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

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:

• 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.

Output

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

Example

Input:
1

3
1 2 1
2 3 2
QUERY 1 2
CHANGE 1 3
QUERY 1 2
DONE

Output:
1
3

树链剖分第一题。

我觉得树链剖分就是lca+线段树，只是引入了重链等新的概念。因此，学树链剖分前把线段树学好，树链剖分就十分简单了！

#include <iostream>
#include <cstring>
#include <algorithm>
#include <cstdlib>
#include <cstdio>
#include <vector>
#define N 10005
#define pb push_back
using namespace std;
int n,T,num,id[N],siz[N],dep[N],top[N],fa[N],son[N],val[N];
vector<int> v[N];
struct edge
{
	int x,y,v;
}e[N*3];
struct segtree
{
	int l,r,ma;
}a[N*3];
void dfs1(int x,int f,int de)  //第一次dfs,得到depth,size,son,fa
{
	dep[x]=de;
	siz[x]=1;
	son[x]=0;
	fa[x]=f;
	for (int i=0;i<v[x].size();i++)
		if (v[x][i]!=fa[x])
		{
			int u=v[x][i];
			dfs1(u,x,de+1);
			siz[x]+=siz[u];
			if (siz[son[x]]<siz[u])
				son[x]=u;
		}
}
void dfs2(int x,int tp)  //第二次dfs,得到每个点所在重链的深度最小点top[x]
{
	top[x]=tp;
	id[x]=++num;
	if (son[x]) dfs2(son[x],tp);
	for (int i=0;i<v[x].size();i++)
	{
		int u=v[x][i];
		if (u==fa[x]||u==son[x]) continue;
		dfs2(u,u);
	}
}
void push_up(int x)
{
	a[x].ma=max(a[x<<1].ma,a[x<<1|1].ma);
}
void build(int x,int l,int r)
{
	a[x].l=l,a[x].r=r;
	if (l==r)
	{
		a[x].ma=val[l];
		return;
	}
	int m=(l+r)>>1;
	build(x<<1,l,m);
	build(x<<1|1,m+1,r);
	push_up(x);
}
void update(int x,int p,int c)  //修改操作
{
	if (a[x].l==a[x].r)
	{
		a[x].ma=c;
		return;
	}
	int m=(a[x].l+a[x].r)>>1;
	if (p<=m) update(x<<1,p,c);
	else update(x<<1|1,p,c);
	push_up(x);
}
int getmax(int x,int l,int r)
{
	if (a[x].l>=l&&a[x].r<=r) return a[x].ma;
	int m=(a[x].l+a[x].r)>>1;
	int ans=0;
	if (l<=m) ans=getmax(x<<1,l,r);
	if (r>m) ans=max(ans,getmax(x<<1|1,l,r));
	return ans;
}
int query(int u,int v)
{
	int tp1=top[u],tp2=top[v];
	int ans=-1;
	while (tp1!=tp2)
	{
		if (dep[tp1]<dep[tp2]) swap(tp1,tp2),swap(u,v);
		ans=max(getmax(1,id[tp1],id[u]),ans);
		u=fa[tp1];
		tp1=top[u];
	}
	if (u==v) return ans;
	if (dep[u]>dep[v]) swap(u,v);
	ans=max(ans,getmax(1,id[son[u]],id[v]));
	return ans;
}
void clea()   //多组数据
{
	for (int i=1;i<=n;i++)
		v[i].clear();
}
int main()
{
    scanf("%d",&T);
	while (T--)
	{
		scanf("%d",&n);
		for (int i=1;i<n;i++)
		{
			scanf("%d%d%d",&e[i].x,&e[i].y,&e[i].v);
			v[e[i].x].pb(e[i].y);
			v[e[i].y].pb(e[i].x);
		}
		num=0;
		dfs1(1,0,1);
		dfs2(1,1);
		for (int i=1;i<n;i++)
		{
			if (dep[e[i].x]<dep[e[i].y]) swap(e[i].x,e[i].y);
			val[id[e[i].x]]=e[i].v;
		}
		build(1,1,num);
		char s[20];
		while (scanf("%s",s)!=EOF)
		{
			if (s[0]=='D') break;
			int x,y;
			scanf("%d%d",&x,&y);
			if (s[0]=='C')
				update(1,id[e[x].x],y);
			else printf("%d\n",query(x,y));
		}
		clea();
	}
	return 0;
}

【树链剖分模板】【SPOJ 375】 Query on a tree

原文：http://blog.csdn.net/regina8023/article/details/41742787