spoj 375 樹鏈剖分 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!

 題意：

一棵樹有修改邊權值操做和詢問兩個節點之間的最大邊權值操做

代碼：

代碼：

//每一個點和他父節點的邊構成一個線段樹上的點。因此線段樹的點實際從2開始
#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
const int MAXN=100000;
int id[MAXN+9],fa[MAXN+9],max_val[MAXN*4+9],head[MAXN+9],son[MAXN+9],top[MAXN+9],lev[MAXN+9],size[MAXN+9];
//id:對應到線段樹上的點編號，son:重兒子，top:重鏈的頭，lev:深度，size:子樹大小
int tot,cnt,val[MAXN+9]; //cnt:線段樹節點數
struct Edge
{
    int u,v,w,next;
}edge[MAXN*2+9];
void init()
{
    for(int i=0;i<=MAXN;i++) fa[i]=top[i]=i;
    memset(size,0,sizeof(size));
    memset(head,-1,sizeof(head));
    memset(val,0,sizeof(val));
    tot=cnt=0;
}
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]=0;
    size[x]=1;
    for(int i=head[x];i!=-1;i=edge[i].next){
        int y=edge[i].v;
        if(y==fa[x]) continue;
        fa[y]=x;
        dfs1(y,d+1);
        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!=-1;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<<1],max_val[rt<<1|1]); }
void build(int l,int r,int rt)
{
    if(l==r) { max_val[rt]=val[l];return; }
    int mid=(l+r)>>1;
    build(l,mid,rt<<1);
    build(mid+1,r,rt<<1|1);
    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)>>1;
    if(id<=mid) update(id,c,l,mid,rt<<1);
    else update(id,c,mid+1,r,rt<<1|1);
    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)>>1,ans=0;
    if(ql<=mid) ans=max(ans,query(ql,qr,l,mid,rt<<1));
    if(qr>mid) ans=max(ans,query(ql,qr,mid+1,r,rt<<1|1));
    return ans;
}
int solve(int l,int r)
{
    int ltp=top[l],rtp=top[r],ans=0;
    while(ltp!=rtp){
        if(lev[rtp]<lev[ltp]){
            swap(ltp,rtp);
            swap(l,r);
        }
        ans=max(ans,query(id[rtp],id[r],1,cnt,1));
        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],1,cnt,1));
    return ans;
}
int main()
{
    //freopen("in.txt","r",stdin);
    int t,n;
    scanf("%d",&t);
    while(t--){
        init();
        scanf("%d",&n);
        for(int i=1;i<n;i++){
            int x,y,z;
            scanf("%d%d%d",&x,&y,&z);
            add(x,y,z);
        }
        dfs1(1,1);
        dfs2(1,1);
        for(int i=0;i<tot;i+=2){
            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(1,cnt,1);
        char ch[20];
        while(scanf("%s",ch)&&ch[0]!='D'){
            int x,y;
            scanf("%d%d",&x,&y);
            if(ch[0]=='C') update(id[edge[x*2-2].v],y,1,cnt,1);
            else printf("%d\n",solve(x,y));
        }
    }
    return 0;
}