You are given a tree (an acyclic undirected connected graph) with N nodes, and edges numbered 1, 2, 3...N-1.
We will 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 on the path from node a to node b
The first line of input contains an integer t, the number of test cases (t <= 20). t test cases follow.
For each test case:
In the first line there is 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 is signified by the string "DONE".
There is one blank line between successive tests.
For each "QUERY" operation, write one integer representing its result.
1
3
1 2 1
2 3 2
QUERY 1 2
CHANGE 1 3
QUERY 1 2
DONE
1
3
树链剖分裸体
md 我一开始看成了加和 打了树上差分 过了样例
被cxy一语点醒
顿时 世界都黑了 mmpmmp
#include <bits/stdc++.h>
using namespace std;
#define maxn (int)(1e5+10)
#define LL long long
pair<int,int>mp[maxn];
inline int read(){
int rtn=0,f=1;char ch=getchar();
while(!isdigit(ch)){if(ch=='-')f=-1;ch=getchar();}
while(isdigit(ch))rtn=(rtn<<1)+(rtn<<3)+ch-'0',ch=getchar();
return rtn*f;
}
int cnt,n,coc,m,p[maxn],size[maxn],son[maxn],dfn[maxn],top[maxn],dep[maxn],id[maxn],fa[maxn],w[maxn];
struct node{
int a,b,nt,w;
}e[maxn<<1];
inline void add(int x,int y,int z){
e[++cnt].a=x;e[cnt].b=y;e[cnt].w=z;
e[cnt].nt=p[x];p[x]=cnt;
}
inline void dfs1(int k){
size[k]=1;
for(int i=p[k];i;i=e[i].nt){
int kk=e[i].b;
if(kk==fa[k])continue;
fa[kk]=k;
dep[kk]=dep[k]+1;
w[kk]=e[i].w;
dfs1(kk);
size[k]+=size[kk];
if(size[kk]>size[son[k]])son[k]=kk;
}
}
inline void dfs2(int x,int y){
top[x]=y;dfn[x]=++coc;id[coc]=x;
if(!son[x])return;
dfs2(son[x],y);
for(int i=p[x];i;i=e[i].nt){
int k=e[i].b;
if(k==fa[x]||k==son[x])continue;
dfs2(k,k);
}
}
namespace Link_Chain_SegmentTree{
LL maxv[maxn<<3];
inline void build(int p,int l,int r){
if(l==r)return (void)(maxv[p]=w[id[l]]);
int mid=l+r>>1;
build(p<<1,l,mid);
build(p<<1|1,mid+1,r);
maxv[p]=max(maxv[p<<1],maxv[p<<1|1]);
}
inline LL query(int p,int lp,int rp,int l,int r){
if(l>r)return 0;
if(l==lp&&r==rp)return maxv[p];
int mid=lp+rp>>1;
if(r<=mid)return query(p<<1,lp,mid,l,r);
else if(l>mid)return query(p<<1|1,mid+1,rp,l,r);
else return max(query(p<<1,lp,mid,l,mid),query(p<<1|1,mid+1,rp,mid+1,r));
}
inline LL query_chain(int x,int y){
LL rtn=0;
while(top[x]!=top[y]){
if(dep[top[x]]<dep[top[y]])swap(x,y);
rtn=max(rtn,query(1,1,n,dfn[top[x]],dfn[x]));
x=fa[top[x]];
}if(dep[x]>dep[y])swap(x,y);
return max(rtn,query(1,1,n,dfn[x]+1,dfn[y]));
}
inline void update(int p,int lp,int rp,int pos,LL val){
if(lp>rp)return;
if(lp==rp)return (void)(maxv[p]=val);
int mid=lp+rp>>1;
if(pos<=mid)update(p<<1,lp,mid,pos,val);
else update(p<<1|1,mid+1,rp,pos,val);
maxv[p]=max(maxv[p<<1],maxv[p<<1|1]);
}
inline void update_chain(int id, LL val){
int from=mp[id].first,to=mp[id].second;
if(dep[from]>dep[to])update(1,1,n,dfn[from],val);
else update(1,1,n,dfn[to],val);
}
}using namespace Link_Chain_SegmentTree;
int main(){
int T;scanf("%d",&T);
while(T--){
n;scanf("%d",&n);coc=0;
memset(p,0,sizeof(p));
memset(fa,0,sizeof(fa));
memset(son,0,sizeof(son));
for(int i=1;i<n;i++){
mp[i].first=read();mp[i].second=read();int w=read();
add(mp[i].first,mp[i].second,w);
add(mp[i].second,mp[i].first,w);
}
dfs1(1);dfs2(1,1) ;
build(1,1,n);
while(true){
char ch[10];scanf("%s",ch);
if(ch[0]=='D')break;
int x=read(),y=read();
if(ch[0]=='Q')printf("%lld\n",query_chain(x,y));
if(ch[0]=='C')update_chain(x,y);
}
}
return 0;
}
/*
1
10
2 1 6824
3 1 21321
4 2 26758
5 1 13610
6 4 19133
7 4 20483
8 7 10438
9 8 19157
10 6 25677
C 2 11799
Q 5 6
Q 6 10
Q 3 1
C 5 9242
C 3 15761
C 2 28270
C 8 8177
C 5 21007
Q 4 8
D
*/ 原文：http://www.cnblogs.com/DexterYsw/p/7942151.html