SPOJ---Query on a tree (树链剖分）

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 

 

题意：输入一个n，给你n-1条边，然后是操作，QUERY a b：查询a，b连点之间边得最大值；CHANGE a b：修改第a条边的值为b：DONE：结束；

这题是树链剖分的模板题，如果树链剖分不懂推荐一篇博客：http://blog.youkuaiyun.com/bobodem/article/details/52330316

刚开始学这个东西是很难理解，不能着急，好好理解每个数组代表的是什么，并且树链剖分会分边权和点权，这题是边权，两个代码有点小差别，这类题没理解透彻很容易写错并且不好找bud，耐心看吧；

AC代码：

 

#include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
#define MAX 10010
using namespace std;
int t,n,tot,pos,head[MAX],
int dep[MAX],top[MAX],son[MAX],sz[MAX],fa[MAX],id[MAX];
//节点在树中的深度，节点的根节点，节点的重儿子，以该节点为根节点的子树的大小，节点的父节点，边在线段树中的位置；
struct Edge//边的结构体
{
    int to;
    int next;
}edge[MAX*2];
struct ee//输入的边
{
    int x;
    int y;
    int val;
}e[MAX];
struct node//线段树
{
    int l;
    int r;
    int val;
}segtree[MAX*4];
void add_edge(int u,int v)
{
    edge[tot].to=v;
    edge[tot].next=head[u];
    head[u]=tot++;
}
void dfs1(int u,int f,int deep)
{
    dep[u]=deep;
    fa[u]=f;
    son[u]=0;
    sz[u]=1;
    for(int i=head[u];i!=-1;i=edge[i].next)//遍历与u相连的所有的点
    {
        int ff=edge[i].to;
        if(ff==f)continue;
        dfs1(ff,u,deep+1);
        sz[u]+=sz[ff];
        if(sz[son[u]]<sz[ff])
            son[u]=ff;
    }
}
void dfs2(int u,int tp)
{
    top[u]=tp;
    id[u]=pos++;
    if(son[u])
        dfs2(son[u],tp);
    for(int i=head[u];i!=-1;i=edge[i].next)
    {
        int ff=edge[i].to;
        if(ff==son[u]||ff==fa[u])continue;
        dfs2(ff,ff);
    }
}

//线段树的操作
void pushup(int root)
{
    segtree[root].val=max(segtree[root<<1].val,segtree[root<<1|1].val);
}
void build(int root,int l,int r)
{
    segtree[root].l=l;
    segtree[root].r=r;
    segtree[root].val=0;
    if(l==r)
        return;
    int mid=(l+r)>>1;
    build(root<<1,l,mid);
    build(root<<1|1,mid+1,r);
}

void update(int root,int p,int val)
{
    int ll=segtree[root].l;
    int rr=segtree[root].r;
    if(ll==rr)
    {
        segtree[root].val=val;
        return;
    }
    int mid=(ll+rr)>>1;
    if(p<=mid) update(root<<1,p,val);
    else update(root<<1|1,p,val);
    pushup(root);
}
int query(int root,int l,int r)
{
    int ll=segtree[root].l;
    int rr=segtree[root].r;
    if(l<=ll&&rr<=r)
        return segtree[root].val;
    int mid=(ll+rr)>>1;
    int ans=-1;
    if(l<=mid)ans=max(ans,query(root<<1,l,r));
    if(r>mid)ans=max(ans,query(root<<1|1,l,r));
    return ans;
}
//
int Yougth(int u,int v)
{
    int tp1=top[u];
    int tp2=top[v];
    int ans=-1;
    while(tp1!=tp2)
    {
        if(dep[tp1]<dep[tp2])
        {
            swap(u,v);
            swap(tp1,tp2);
        }
        ans=max(ans,query(1,id[tp1],id[u]));
        u=fa[tp1];
        tp1=top[u];
    }
    if(u==v)return ans;
    if(dep[u]>dep[v])swap(u,v);
    ans=max(ans,query(1,id[son[u]],id[v]));
    return ans;
}
void init()
{
    tot=0;
    pos=0;
    memset(head,-1,sizeof(head));
}
int main()
{
    scanf("%d",&t);
    while(t--)
    {
        init();
        scanf("%d",&n);
        for(int i=1;i<n;i++)
        {
            scanf("%d%d%d",&e[i].x,&e[i].y,&e[i].val);
            add_edge(e[i].x,e[i].y);
            add_edge(e[i].y,e[i].x);
        }
        dfs1(1,0,0);
        dfs2(1,1);
        build(1,1,pos);
        for(int i=1;i<n;i++)
        {
            if(dep[e[i].x]<dep[e[i].y])
                swap(e[i].x,e[i].y);
            update(1,id[e[i].x],e[i].val);
        }
        char op[30];
        int u,v;
        while(scanf("%s",op)==1)
        {
            if(op[0]=='D')break;
            scanf("%d%d",&u,&v);
            if(op[0]=='Q')printf("%d\n",Yougth(u,v));
            if(op[0]=='C')update(1,id[e[u].x],v);
        }
    }
    return 0;
}