SPOJ QTREE Query on a tree

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:131 2 12 3 2QUERY 1 2CHANGE 1 3QUERY 1 2DONEOutput:13

说多皆泪。

树链剖分的操作对象是点，所以可以把对边的询问和修改映射到深度大的点上，注意询问的时候不能去lca。

然而可能是昨天晚上太晚了，各种出错，今天的迷之调试，根本查不出来，窝的青春啊。

#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;
const int N=10005;
const int inf=1e9+7;
int T,n,cnt,dcnt,hd[N],c[N],mp[N];
struct edge
{
	int to,nxt,val;
}v[2*N];
struct node
{
	int sz,fa,son,dep,tp,w;
}tr[N];
struct segtree
{
	int l,r,mx;
}st[4*N];
void addedge(int x,int y,int z)
{
	v[++cnt].to=y;
	v[cnt].val=z;
	v[cnt].nxt=hd[x];
	hd[x]=cnt;
}
void dfs1(int u)
{
	tr[u].sz=1;
	tr[u].son=0;
	for(int i=hd[u];i;i=v[i].nxt)
		if(v[i].to!=tr[u].fa)
		{
			tr[v[i].to].fa=u;
			tr[v[i].to].dep=tr[u].dep+1;
			dfs1(v[i].to);
			tr[u].sz+=tr[v[i].to].sz;
			if(tr[v[i].to].sz>tr[tr[u].son].sz)
				tr[u].son=v[i].to;
		}
}
void dfs2(int u,int top)
{
	tr[u].tp=top;
	tr[u].w=++dcnt;
	mp[dcnt]=u;
	if(tr[u].son)
	{
		dfs2(tr[u].son,top);
		for(int i=hd[u];i;i=v[i].nxt)
			if(v[i].to!=tr[u].fa&&v[i].to!=tr[u].son)
				dfs2(v[i].to,v[i].to);
	}
}
void pushup(int num)
{
	st[num].mx=max(st[2*num].mx,st[2*num+1].mx);
}
void build(int num,int l,int r)
{
	st[num].l=l,st[num].r=r;
	if(l==r)
	{
		st[num].mx=c[mp[l]];
		return ;
	}
	int mid=(l+r)/2;
	build(2*num,l,mid),build(2*num+1,mid+1,r);
	pushup(num);
}
void change(int num,int x,int y)//单点修改 
{
	if(st[num].l>x||st[num].r<x)
		return ;
	if(st[num].l==st[num].r)
	{
		if(st[num].l==x)
			st[num].mx=y;
		return ;
	}
	change(2*num,x,y),change(2*num+1,x,y);
	pushup(num);
}
int query(int num,int x,int y)//询问区间最大值 
{
	if(st[num].l>y||st[num].r<x)
		return -inf;
	if(st[num].l>=x&&st[num].r<=y)
		return st[num].mx;
	return max(query(2*num,x,y),query(2*num+1,x,y));
}
int ask(int x,int y)
{
	int ans=-inf;
	while(tr[x].tp!=tr[y].tp)
	{
		if(tr[tr[x].tp].dep>tr[tr[y].tp].dep)
		{
			ans=max(ans,query(1,tr[tr[x].tp].w,tr[x].w));
			x=tr[tr[x].tp].fa;
		}
		else
		{
			ans=max(ans,query(1,tr[tr[y].tp].w,tr[y].w));
			y=tr[tr[y].tp].fa;
		}
	}
	if(tr[x].dep<tr[y].dep)
		ans=max(ans,query(1,tr[tr[x].son].w,tr[y].w));
	else if(tr[x].dep>tr[y].dep)
		ans=max(ans,query(1,tr[tr[y].son].w,tr[x].w));
	return ans;
}
int main()
{
	scanf("%d",&T);
	while(T--)
	{
		cnt=dcnt=0;
		memset(hd,0,sizeof(hd));
		memset(c,0,sizeof(c));
		memset(mp,0,sizeof(mp));
		memset(tr,0,sizeof(tr));
		memset(st,0,sizeof(st));
		scanf("%d",&n);
		for(int i=1;i<=n-1;i++)
		{
			int x,y,z;
			scanf("%d%d%d",&x,&y,&z);
			addedge(x,y,z),addedge(y,x,z);
		}
		dfs1(1);
		dfs2(1,1);
		for(int i=1;i<=n-1;i++)
			if(tr[v[2*i-1].to].dep>tr[v[2*i].to].dep)
				c[v[2*i-1].to]=v[2*i].val;
			else
				c[v[2*i].to]=v[2*i].val;
		build(1,1,n);
		char ch[11];
		int x,y,z;
		while(1)
		{
			scanf("%s",ch);
			if(ch[0]=='C')
			{
				scanf("%d%d",&x,&z);
				if(tr[v[2*x].to].dep>tr[v[2*x-1].to].dep)
					change(1,tr[v[2*x].to].w,z);
				else
					change(1,tr[v[2*x-1].to].w,z);
			}
			else if(ch[0]=='Q')
			{
				scanf("%d%d",&x,&y);
				printf("%d\n",ask(x,y));
			}
			else
				break;
		}
	}
	return 0;
}