COJ966 WZJ data structure (minus 34)

WZJ data structure (minus 34)

Difficulty level: C; operating time limit: 20000ms; operating space limit: 262144KB; code length limit: 2000000B

Question Description

to a tree of n nodes, for the shape of "u r" q query, the answer is centered on the u node, radius r within the node, the maximum weight of the node number is. If there are multiple nodes, returns the smallest number.

Input

There is a set of test data. The first line contains an integer n (1≤n≤10^5) that represents the total number of nodes. The next line, containing n numbers, represents the Weight VI (1≤vi≤10^6) of each node. The next n-1 line, with three integers per line (AI, Bi, WI), represents an edge that connects the AI, the BI node, and the edge length to WI (1≤ai, Bi≤n, 1≤wi≤3). The next line contains an integer q, which indicates the total number of queries (1≤Q≤10^5). The next Q line, each row contains two integers u, r (1≤u≤n, 0≤r≤300), which indicates the number of nodes with the largest weight in the node with the U node as the center and radius R. If there is more than one solution returns the smallest number.

Output

For each set of queries, the output line represents the answer.

Input example

7 1 2 3 4 5 6 7 1 2 1 2 3 1 2 4 1 1 5 1 5 6 1 5 7 1 4 1 1 1 2 2 1 2 2

Output example

5 7 4 5

Other Notes

The example is evil.

Consider using point division to solve the problem offline, then the question is converted to how to solve the overweight x question Max (val[y]|depx+depy<=r).

We can use the usual practice of maintaining DEP, Val simultaneously incrementing the decision sequence, which can be done with a balanced tree, and then swept back and forward.

However, it is not necessary to note that X, y in the same subtrees tree does not affect the answer (think, why), so just offline to construct the decision sequence and then two points on the line.

#include <cstdio>#include<cctype>#include<queue>#include<stack>#include<cstring>#include<algorithm>#defineRep (i,s,t) for (int i=s;i<=t;i++)#defineDwn (i,s,t) for (int i=s;i>=t;i--)#defineren for (int i=first[x];i;i=next[i])using namespaceStd;inlineintRead () {intx=0, f=1;CharC=GetChar ();  for(;! IsDigit (c); C=getchar ())if(c=='-') f=-1;  for(; IsDigit (c); C=getchar ()) x=x*Ten+c-'0'; returnx*F;}Const intmaxn=100010;intfirst[maxn],next[maxn<<1],to[maxn<<1],dis[maxn<<1],e;voidAddedge (intWintVintu) {to[++e]=v;dis[e]=w;next[e]=first[u];first[u]=e; to[++e]=u;dis[e]=w;next[e]=first[v];first[v]=e;}intN,Q,VAL[MAXN],ANS[MAXN];intBetterintXinty) {if(val[x]<val[y]| | (Val[x]==val[y]&&x>y))return 0; return 1;}voidRelaxint& X,intY) {if(Better (y,x)) x=y;}structQuery {intX,r,id,next;} Q[MAXN];intfirst2[maxn],cnt;voidAddQuery (intIdintRintx) {q[++cnt]= (Query) {x,r,id,first2[x]};first2[x]=CNT;}intVis[maxn],f[maxn],s[maxn],size,root;voidGetroot (intXintFA) {S[x]=1;intmaxs=0; renif(to[i]!=fa&&!Vis[to[i]])        {getroot (to[i],x); S[X]+=s[to[i]];maxs=Max (Maxs,s[to[i]]); } F[x]=max (maxs,size-s[x]); if(F[root]>f[x]) root=x;}intTOT,NUM[MAXN],DEP[MAXN],ID[MAXN],A[MAXN],B[MAXN];voidDfsintXintFaintD) {num[++tot]=x;dep[tot]=e; renif(To[i]!=fa&&!vis[to[i]]) DFS (to[i],x,d+dis[i]);}intcmpintXintY) {returndep[x]<dep[y]| | (dep[x]==dep[y]&&val[num[x]]>Val[num[y]]);}voidSolveintx) {vis[x]=1; tot=0;d FS (x,0,0); Rep (I,1, tot) id[i]=i; Sort (ID+1, id+tot+1, CMP); inttmp=tot;tot=0; Rep (I,1, TMP)if(Better (Num[id[i]],a[tot])) a[++tot]=num[id[i]],b[tot]=Dep[id[i]]; Rep (I,1, TMP) for(intj=first2[num[i]];j;j=Q[j].next) {        intL=1, r=tot+1;  while(L +1<r) {intMid=l+r>>1; if(B[mid]<=q[j].r-dep[i]) l=mid; ElseR=mid; }        if(b[l]<=q[j].r-Dep[i]) relax (ans[q[j].id],a[l]); } renif(!Vis[to[i]]) {Size=f[0]=s[to[i]];getroot (to[i],root=0);    Solve (root); }}intMain () {n=read (); Rep (I,1, N) val[i]=read (); Rep (I,2, N)    Addedge (read (), read (), read ()); Q=read (); Rep (I,1, Q)    AddQuery (I,read (), read ()); Size=f[0]=n;getroot (1,0);    Solve (root); Rep (I,1, q) printf ("%d\n", Ans[i]); return 0;}

View Code

COJ966 WZJ data structure (minus 34)