SPOJ Query on a tree

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本文介绍了一种处理树状结构中边权变更与路径最大边权查询的问题。通过使用线段树和重链剖分等高级数据结构，实现高效地变更边权和查询路径上最大边权的功能。适用于解决涉及大量查询与更新操作的复杂问题。

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

#include<cstdio>
#include<iostream>
#include<cstring>
#include<algorithm>
using namespace std;
#define maxn 10005
int g[maxn];
struct node{
	int to,nxt,vl;
}ed[maxn<<1];
int head[maxn],cnt;
void addedge(int u,int v,int vl){
	ed[cnt].to=v;
	ed[cnt].vl=vl;
	ed[cnt].nxt=head[u];
	head[u]=cnt++;
}

int sz[maxn],top[maxn],son[maxn],fa[maxn],p[maxn],dep[maxn];
int pos;
void dfs1(int u,int f,int d){
	//printf("%d %d %d\n",u,f,d);
	dep[u]=d;	
	sz[u]=1;
	fa[u]=f;
	for(int i=head[u];~i;i=ed[i].nxt){
		int v=ed[i].to;
		if(v==f)continue;
		dfs1(v,u,d+1);
		sz[u]+=sz[v];
		if(son[u]==-1||sz[son[u]]<sz[v])son[u]=v;
	}
}
void dfs2(int u,int tp){
	//printf("%d %d\n",u,tp);
	top[u]=tp;
	p[u]=pos++;
	if(son[u]==-1)return;
	dfs2(son[u],tp);
	for(int i=head[u];~i;i=ed[i].nxt){
		int v=ed[i].to;
		if(v!=son[u]&&v!=fa[u]){
			dfs2(v,v);
		}
	}
}

struct T{
	int l,r,ma;
}stu[maxn*4];
void pushup(int rt){
	if(stu[rt].l==stu[rt].r)return ;
	stu[rt].ma=max(stu[rt<<1].ma,stu[(rt<<1)+1].ma);
	return ;
}
void build(int l,int r,int rt){
	stu[rt].l=l,stu[rt].r=r;
	if(l==r){
		stu[rt].ma=g[l];
		return ;
	}
	int mid=(l+r)>>1;
	build(l,mid,rt<<1);
	build(mid+1,r,(rt<<1)+1);
	pushup(rt);
	return ;
}
void updata(int a,int b,int r){
	if(stu[r].l==stu[r].r)
	{stu[r].ma=b,g[a]=b;
	return ;}
	int mid=(stu[r].l+stu[r].r)>>1;
	if(a>mid)updata(a,b,(r<<1)+1);
	else updata(a,b,r<<1);
	pushup(r);
	return ;
}
int q(int l,int r,int rt){
	if(stu[rt].l==l&&stu[rt].r==r)return stu[rt].ma;
	int mid=(stu[rt].l+stu[rt].r)>>1;
	if(mid>=r)return q(l,r,rt<<1);
	else if(mid<l)return q(l,r,(rt<<1)+1);
	else return max(q(l,mid,rt<<1),q(mid+1,r,(rt<<1)+1));
}
int getma(int u,int v){
	int f1=top[u],f2=top[v];
	int ans=0;
	while(f1!=f2){
		if(dep[f1]<dep[f2]){
			swap(f1,f2);
			swap(u,v);
		}
		ans=max(ans,q(p[f1],p[u],1));
		u=fa[f1];
		f1=top[u];
	}
	if(u==v)return ans;
	if(dep[u]<dep[v])swap(u,v);
	ans=max(ans,q(p[son[v]],p[u],1));
	return ans;
}
int uu[maxn];
void init(){
	memset(head,-1,sizeof(head));
	cnt=0;
	memset(son,-1,sizeof(son));
	pos=0;
}
int main(){
	int t;
	scanf("%d",&t);
	while(t--){
		init();
		int n;
		scanf("%d",&n);
		for(int i=1;i<n;i++){
			int u,v,vl;
			scanf("%d%d%d",&u,&v,&vl);
			addedge(u,v,vl);
			addedge(v,u,vl);
		}
		dfs1(1,0,0);
		dfs2(1,1);
		for(int i=0;i<cnt;i+=2){
			int v=ed[i].to,vl=ed[i].vl;
			int u=ed[i^1].to;
			if(dep[u]<dep[v])swap(u,v);
			g[p[u]]=vl;
			uu[i/2+1]=u;
		}
		build(0,n-1,1);
		char str[10];
		while(~scanf("%s",str)){
			if(str[0]=='D')break;
			int a,b;
			scanf("%d%d",&a,&b);
			if(str[0]=='C'){
				updata(p[uu[a]],b,1);
			}
			else printf("%d\n",getma(a,b));
		}
	}
	return 0;
}