bzoj3531 树刨+线段树动态开点

题意：一棵树，每个点有颜色，4种操作，1.单点修改权值 2.单点修改颜色 3.查询路径颜色相同的点权和 4.查询路径颜色相同的点权最大值。
思路：树刨+线段树动态开点，开1e5个线段树。
代码：

#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define forn(i,n) for(int i=0;i<n;++i)
#define for1(i,n) for(int i=1;i<=n;++i)
#define IO ios::sync_with_stdio(false);cin.tie(0)
const int maxn = 2e5+5;

int a[maxn],b[maxn],head[maxn],deep[maxn],sz[maxn],par[maxn],son[maxn],p[maxn],fp[maxn],top[maxn],root[maxn];
int n,q,tot = 1,id,cnt;
class dsegment_tree{public:
    struct dsegment_node{
        int son[2],sum,maxv;
    }node[maxn*50];
    void update(int pos,int v,int &now,int l = 1,int r = n){
        if(!now) now = ++cnt;
        if(l==r){
            node[now].sum = node[now].maxv = v;
            return;
        }
        int mid = l+r>>1;
        if(pos<=mid) update(pos,v,node[now].son[0],l,mid);
        else update(pos,v,node[now].son[1],mid+1,r);
        node[now].sum = node[node[now].son[0]].sum+node[node[now].son[1]].sum;
        node[now].maxv = max(node[node[now].son[0]].maxv,node[node[now].son[1]].maxv);
    }
    int query(int L,int R,int now,int l = 1,int r = n){
        if(L<=l&&R>=r) return node[now].sum;
        int mid = l+r>>1,res = 0;
        if(L<=mid) res+=query(L,R,node[now].son[0],l,mid);
        if(R>mid) res+=query(L,R,node[now].son[1],mid+1,r);
        return res;
    }
    int query2(int L,int R,int now,int l = 1,int r = n){
        if(L<=l&&R>=r) return node[now].maxv;
        int mid = l+r>>1,res = 0;
        if(L<=mid) res = max(res,query2(L,R,node[now].son[0],l,mid));
        if(R>mid) res = max(res,query2(L,R,node[now].son[1],mid+1,r));
        return res;
    }
}tree;
struct edge{
    int v,nex;
}e[maxn];
void add(int u,int v){
    e[++tot] = {v,head[u]},head[u] = tot;
    e[++tot] = {u,head[v]},head[v] = tot;
}
void dfs(int u,int pre,int d){
    deep[u] = d,sz[u] = 1,par[u] = pre;
    for(int i = head[u];i;i = e[i].nex){
        int v = e[i].v;
        if(v==pre) continue;
        dfs(v,u,d+1);
        sz[u]+=sz[v];
        if(sz[son[u]]<sz[v]) son[u] = v;
    }
}
void getpos(int u,int gg){
    top[u] = gg,p[u] = ++id,fp[id] = u;
    if(son[u]) getpos(son[u],gg);
    for(int i = head[u];i;i = e[i].nex){
        int v = e[i].v;
        if(v==par[u]||v == son[u]) continue;
        getpos(v,v);
    }
}
int calsum(int x,int y){
    int z = b[x];
    int fx = top[x],fy = top[y];
    int res = 0;
    while(fx!=fy){
        if(deep[fx]<deep[fy]) swap(fx,fy),swap(x,y);
        res+=tree.query(p[fx],p[x],root[z]);
        x = par[fx],fx = top[x];
    }
    if(deep[x]>deep[y]) swap(x,y);
    res+=tree.query(p[x],p[y],root[z]);
    return res;
}
int calmax(int x,int y){
    int z = b[x];
    int fx = top[x],fy = top[y];
    int res = 0;
    while(fx!=fy){
        if(deep[fx]<deep[fy]) swap(fx,fy),swap(x,y);
        res = max(res,tree.query2(p[fx],p[x],root[z]));
        x = par[fx],fx = top[x];
    }
    if(deep[x]>deep[y]) swap(x,y);
    res= max(res,tree.query2(p[x],p[y],root[z]));
    return res;
}
int main(){
    IO;
    cin>>n>>q;
    for1(i,n) cin>>a[i]>>b[i];
    for1(i,n-1){
        int x,y;cin>>x>>y;
        add(x,y);
    }
    dfs(1,0,0);
    getpos(1,1);
    for1(i,n) tree.update(p[i],a[i],root[b[i]]);
    for1(i,q){
        string s;int x,y;cin>>s>>x>>y;
        if(s[1] == 'C'){
            tree.update(p[x],0,root[b[x]]);
            tree.update(p[x],a[x],root[y]);
            b[x] = y;
        }else if(s[1]=='W'){
            tree.update(p[x],y,root[b[x]]);
            a[x] = y;
        }else if(s[1]=='S') cout<<calsum(x,y)<<'
';
        else cout<<calmax(x,y)<<'
';
    }
    return 0;
}