[SPOJ 375 ] Query on a tree

375. 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:
1

3
1 2 1
2 3 2
QUERY 1 2
CHANGE 1 3
QUERY 1 2
DONE

Output:
1
3

树链剖分边权、= =
这题告诉我、宏定义可能导致超时、姿势不对可能导致超时

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <queue>
#include <cmath>
#include <map>
#include <iterator>
#include <cstring>
#include <string>
using namespace std;
#pragma comment(linker, "/STACK:1024000000,1024000000") //手动加栈、windows系统容易爆栈
#define INF 0x7fffffff
#define ll long long
#define N 50005

struct Edge2
{
int a,b,c;
}s[N<<1];

struct Edge
{
int to,next;
}edge[N<<1];
int head
,tot;

int num
;
int pos;
int fa
;
int son
;
int p
;
int fp
;
int deep
;
int size
;
int top
;
int n;

void init()
{
tot=0;
pos=1;
memset(head,-1,sizeof(head));
memset(son,-1,sizeof(son));
}

void add(int x,int y)
{
edge[tot].to=y;
edge[tot].next=head[x];
head[x]=tot++;
}

void dfs1(int now,int pre,int d)
{
deep[now]=d;
fa[now]=pre;
size[now]=1;
for(int i=head[now];i!=-1;i=edge[i].next)
{
int next=edge[i].to;
if(next!=pre)
{
dfs1(next,now,d+1);
size[now]+=size[next];
if(son[now]==-1 || size[next]>size[son[now]])
{
son[now]=next;
}
}
}
}

void dfs2(int now,int tp)
{
top[now]=tp;
p[now]=pos++;
fp[p[now]]=now;
if(son[now]==-1) return;
dfs2(son[now],tp);
for(int i=head[now];i!=-1;i=edge[i].next)
{
int next=edge[i].to;
if(next!=son[now]&&next!=fa[now])
{
dfs2(next,next);
}
}
}

/* 线段树、单点更新 */

int mx[N<<2];

void pushup(int rt)
{
mx[rt]=max(mx[rt<<1],mx[rt<<1|1]);
}

void build(int l,int r,int rt)
{
if(l==r)
{
mx[rt]=num[fp[l]];
return;
}
int m=(l+r)>>1;
build(l,m,rt<<1);
build(m+1,r,rt<<1|1);
pushup(rt);
}

void update(int l,int r,int rt,int pos,int val)
{
if(l==r)
{
mx[rt]=val;
return;
}
int m=(l+r)>>1;
if(pos<=m) update(l,m,rt<<1,pos,val);
else update(m+1,r,rt<<1|1,pos,val);
pushup(rt);
}

int query(int l,int r,int rt,int L,int R)
{

if(L<=l && R>=r)
{
return mx[rt];
}
int m=(l+r)>>1;
int res=-1;
if(L<=m) res=max(res,query(l,m,rt<<1,L,R));
if(R>m) res=max(res,query(m+1,r,rt<<1|1,L,R));
return res;

/*
这样超时、不明就里、看出来的大神请指导。。。
if(l==L && r==R)
{
return mx[rt];
}
int m=(l+r)>>1;
if(L>m) return query(m+1,r,rt<<1|1,L,R);
else if(R<=m) return query(l,m,rt<<1,L,R);
else return max(query(l,m,rt<<1,L,m),query(m+1,r,rt<<1|1,m+1,R));
*/
}

int convert(int pos)
{
int a=s[pos].a;
int b=s[pos].b;
if(deep[a]>deep[b]) return a;
return b;
}

void pre_solve()
{
memset(num,-1,sizeof(num));
for(int i=1;i<n;i++)
{
num[convert(i)]=s[i].c;
}
}

int change(int x,int y)
{
int ans=-INF;
int f1=top[x];
int f2=top[y];
while(f1!=f2)
{
if(deep[f1]<deep[f2])
{
swap(x,y);
swap(f1,f2);
}
ans=max(ans,query(1,n,1,p[f1],p[x]));
x=fa[f1];
f1=top[x];
}
if(deep[x]>deep[y]) swap(x,y);
ans=max(ans,query(1,n,1,p[x]+1,p[y]));
return ans;
}

int main()
{
int T;
scanf("%d",&T);
while(T--)
{
init();
scanf("%d",&n);
for(int i=1;i<n;i++)
{
scanf("%d%d%d",&s[i].a,&s[i].b,&s[i].c);
add(s[i].a,s[i].b);
add(s[i].b,s[i].a);
}
dfs1(1,0,0);
dfs2(1,1);
pre_solve();
build(1,n,1);
while(1)
{
char op[10];
scanf("%s",op);
if(op[0]=='D') break;
else if(op[0]=='C')
{
int a,b;
scanf("%d%d",&a,&b);
a=convert(a);
update(1,n,1,p[a],b);
}
else
{
int a,b;
scanf("%d%d",&a,&b);
printf("%d\n",change(a,b));
}
}
}
return 0;
}