SPOJ375 Query on a tree chain split

SPOJ375 Query on a tree chain split No Tags

You is given a tree (an acyclic undirected connected graph) with N nodes, and edges numbered 1, 2, 3...n-1.

We'll ask you to perfrom some instructions of the following form:

• Change I ti:change the cost of the i-th edge to Ti
  Or
• QUERY a b:ask for the maximum edge cost in the path from Node A to Node B

Input

The first line of input contains an integer t, the number of test cases (T <= 20). T test cases follow.

For each test case:

• The first line there are an integer n (n <= 10000),
• In the next N-1 lines, the i-th line describes the i-th edge:a line with three integers a b c denotes an edge between a, B of cost C (C <= 1000000),
• The next lines contain instructions "Change I ti" or "QUERY a B",
• The end of each test case was signified by the string "done".

There is one blank line between successive tests.

Output

For each "QUERY" operation, the write one integer representing its result.

Example

input:131 2 3 2QUERY 1 2CHANGE 1 3QUERY 1 2doneoutput:13
Test instructions: to a tree, n points, n-1 edge, with edge right, multiple operations, there are two kinds of operation, change the Benquan of an edge, or ask two points on the path of the maximum edge right.
First tree chain split.
Tree chain is to split the tree into a number of chains, end-to-end, and then use data structure (line tree, etc.).

#include <bits/stdc++.h>#defineLson l,m,rt<<1#defineRson m+1,r,rt<<1|1using namespacestd;Const intmaxn=1000100;Const intInf= (1<< in);intN;Charop[ -];intU,v,w;vector<int>G[MAXN];intDEP[MAXN],SON[MAXN],FA[MAXN],SIZ[MAXN];intTOP[MAXN];intID[MAXN];intnum;intVAL[MAXN];structtree{intU,v,val;}; Tree E[MAXN];structsegtree{intVal;}; Segtree T[MAXN<<2];voidDFS1 (intUintFintd) {Fa[u]=F; Dep[u]=D; Son[u]=0; Siz[u]=1;  for(intI=0; I<g[u].size (); i++){        intv=G[u][i]; if(v==f)Continue; DFS1 (V,u,d+1); if(Siz[v]>siz[son[u]]) son[u]=v; Siz[u]+=Siz[v]; }}voidDFS2 (intUintTP) {Top[u]=TP; Id[u]=++num; if(Son[u]) DFS2 (SON[U],TP);  for(intI=0; I<g[u].size (); i++){        intv=G[u][i]; if(v==fa[u]| | V==son[u])Continue;    DFS2 (V,V); }}voidPUSH_UP (intRT) {T[rt].val=max (t[rt<<1].val,t[rt<<1|1].val);}voidBuildintLintRintRT) {    if(l==S) {T[rt].val=Val[l]; return; }    intM= (l+r) >>1;    Build (Lson);    Build (Rson); PUSH_UP (RT);}voidUpdateintPintCintLintRintRT) {    if(l==R) {T[rt].val=C; return; }    intM= (l+r) >>1; if(p<=m) update (P,c,lson); ElseUpdate (P,c,rson); PUSH_UP (RT);}intQueryintLintRintLintRintRT) {    if(L&LT;=L&AMP;&AMP;R&LT;=R)returnT[rt].val; intM= (l+r) >>1; intres=-INF; if(l<=m) res=Max (Res,query (L,r,lson)); if(r>m) res=Max (Res,query (L,r,rson)); returnRes;}intChaintUintv) {    intres=-INF;  while(top[u]!=Top[v]) {        if(dep[top[u]]<Dep[top[v]) swap (U,V); Res=max (Res,query (Id[top[u]],id[u),1Num1)); U=Fa[top[u]]; }    if(u!=v) {        if(dep[u]>Dep[v]) swap (U,V); Res=max (Res,query (ID[SON[U]],ID[V),1Num1)); }    returnRes;}intMain () {//freopen ("In.txt", "R", stdin);    intCasen; CIN>>Casen;  while(casen--) {cin>>O;  for(intI=1; i<=n;i++) g[i].clear ();  for(intI=1; i<n;i++) {scanf ("%d%d%d",&u,&v,&W); E[i]={u,v,w};            G[u].push_back (v);        G[v].push_back (U); } num=0; DFS1 (1,0,1); DFS2 (1,1);  for(intI=1; i<n;i++){            if(dep[e[i].u]>DEP[E[I].V]) swap (E[I].U,E[I].V); VAL[ID[E[I].V]]=E[i].val; } Build (1Num1);  while(~SCANF ("%s", op)) {            if(op[0]=='D') Break; if(op[0]=='C') {scanf ("%d%d",&u,&W); intp=ID[E[U].V]; Update (P,W,1Num1); }            Else{scanf ("%d%d",&u,&v); printf ("%d\n", Cha (u,v)); }        }    }    return 0;}

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SPOJ375 Query on a tree chain split