SPOJ375 Query on a tree 树链剖分

SPOJ375  Query on a tree   树链剖分

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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：给一颗树，n个点，n-1条边，带边权，多次操作，操作有两种，改变某条边的边权，或者询问两个点的路径上的最大边权。
第一道树链剖分。
树链剖分就是把树拆成若干条链，首尾相接，然后用数据结构搞(线段树等).

#include<bits/stdc++.h>
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1

using namespace std;

const int maxn=1000100;
const int INF=(1<<29);

int n;
char op[20];
int u,v,w;
vector<int> G[maxn];
int dep[maxn],son[maxn],fa[maxn],siz[maxn];
int top[maxn];
int id[maxn];
int num;
int val[maxn];
struct Tree
{
    int u,v,val;
};
Tree e[maxn];
struct SegTree
{
    int val;
};
SegTree T[maxn<<2];

void dfs1(int u,int f,int d)
{
    fa[u]=f;
    dep[u]=d;
    son[u]=0;
    siz[u]=1;
    for(int i=0;i<G[u].size();i++){
        int v=G[u][i];
        if(v==f) continue;
        dfs1(v,u,d+1);
        if(siz[v]>siz[son[u]]) son[u]=v;
        siz[u]+=siz[v];
    }
}

void dfs2(int u,int tp)
{
    top[u]=tp;
    id[u]=++num;
    if(son[u]) dfs2(son[u],tp);
    for(int i=0;i<G[u].size();i++){
        int v=G[u][i];
        if(v==fa[u]||v==son[u]) continue;
        dfs2(v,v);
    }
}

void push_up(int rt)
{
    T[rt].val=max(T[rt<<1].val,T[rt<<1|1].val);
}

void build(int l,int r,int rt)
{
    if(l==r){
        T[rt].val=val[l];
        return;
    }
    int m=(l+r)>>1;
    build(lson);
    build(rson);
    push_up(rt);
}

void update(int p,int c,int l,int r,int rt)
{
    if(l==r){
        T[rt].val=c;
        return;
    }
    int m=(l+r)>>1;
    if(p<=m) update(p,c,lson);
    else update(p,c,rson);
    push_up(rt);
}

int query(int L,int R,int l,int r,int rt)
{
    if(L<=l&&r<=R) return T[rt].val;
    int m=(l+r)>>1;
    int res=-INF;
    if(L<=m) res=max(res,query(L,R,lson));
    if(R>m) res=max(res,query(L,R,rson));
    return res;
}

int cha(int u,int v)
{
    int res=-INF;
    while(top[u]!=top[v]){
        if(dep[top[u]]<dep[top[v]]) swap(u,v);
        res=max(res,query(id[top[u]],id[u],1,num,1));
        u=fa[top[u]];
    }
    if(u!=v){
        if(dep[u]>dep[v]) swap(u,v);
        res=max(res,query(id[son[u]],id[v],1,num,1));
    }
    return res;
}

int main()
{
    //freopen("in.txt","r",stdin);
    int casen;
    cin>>casen;
    while(casen--){
        cin>>n;
        for(int i=1;i<=n;i++) G[i].clear();
        for(int i=1;i<n;i++){
            scanf("%d%d%d",&u,&v,&w);
            e[i]={u,v,w};
            G[u].push_back(v);
            G[v].push_back(u);
        }
        num=0;
        dfs1(1,0,1);
        dfs2(1,1);
        for(int i=1;i<n;i++){
            if(dep[e[i].u]>dep[e[i].v]) swap(e[i].u,e[i].v);
            val[id[e[i].v]]=e[i].val;
        }
        build(1,num,1);
        while(~scanf("%s",op)){
            if(op[0]=='D') break;
            if(op[0]=='C'){
                scanf("%d%d",&u,&w);
                int p=id[e[u].v];
                update(p,w,1,num,1);
            }
            else{
                scanf("%d%d",&u,&v);
                printf("%d\n",cha(u,v));
            }
        }
    }
    return 0;
}