[SPOJ 375]Query on a tree(树剖模板)

本文介绍了一种基于树状结构的数据处理算法,通过构建树形结构并利用线段树进行高效的边权更新与路径最大边权查询。该算法适用于解决特定类型的图论问题,例如在给定的树上执行边权变更及查询两个节点间路径上的最大边权。

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Description


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

Solution


存一下模板QvQ
代码能力太渣。。

#include<iostream>
#include<cstring>
#include<cstdio>
#define Max(a,b) (a>b?a:b)
#define MAXN 10005
using namespace std;
int T,n;
int head[MAXN],cnt,e[MAXN][3],sz;
int top[MAXN],siz[MAXN],dep[MAXN],father[MAXN],ans,pos[MAXN];
char opt[15];
struct Node{
    int next,to;
}Edges[MAXN*2];
void addedge(int u,int v)
{
    Edges[++cnt].next=head[u];
    head[u]=cnt;
    Edges[cnt].to=v;
}
void dfs1(int u)
{
    siz[u]=1;
    for(int i=head[u];~i;i=Edges[i].next)
    {
        int v=Edges[i].to;
        if(v!=father[u])
        {
            father[v]=u;
            dep[v]=dep[u]+1;
            dfs1(v);
            siz[u]+=siz[v];
        }
    }
}
void dfs2(int u,int t)
{
    int k,maxsiz=0;
    sz++;pos[u]=sz;top[u]=t;
    for(int i=head[u];~i;i=Edges[i].next)
    {
        int v=Edges[i].to;
        if(v!=father[u]&&siz[v]>maxsiz)
        k=v,maxsiz=siz[v];
    }
    if(!maxsiz)return;
    dfs2(k,t);
    for(int i=head[u];~i;i=Edges[i].next)
    {
        int v=Edges[i].to;
        if(v!=k&&v!=father[u])
        dfs2(v,v);
    }
}
struct Segtree{
    int l,r,max;
}t[MAXN*4];
void build(int idx,int l,int r)
{
    t[idx].l=l,t[idx].r=r;
    if(l==r)return;
    int mid=(l+r)/2;
    build(idx*2,l,mid);
    build(idx*2+1,mid+1,r);
}
void change(int idx,int a,int b)
{
    if(t[idx].l==t[idx].r)
    {
        t[idx].max=b;
        return;
    }
    int mid=(t[idx].l+t[idx].r)/2;
    if(a<=mid)change(idx*2,a,b);
    else change(idx*2+1,a,b);
    t[idx].max=Max(t[idx*2].max,t[idx*2+1].max);
}
void query(int idx,int a,int b)
{
    if(t[idx].l>=a&&t[idx].r<=b)
    {
        ans=Max(ans,t[idx].max);
        return;
    }
    int mid=(t[idx].l+t[idx].r)/2;
    if(a>mid)query(idx*2+1,a,b);
    else if(b<=mid)query(idx*2,a,b);
    else
    {
        query(idx*2,a,b);
        query(idx*2+1,a,b);
    }
}
int main()
{
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d",&n);
        cnt=sz=0;
        memset(head,-1,sizeof(head));
        for(int i=1;i<n;i++)
        {
            int a,b,c;
            scanf("%d%d%d",&a,&b,&c);
            addedge(a,b);
            addedge(b,a);
            e[i][0]=a,e[i][1]=b,e[i][2]=c;
        }
        dfs1(1);dfs2(1,1);
        build(1,1,sz);
        for(int i=1;i<n;i++)
        {
            if(dep[e[i][0]]<dep[e[i][1]])swap(e[i][0],e[i][1]);
            change(1,pos[e[i][0]],e[i][2]);
        }
        for(;;)
        {
            int a,b;
            scanf("%s",opt);
            if(opt[0]=='D')break;
            if(opt[0]=='C')
            {
                scanf("%d%d",&a,&b);
                if(dep[e[a][0]]<dep[e[a][1]])swap(e[a][0],e[a][1]);
                change(1,pos[e[a][0]],b);
            }
            else if(opt[0]=='Q')
            {
                ans=0;
                scanf("%d%d",&a,&b);
                while(top[a]!=top[b])
                {
                    if(dep[top[a]]<dep[top[b]])swap(a,b);
                    query(1,pos[top[a]],pos[a]);
                    a=father[top[a]];
                }
                if(dep[a]>dep[b])swap(a,b);
                if(a!=b)query(1,pos[a]+1,pos[b]);
                printf("%d\n",ans);
            }
        }
    }
    return 0;
 } 
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