bzoj 1036 树的统计Count (树链剖分+线段树)

题目大意：给你一棵树，每个点都有点权

有3种操作，修改某节点的权值，求树链上节点的权值的最大值，求树链上节点的权值和

树剖裸题，搜一个树链剖分序，用线段树维护一下即可，总时间

#include <cstdio>
#include <algorithm>
#include <cstring>
#include <queue>
#define inf 0x3f3f3f3f
#define ll long long 
#define N 30100
using namespace std;

char str[10];
int n,num,cte;
ll a[N],sum[N<<2],ma[N<<2];
int dep[N],head[N],son[N],sz[N],que[N],pos[N],tp[N],fa[N];
struct Chain{
    int up,dw;
    Chain(int up,int dw):up(up),dw(dw){}
};
struct EDGE{
    int to,nxt;
}edge[N<<1];
void edge_add(int u,int v)
{
    cte++;
    edge[cte].to = v;
    edge[cte].nxt=head[u];
    head[u]=cte;
}
void dfs1(int x,int pre)
{
    for(int j=head[x];j!=-1;j=edge[j].nxt)
    {
        int v=edge[j].to;
        if(v==pre) continue;
        dep[v]=dep[x]+1;
        fa[v]=x;
        dfs1(v,x);
        son[x]=(sz[v]>sz[son[x]])?v:son[x];
        sz[x]+=sz[v];
    }
    sz[x]++;
}
void dfs2(int x)
{
    que[++num]=x,pos[x]=num;
    if(son[x]) tp[son[x]]=tp[x],dfs2(son[x]);
    for(int j=head[x];j!=-1;j=edge[j].nxt)
    {
        int v=edge[j].to;
        if(v==fa[x]||v==son[x]) continue;
        tp[v]=v,dfs2(v);
    }
}
void pushup(int rt){
    sum[rt]=sum[rt<<1]+sum[rt<<1|1];
    ma[rt]=max(ma[rt<<1],ma[rt<<1|1]);
}
void build(int l,int r,int rt)
{
    if(l==r){sum[rt]=ma[rt]=a[que[l]];return;}
    int mid=(l+r)>>1;
    build(l,mid,rt<<1);
    build(mid+1,r,rt<<1|1);
    pushup(rt);
}
void update(int x,int l,int r,int rt,ll w)
{
    if(l==r){sum[rt]=ma[rt]=a[que[l]];return;}
    int mid=(l+r)>>1;
    if(x<=mid) update(x,l,mid,rt<<1,w);
    else  update(x,mid+1,r,rt<<1|1,w);
    pushup(rt);
}
ll qmax1(int L,int R,int l,int r,int rt)
{
    if(L<=l&&r<=R) return ma[rt];
    int mid=(l+r)>>1;ll ans=-inf;
    if(L<=mid) ans=max(ans,qmax1(L,R,l,mid,rt<<1));
    if(R>mid) ans=max(ans,qmax1(L,R,mid+1,r,rt<<1|1));
    return ans;
}
ll qsum1(int L,int R,int l,int r,int rt)
{
    if(L<=l&&r<=R) return sum[rt];
    int mid=(l+r)>>1;ll ans=0;
    if(L<=mid) ans+=qsum1(L,R,l,mid,rt<<1);
    if(R>mid) ans+=qsum1(L,R,mid+1,r,rt<<1|1);
    return ans;
}
int LCA(int x,int y)
{
    while(tp[x]!=tp[y]){
        if(dep[tp[x]]<dep[tp[y]]) swap(x,y);
        x=fa[tp[x]];
    }return dep[x]<dep[y]?x:y;
}
queue<Chain>q;
ll Qmax(int x,int y)
{
    while(tp[x]!=tp[y]){
        if(dep[tp[x]]<dep[tp[y]]) swap(x,y);
        q.push(Chain(tp[x],x));
        x=fa[tp[x]];    
    } if(dep[x]<dep[y]) swap(x,y);
    q.push(Chain(y,x));ll ans=-inf;
    while(!q.empty()){
        Chain k=q.front();q.pop();
        ans=max(ans,qmax1(pos[k.up],pos[k.dw],1,n,1));
    }return ans;
}
ll Qsum(int x,int y)
{
    while(tp[x]!=tp[y]){
        if(dep[tp[x]]<dep[tp[y]]) swap(x,y);
        q.push(Chain(tp[x],x));
        x=fa[tp[x]];
    } if(dep[x]<dep[y]) swap(x,y);
    q.push(Chain(y,x));ll ans=0;
    while(!q.empty()){
        Chain k=q.front();q.pop();
        ans+=qsum1(pos[k.up],pos[k.dw],1,n,1);
    }return ans;
}

int main()
{
    //freopen("data.in","r",stdin);
    scanf("%d",&n);
    int x,y;ll z;
    memset(head,-1,sizeof(head));
    for(int i=1;i<n;i++)
    {
        scanf("%d%d",&x,&y);
        edge_add(x,y);
        edge_add(y,x);
    } 
    for(int i=1;i<=n;i++) scanf("%lld",&a[i]);
    dep[1]=1,dfs1(1,-1);
    tp[1]=1,dfs2(1);
    build(1,n,1);
    int q=0;
    scanf("%d",&q);
    for(int i=1;i<=q;i++)
    {
        scanf("%s",str);
        if(str[1]=='H'){
            scanf("%d%lld",&x,&z),a[x]=z;
            update(pos[x],1,n,1,z);
        }else if(str[1]=='M'){
            scanf("%d%d",&x,&y);
            printf("%lld
",Qmax(x,y));
        }else{
            scanf("%d%d",&x,&y);
            printf("%lld
",Qsum(x,y));
        }
    }
    return 0;
}