SPOJ 375 树链剖分

树状结构查询问题

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本文介绍了一种在树形结构中进行高效边权查询的方法，包括树的预处理、线段树构建及更新操作等关键步骤。通过实现特定的数据结构和算法，解决了SPOJ平台上QTREE问题的求解。

SPOJ Problem Set (classical)

375. Query on a tree

Problem code: QTREE

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

边权，单点修改，区间最值。

代码：

/* ***********************************************
Author :xianxingwuguan
Created Time :2014/3/1 10:06:18
File Name :1.cpp
************************************************ */
#pragma comment(linker, "/STACK:102400000,102400000")
#include <stdio.h>
#include <iostream>
#include <algorithm>
#include <sstream>
#include <stdlib.h>
#include <string.h>
#include <limits.h>
#include <string>
#include <time.h>
#include <math.h>
#include <queue>
#include <stack>
#include <set>
#include <map>
using namespace std;
#define INF 0x3f3f3f3f
#define eps 1e-8
#define pi acos(-1.0)
typedef long long ll;
const int maxn=60010;
struct Edge{
	int next,to;
}edge[2*maxn];
int head[maxn],tot;
int top[maxn];
int fa[maxn];
int deep[maxn];
int num[maxn];
int p[maxn],fp[maxn],son[maxn],pos;
void init(){
	tot=0;
	memset(head,-1,sizeof(head));
	pos=0;
	memset(son,-1,sizeof(son));
}
void addedge(int u,int v){
	edge[tot].to=v;
	edge[tot].next=head[u];
	head[u]=tot++;
}
void dfs1(int u,int pre,int d){
	deep[u]=d;
	fa[u]=pre;
	num[u]=1;
	for(int i=head[u];i!=-1;i=edge[i].next){
		int v=edge[i].to;
		if(v==pre)continue;
		dfs1(v,u,d+1);
		num[u]+=num[v];
		if(son[u]==-1||num[v]>num[son[u]])
			son[u]=v;
	}
}
void getpos(int u,int sp){
	top[u]=sp;
	p[u]=pos++;
	fp[p[u]]=u;
	if(son[u]==-1)return;
	getpos(son[u],sp);
	for(int i=head[u];i!=-1;i=edge[i].next){
		int v=edge[i].to;
		if(v!=son[u]&&v!=fa[u])
			getpos(v,v);
	}
}
struct Node{
	int l,r,sum;
}tree[7*maxn];
void build(int i,int l,int r){
	tree[i].l=l;
	tree[i].r=r;
	tree[i].sum=0;
	if(l==r)return;
	int mid=(l+r)/2;
	build(2*i,l,mid);
	build(2*i+1,mid+1,r);
}
void pushup(int i){
	tree[i].sum=max(tree[2*i].sum,tree[2*i+1].sum);
}
void update(int i,int k,int val){
	if(tree[i].l==tree[i].r){
		tree[i].sum=val;
		return;
	}
	int mid=(tree[i].l+tree[i].r)>>1;
	if(k<=mid)update(2*i,k,val);
	else update(2*i+1,k,val);
	pushup(i);
}
int query(int i,int l,int r){
	if(tree[i].l>=l&&tree[i].r<=r)return tree[i].sum;
	int mid=(tree[i].l+tree[i].r)/2;
	int ans=0;
	if(l<=mid)ans=max(ans,query(2*i,l,r));
	if(r>mid)ans=max(ans,query(2*i+1,l,r));
	return ans;
}
int find(int u,int v){
	int f1=top[u],f2=top[v];
	int tmp=0;
	while(f1!=f2){
		if(deep[f1]<deep[f2])
			swap(f1,f2),swap(u,v);
		tmp=max(tmp,query(1,p[f1],p[u]));
		u=fa[f1];f1=top[u];
	}
	if(u==v)return tmp;
	if(deep[u]>deep[v])swap(u,v);
	return max(tmp,query(1,p[son[u]],p[v]));
}
int e[maxn][3];
int main()
{
     //freopen("data.in","r",stdin);
     //freopen("data.out","w",stdout);
 	 int n,m,T;
	 scanf("%d",&T);
	 while(T--){
		 init();
		 scanf("%d",&n);
		 for(int i=0;i<n-1;i++){
			 scanf("%d%d%d",&e[i][0],&e[i][1],&e[i][2]);
			 addedge(e[i][0],e[i][1]);
			 addedge(e[i][1],e[i][0]);
		 }
		 dfs1(1,0,0);
		 getpos(1,1);
		// cout<<"111111"<<endl;
		 build(1,0,pos-1);
		// cout<<"222222"<<endl;
		 for(int i=0;i<n-1;i++){
			 if(deep[e[i][0]]>deep[e[i][1]])
				 swap(e[i][0],e[i][1]);
				 update(1,p[e[i][1]],e[i][2]);
		 }
		 char op[33];
		 while(~scanf("%s",op)){
			 int a,b;
			 if(op[0]=='D')break;
			 scanf("%d%d",&a,&b);
			 if(op[0]=='C')update(1,p[e[a-1][1]],b);
			 else printf("%d\n",find(a,b));
		 }
	 }
     return 0;
}