SPOJ QTREE- Query on a tree (树链剖分)
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
题意：给你一棵树n个点n-1条边，有两个操作，一个操作是把改变某一条边的值，还有一个操作是询问书上两点间的路径中边的最大值。
思路：熟练剖分模板题，wa了很多发，最后发现是线段树写错了= =.
#include<iostream>
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<math.h>
#include<vector>
#include<map>
#include<set>
#include<string>
#include<bitset>
#include<algorithm>
using namespace std;
#define lson th<<1
#define rson th<<1|1
typedef long long ll;
typedef long double ldb;
#define inf 999999999
#define pi acos(-1.0)
#define maxn 10010
struct edge{
int next,to,len;
}e[2*maxn];
int first[maxn];
int top[maxn]; //top[u]表示u所在的链的顶端节点
int son[maxn]; //son[u]表示u的重儿子
int fa[maxn],dep[maxn];//fa[u]表示u的父亲节点，dep[u]表示u在树中的深度
int num[maxn]; //num[u]表示u的子树的节点总个数
int p[maxn]; //p[u]表示u和其父亲节点的连边所在的线段树中区间的位置，这里先把重链都放在一起，然后再处理轻链。
void dfs1(int u,int pre,int deep)
{
int i,j,v;
dep[u]=deep;
num[u]=1;
fa[u]=pre;
for(i=first[u];i!=-1;i=e[i].next)
{
v=e[i].to;
if(v==pre)continue;
dfs1(v,u,deep+1);
num[u]+=num[v];
if(son[u]==-1 || num[son[u] ]<num[v] ){
son[u]=v;
}
}
}
int pos;
void dfs2(int u,int tp)
{
int i,j,v;
top[u]=tp;
if(son[u]!=-1){
p[u]=++pos;
dfs2(son[u],tp);
}
else{
p[u]=++pos;
return;
}
for(i=first[u];i!=-1;i=e[i].next){
v=e[i].to;
if(v==fa[u] || v==son[u] )continue;
dfs2(v,v);
}
}
struct node{
int maxnum;
}b[4*maxn];
void build(int L,int R,int th)
{
int i,j,mid;
b[th].maxnum=0;
if(L==R)return;
mid=(L+R)/2;
build(L,mid,lson);
build(mid+1,R,rson);
}
void update(int idx,int num,int L,int R,int th)
{
int i,j,mid;
if(idx==L && idx==R){
b[th].maxnum=num;
return;
}
mid=(L+R)/2;
if(idx<=mid)update(idx,num,L,mid,lson);
else update(idx,num,mid+1,R,rson);
b[th].maxnum=max(b[lson].maxnum,b[rson].maxnum);
}
int question(int l,int r,int L,int R,int th)
{
int i,j,mid;
if(l==L && r==R){
return b[th].maxnum;
}
mid=(L+R)/2;
if(r<=mid)return question(l,r,L,mid,lson);
else if(l>mid)return question(l,r,mid+1,R,rson);
else{
return max(question(l,mid,L,mid,lson),question(mid+1,r,mid+1,R,rson) );
}
}
int ed[maxn][3];
int fd(int u,int v)
{
int i,j;
int f1,f2;
int num=0;
f1=top[u];f2=top[v];
while(f1!=f2){
if(dep[f1]<dep[f2]){
swap(f1,f2);
swap(u,v);
}
num=max(num,question(p[f1],p[u],1,pos,1) );
u=fa[f1];
f1=top[u];
}
if(u==v)return num;
if(dep[u]<dep[v])swap(u,v);
num=max(num,question(p[son[v] ],p[u],1,pos,1));
return num;
}
int main()
{
int m,i,j,T,c,d,g,tot,f,n;
char s[10];
scanf("%d",&T);
while(T--)
{
memset(first,-1,sizeof(first));
memset(son,-1,sizeof(son));
tot=0;
pos=0;
scanf("%d",&n);
for(i=1;i<=n-1;i++){
scanf("%d%d%d",&c,&d,&g);
ed[i][0]=c;ed[i][1]=d;ed[i][2]=g;
tot++;
e[tot].next=first[c];e[tot].to=d;e[tot].len=g;
first[c]=tot;
tot++;
e[tot].next=first[d];e[tot].to=c;e[tot].len=g;
first[d]=tot;
}
dfs1(1,0,1);
dfs2(1,1);
build(1,pos,1);
for(i=1;i<=n-1;i++){
c=ed[i][0];d=ed[i][1];
if(dep[c ]<dep[d ]){
swap(c,d);
}
update(p[c],ed[i][2],1,pos,1);
}
while(1)
{
scanf("%s",s);
if(s[0]=='D'){
break;
}
scanf("%d%d",&c,&d);
if(s[0]=='C'){
f=ed[c][0];g=ed[c][1];
if(dep[f]<dep[g ]){
swap(f,g);
}
update(p[f],d,1,pos,1);
}
else if(s[0]=='Q'){
printf("%d\n",fd(c,d) );
}
}
}
return 0;
}
