spoj 375 树链剖分 模板

QTREE - Query on a tree

#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

Submit solution!

题意：

一棵树有修改边权值操作和询问两个节点之间的最大边权值操作

代码：

代码：

//每个点和他父节点的边构成一个线段树上的点。所以线段树的点实际从2开始
#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
const int MAXN=;
int id[MAXN+],fa[MAXN+],max_val[MAXN*+],head[MAXN+],son[MAXN+],top[MAXN+],lev[MAXN+],size[MAXN+];
//id:对应到线段树上的点编号，son:重儿子，top:重链的头，lev:深度，size:子树大小
int tot,cnt,val[MAXN+]; //cnt:线段树节点数
struct Edge
{
    int u,v,w,next;
}edge[MAXN*+];
void init()
{
    for(int i=;i<=MAXN;i++) fa[i]=top[i]=i;
    memset(size,,sizeof(size));
    memset(head,-,sizeof(head));
    memset(val,,sizeof(val));
    tot=cnt=;
}
void add(int x,int y,int z)
{
    edge[tot].u=x;edge[tot].v=y;edge[tot].w=z;
    edge[tot].next=head[x];
    head[x]=tot++;
    edge[tot].u=y;edge[tot].v=x;edge[tot].w=z;
    edge[tot].next=head[y];
    head[y]=tot++;
}
void dfs1(int x,int d)
{
    lev[x]=d;
    son[x]=;
    size[x]=;
    for(int i=head[x];i!=-;i=edge[i].next){
        int y=edge[i].v;
        if(y==fa[x]) continue;
        fa[y]=x;
        dfs1(y,d+);
        size[x]+=size[y];
        if(size[son[x]]<size[y]) son[x]=y;
    }
}
void dfs2(int x,int tp)
{
    top[x]=tp;
    id[x]=++cnt;
    if(son[x]) dfs2(son[x],tp);
    for(int i=head[x];i!=-;i=edge[i].next){
        int y=edge[i].v;
        if(y==fa[x]||y==son[x]) continue;
        dfs2(y,y);
    }
}
void pushup(int rt) { max_val[rt]=max(max_val[rt<<],max_val[rt<<|]); }
void build(int l,int r,int rt)
{
    if(l==r) { max_val[rt]=val[l];return; }
    int mid=(l+r)>>;
    build(l,mid,rt<<);
    build(mid+,r,rt<<|);
    pushup(rt);
}
void update(int id,int c,int l,int r,int rt)
{
    if(l==r){
        max_val[rt]=c;
        return;
    }
    int mid=(l+r)>>;
    if(id<=mid) update(id,c,l,mid,rt<<);
    else update(id,c,mid+,r,rt<<|);
    pushup(rt);
}
int query(int ql,int qr,int l,int r,int rt)
{
    if(ql<=l&&qr>=r) return max_val[rt];
    int mid=(l+r)>>,ans=;
    if(ql<=mid) ans=max(ans,query(ql,qr,l,mid,rt<<));
    if(qr>mid) ans=max(ans,query(ql,qr,mid+,r,rt<<|));
    return ans;
}
int solve(int l,int r)
{
    int ltp=top[l],rtp=top[r],ans=;
    while(ltp!=rtp){
        if(lev[rtp]<lev[ltp]){
            swap(ltp,rtp);
            swap(l,r);
        }
        ans=max(ans,query(id[rtp],id[r],,cnt,));
        r=fa[rtp];
        rtp=top[r];
    }
    if(lev[r]>lev[l]) swap(r,l);
    if(l!=r) ans=max(ans,query(id[son[r]],id[l],,cnt,));
    return ans;
}
int main()
{
    //freopen("in.txt","r",stdin);
    int t,n;
    scanf("%d",&t);
    while(t--){
        init();
        scanf("%d",&n);
        for(int i=;i<n;i++){
            int x,y,z;
            scanf("%d%d%d",&x,&y,&z);
            add(x,y,z);
        }
        dfs1(,);
        dfs2(,);
        for(int i=;i<tot;i+=){
            if(lev[edge[i].u]>lev[edge[i].v]) swap(edge[i].u,edge[i].v);
            val[id[edge[i].v]]=edge[i].w;
        }
        build(,cnt,);
        char ch[];
        while(scanf("%s",ch)&&ch[]!='D'){
            int x,y;
            scanf("%d%d",&x,&y);
            if(ch[]=='C') update(id[edge[x*-].v],y,,cnt,);
            else printf("%d\n",solve(x,y));
        }
    }
    return ;
}