=e[i].next)if(!Dep[e[i].to]) {anc[e[i].to][0]=N

Lca. 3 points respectively LCA, you can know, depth of the most profound LCA is the answer. #include #include#includeusing namespacestd;Const intMAXN =500000+Ten;Const intMAXM =1000000+Ten;Const intMAXL = A;intG[maxn],v[maxm],next[maxm],eid;intANC[MAXN][MAXL],H[MAXN];intQ[MAXN];inta,b,c,n,m;voidAddedge (intAintb) {V[eid]=b; Next[eid]=g[a]; g[a]=eid++; V[eid]=a; NEXT[EID]=G[B]; g[b]=eid++;}voidbuild () {scanf ("%d%d",n,m); Memset (g,-1,sizeof(g)); for(intI=1, u,v;i) {scanf ("%d%d",u,v); Added

The details are much too big.Multiply.#include #include#include#include#includeSet>#include#include#defineMAXV 200500#defineMAXN 100500#defineEPS 1e-7#defineINF 0X7F7F7F7F7F7F7F7FLLusing namespacestd;Long Longn,m,h[maxn],x,y,aft[maxn][3],anc[maxv][ -],dis1[maxv][ -],dis2[maxv][ -],HASHED[MAXN],POS[MAXN];Long Longret1=0, ret2=0, ans;structpnt{Long LongRet,dot,rets;} p[5];SetLong Long>s;SetLong Long>:: iterator it;Doubleans1=inf;BOOLcmp (PNT x,pnt y) {i