bzoj1036点权模板题

/*
HYSBZ1036
树上有1-n个结点，每个节点都有一个权值w
操作 CHANGE u t:把结点u的权值改为t
     QMAX u v:询问从点u到v的路径上的节点的最大权值
     QSUM u v:询问从点u到v的路径上的结点的权值和
从点u到点v路径上的结点包括u，v本身
 
*/
#include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
#define MAXN 30010
using namespace std;
struct Edge{
    int to, next;
}edge[MAXN*2];
int head[MAXN], tot;
int top[MAXN];//重链顶端 
int fa[MAXN];//父亲 
int deep[MAXN];//深度，root=1 
int num[MAXN];//子节点数 
int p[MAXN];//v在线段树中的位置 
int fp[MAXN];
int son[MAXN];//重儿子 
int pos;
void init(){
    tot = 0;
    memset(head, -1, sizeof(head));
    pos = 0;
    memset(son, -1, sizeof(son));
}
void addedge(int u, int v){
    edge[tot].to = v;
    edge[tot].next = head[u];
    head[u] = tot++;
}
//第一次dfs求fa,deep,son,num 
void dfs1(int u, int pre, int d){
    deep[u] = d;
    fa[u] = pre;
    num[u] = 1;
    for(int i = head[u]; i != -1; i = edge[i].next){
        int v = edge[i].to;
        if (v != pre){
            dfs1(v, u, d+1);
            num[u]+=num[v];
            if (son[u]==-1||num[v]>num[son[u]])
                son[u] = v;
        }
    }    
}
//第二次dfs求top，p,用pos记数 
void getpos(int u, int sp){//sp表示重链顶端 
    top[u] = sp;
    p[u] = pos++;
    fp[p[u]] = u;
    if (son[u]==-1)
        return;
    getpos(son[u], sp);//先把重儿子查到底
    for(int i = head[u]; i != -1; i = edge[i].next){
        int v = edge[i].to;//轻儿子 
        if (v != son[u] && v != fa[u])
            getpos(v, v);
    } 
}


//线段树部分
struct Node{
    int l, r;
    int sum;
    int Max;
}segTree[MAXN*3];
inline void push_up(int i){
    segTree[i].sum = segTree[i<<1].sum + segTree[i<<1|1].sum;
    segTree[i].Max = max(segTree[i<<1].Max, segTree[i<<1|1].Max);
}
int s[MAXN];
void build(int i, int l, int r){
    segTree[i].l = l;
    segTree[i].r = r;
    
    if (l==r){
        segTree[i].sum = segTree[i].Max = s[fp[l]];
        return;
    }
    
    int mid = l+r >> 1;
    build(i<<1, l, mid);
    build(i<<1|1, mid+1, r);
    push_up(i);
}
//更新线段树的第k个值为val 
void update(int i, int k, int val){
    if(segTree[i].l == k && segTree[i].r == k){
        segTree[i].Max = segTree[i].sum = val; 
        return;
    }
    int mid = segTree[i].l + segTree[i].r >> 1;
    if (k <= mid)
        update(i<<1, k, val);
    else 
        update(i<<1|1, k, val);
    push_up(i);//单点更新后要重新计算Max和Sum 
}
//查询[l,r]区间的最大值 
int queryMax(int i, int l, int r){
    if(segTree[i].l == l && segTree[i].r == r)
        return segTree[i].Max;
    int mid = segTree[i].l+segTree[i].r >> 1;
    if (r <= mid)
        return queryMax(i<<1, l, r);
    else if (l > mid)
        return queryMax(i<<1|1, l, r);
    else 
        return max(queryMax(i<<1, l, mid), queryMax(i<<1|1, mid+1, r));
} 
//查询[l,r]区间的和 
int querySum(int i, int l, int r){
    if(segTree[i].l==l && segTree[i].r == r)
        return segTree[i].sum;
    int mid = segTree[i].l+segTree[i].r>>1;
    if(r <= mid)
        return querySum(i<<1,l,r);
    else if(l > mid)
        return querySum((i<<1)|1,l,r);
    else 
        return querySum(i<<1, l, mid)+querySum(i<<1|1, mid+1, r); 
}
//查询u->v路径上结点的最大权值
int findMax(int u, int v){
    int f1 = top[u], f2 = top[v];
    int tmp = -1000000000;
    while(f1!=f2){//不在同一条重链上时 
        if(deep[f1]<deep[f2]){
            swap(u, v);
            swap(f1, f2);
        }
        tmp = max(tmp, queryMax(1, p[f1], p[u]));//把这一条链上的求出来 
        u = fa[f1];
        f1 = top[u];
    }
    if (deep[u]>deep[v])
        swap(u,v);
    return 
        max(tmp, queryMax(1, p[u], p[v]));
    
}
//查询u->v路径上结点的权值 
int findSum(int u, int v){
    int f1 = top[u];
    int f2 = top[v];
    int tmp = 0;
    while(f1 != f2){
        if (deep[f1] < deep[f2]){
            swap(f1, f2);
            swap(u, v);
        }
        tmp += querySum(1, p[f1], p[u]);
        u = fa[f1];
        f1 = top[u];
    }
    if(deep[u]>deep[v])
        swap(u, v);
    return tmp + querySum(1, p[u], p[v]); 
}


int main(){
    int n;
    int q;
    char op[20];
    int u, v;
    while(scanf("%d", &n)==1){
        init();
        for(int i = 1; i < n; i++){
            scanf("%d%d", &u, &v);
            addedge(u, v);
            addedge(v, u);
        }
        for(int i = 1; i <= n; i++)
            scanf("%d", &s[i]);
        
        dfs1(1,0,0);
        getpos(1,1);
        build(1, 0, pos-1);
        
        scanf("%d", &q);
        while(q--){
            scanf("%s%d%d", op, &u, &v);
            if (op[0]=='C')
                update(1, p[u], v);//单点修改 
            else if (strcmp(op, "QMAX")==0)
                printf("%d\n", findMax(u, v));//查询u->v路径上点权的最大值
            else
                printf("%d\n", findSum(u, v));//查询路径上点权的和 
        }
    }
    return 0; 
}