SPOJ375 QTREE

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本文作者：ljh2000
作者博客：http://www.cnblogs.com/ljh2000-jump/
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Description

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


正解：树链剖分
解题报告：
　　链剖裸题，注意清空数组。

//It is made by ljh2000
#include <iostream>
#include <cstdlib>
#include <cstring>
#include <cstdio>
#include <cmath>
#include <algorithm>
#include <ctime>
#include <vector>
#include <queue>
#include <map>
#include <set>
#include <string>
#include <complex>
using namespace std;
#define lc root<<1
#define rc root<<1|1
typedef long long LL;
const int MAXN = 20011;
const int MAXM = 40011;
const int inf = (1<<30);
int n,ecnt,first[MAXN],to[MAXM],next[MAXM],w[MAXM],father[MAXN],quan[MAXN],quanv[MAXN];
int deep[MAXN],id[MAXN],pre[MAXN],son[MAXN],size[MAXN],top[MAXN],match[MAXN],ans,ql,qr,CC;
char ch[12];
struct node{ int maxl; }a[MAXN*3];
inline int getint(){
    int w=0,q=0; char c=getchar(); while((c<'0'||c>'9') && c!='-') c=getchar();
    if(c=='-') q=1,c=getchar(); while (c>='0'&&c<='9') w=w*10+c-'0',c=getchar(); return q?-w:w;
}

inline void dfs(int x,int fa){
	size[x]=1;
	for(int i=first[x];i;i=next[i]) {
		int v=to[i]; if(v==fa) continue;
		father[v]=x; deep[v]=deep[x]+1; quanv[v]=(i+1)>>1; 
		quan[v]=w[i]; match[(i+1)>>1]=v;
		dfs(v,x); size[x]+=size[v];
		if(size[v]>=size[son[x]]) son[x]=v;
	}
}

inline void dfs2(int x,int fa){
	id[x]=++ecnt; pre[ecnt]=x;
	if(son[x]) top[son[x]]=top[x],dfs2(son[x],x);
	for(int i=first[x];i;i=next[i]) {
		int v=to[i]; if(v==fa || v==son[x]) continue;
		top[v]=v; dfs2(v,x);
	}
}

inline void build(int root,int l,int r){
	if(l==r) { a[root].maxl=quan[pre[l]]; return ; }
	int mid=(l+r)>>1; build(lc,l,mid); build(rc,mid+1,r);
	a[root].maxl=max(a[lc].maxl,a[rc].maxl);
}

inline void query(int root,int l,int r){
	if(ql<=l && r<=qr) { ans=max(ans,a[root].maxl); return ; }
	int mid=(l+r)>>1; if(ql<=mid) query(lc,l,mid); if(qr>mid) query(rc,mid+1,r);
}

inline void lca(int x,int y){
	ans=-inf; int f1=top[x],f2=top[y];
	while(f1!=f2) {
		if(deep[f1]<deep[f2]) swap(f1,f2),swap(x,y);
		ql=id[f1]; qr=id[x]; query(1,1,n);
		x=father[f1]; f1=top[x];
	}
	if(deep[x]<deep[y]) swap(x,y);
	ql=id[son[y]]; qr=id[x]; 
	if(ql<=qr) query(1,1,n);
	printf("%d
",ans);
}

inline void update(int root,int l,int r){
	if(l==r) { a[root].maxl=CC; return ; }
	int mid=(l+r)>>1;
	if(ql<=mid) update(lc,l,mid); else update(rc,mid+1,r);
	a[root].maxl=max(a[lc].maxl,a[rc].maxl);
}

inline void work(){
	int T=getint(); int x,y,z;
	while(T--) {
		n=getint(); ecnt=0; memset(first,0,sizeof(first));
		memset(son,0,sizeof(son));
		for(int i=1;i<n;i++) {
			x=getint(); y=getint(); z=getint();
			next[++ecnt]=first[x]; first[x]=ecnt; to[ecnt]=y; w[ecnt]=z;
			next[++ecnt]=first[y]; first[y]=ecnt; to[ecnt]=x; w[ecnt]=z;
		}
		deep[1]=1; dfs(1,0);
		ecnt=0; top[1]=1; dfs2(1,0);
		build(1,1,n);
		while(1) {
			scanf("%s",ch); if(ch[0]=='D') break;
			if(ch[0]=='Q') {
				x=getint(); y=getint();
				lca(x,y);
			}
			else {
				x=getint(); y=getint(); CC=y;
				z=match[x];//边对应的连接的儿子节点
				quan[z]=y; ql=id[z]; update(1,1,n);
			}
		}
	}
}

int main()
{
    work();
    return 0;
}