【SPOJ】QTREE 1 树链剖分裸题

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

树链剖分裸题……边权赋给点上，注意边界。

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstring>
using namespace std;
const int size=200010;

int head[size],nxt[size],tot=0;

struct edge{
    int t,d;
}l[size];

void build(int f,int t,int d)
{
    l[++tot].t=t;
    l[tot].d=d;
    nxt[tot]=head[f];
    head[f]=tot;
}

int deep[size],top[size],inseg[size];
int fa[size],son[size],sz[size];

void dfs_1(int u,int f)
{
    fa[u]=f;
    deep[u]=deep[f]+1;
    sz[u]=1;
    for(int i=head[u];i;i=nxt[i])
    {
        int v=l[i].t;
        if(v==f) continue;
        dfs_1(v,u);
        sz[u]+=sz[v];
        if(!son[u]||sz[v]>sz[son[u]]) son[u]=v;
    }
}

int totp=0;
void dfs_2(int u,int topu)
{
    top[u]=topu;
    inseg[u]=++totp;
    if(!son[u]) return ;
    dfs_2(son[u],topu);
    for(int i=head[u];i;i=nxt[i])
    {
        int v=l[i].t;
        if(v==fa[u]||v==son[u]) continue;
        dfs_2(v,v);
    }
}

struct segment{
    int l,r;
    int maxx;
}tree[size*4];

void update(int p)
{
    tree[p].maxx=max(tree[p<<1].maxx,tree[p<<1|1].maxx);
}

void build_tree(int p,int l,int r)
{
    tree[p].l=l;    tree[p].r=r;
    if(l==r)
    {
        tree[p].maxx=0;
        return ;
    }
    int mid=(l+r)>>1;
    build_tree(p<<1,l,mid);
    build_tree(p<<1|1,mid+1,r);
    update(p);
}

void change(int p,int x,int d)
{
    if(tree[p].l==tree[p].r)
    {
        tree[p].maxx=d;
        return ;
    }
    int mid=(tree[p].l+tree[p].r)>>1;
    if(x<=mid) change(p<<1,x,d);
    else change(p<<1|1,x,d);
    update(p);
}

int ask_max(int p,int l,int r)
{
    if(l<=tree[p].l&&tree[p].r<=r)
    {
        return tree[p].maxx;
    }
    int mid=(tree[p].l+tree[p].r)>>1;
    int ans=0;
    if(l<=mid) ans=max(ans,ask_max(p<<1,l,r));
    if(mid<r) ans=max(ans,ask_max(p<<1|1,l,r));
    return ans;
}

int find_max(int x,int y)
{
    int fx=top[x],fy=top[y];
    int ans=0;
    while(fx!=fy)
    {
        if(deep[fx]<deep[fy]) swap(fx,fy),swap(x,y);
        ans=max(ans,ask_max(1,inseg[fx],inseg[x]));
        x=fa[fx];fx=top[x];
    }
    if(x!=y)
    {
        if(deep[x]>deep[y]) swap(x,y);
        ans=max(ans,ask_max(1,inseg[x]+1,inseg[y]));
    }
    return ans;
}

int n;

int ff[size],tt[size],dd[size];
void init()
{
    memset(head,0,sizeof(head));
    memset(nxt,0,sizeof(nxt));
    memset(l,0,sizeof(l));
    memset(deep,0,sizeof(deep));
    memset(top,0,sizeof(top));
    memset(inseg,0,sizeof(inseg));
    memset(fa,0,sizeof(fa));
    memset(son,0,sizeof(son));
    memset(sz,0,sizeof(sz));
    memset(ff,0,sizeof(ff));
    memset(tt,0,sizeof(tt));
    memset(dd,0,sizeof(dd));
    memset(tree,0,sizeof(tree));
    tot=0;
    totp=0;
}

char in[233];
int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        init();
        scanf("%d",&n);
        for(int i=1;i<=n-1;i++)
        {
            scanf("%d%d%d",&ff[i],&tt[i],&dd[i]);
            build(ff[i],tt[i],dd[i]);
            build(tt[i],ff[i],dd[i]);
        }
        dfs_1(1,0);
        dfs_2(1,1);
        build_tree(1,1,n);

        for(int i=1;i<=n-1;i++)
        {
            if(deep[ff[i]]>deep[tt[i]]) swap(ff[i],tt[i]);
            change(1,inseg[tt[i]],dd[i]);
        }
        while(233)
        {
            int a,b;
            scanf("%s",in);
            if(in[0]=='D') break;
            scanf("%d%d",&a,&b);
            if(in[0]=='C') change(1,inseg[tt[a]],b);
            else printf("%d\n",find_max(a,b));
            //for(int i=1;i<=n;i++) printf("%d ",ask_max(1,inseg[i],inseg[i]));puts("");
        }
        puts("");
    }
    return 0;
}
/* 1 7 1 3 5 5 3 2 1 2 1 1 4 3 5 7 7 3 6 4 QUERY 7 4 QUERY 5 6 QUERY 2 7 CHANGE 5 2 */