HYSBZ1036 树链剖分

　　这题我建了2棵线段树，这样来处理 最值和和值，简单的题目。

#include<queue>
#include<stack>
#include<cmath>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define INF 99999999
#define ll __int64
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
using namespace std;
const int MAXN = 50010;
struct node
{
    int to;
    int v;
    int next;
}edge[MAXN*3];
int pre[MAXN],ind,top[MAXN],fa[MAXN],son[MAXN],w[MAXN],deq[MAXN],siz[MAXN],fn;
int tree1[MAXN<<2],tree2[MAXN<<2];
int n,val[MAXN];
void add(int x,int y)
{
    edge[ind].to = y;
    edge[ind].next = pre[x];
    pre[x] = ind++;
}
void dfs1(int rt,int pa,int d)
{
    deq[rt] = d;
    son[rt] = 0;
    fa[rt] = pa;
    siz[rt] = 1;
    for(int i=pre[rt]; i!=-1; i=edge[i].next){
        int t = edge[i].to;
        if(t != fa[rt]){
            dfs1(t,rt,d+1);
            siz[rt] += siz[t];
            if(siz[son[rt]] < siz[t]){
                son[rt] = t;
            }
        }
    }
}
void dfs2(int rt,int tp)
{
    w[rt] = ++ fn;
    top[rt] = tp;
    if(son[rt] != 0)
        dfs2(son[rt],tp);
    for(int i=pre[rt]; i!=-1; i=edge[i].next){
        int t = edge[i].to;
        if(son[rt] != t && t != fa[rt]){
            dfs2(t,t);
        }
    }
}
void pushup1(int rt)
{
    tree1[rt] = max(tree1[rt<<1],tree1[rt<<1|1]);
}
void pushup2(int rt)
{
    tree2[rt] = tree2[rt<<1] + tree2[rt<<1|1];
}
void Insert(int p,int v,int l,int r,int rt)
{
    if(l == r){
        tree1[rt] = tree2[rt] = v;
        return ;
    }
    int m = (l+r)/2;
    if(m >= p){
        Insert(p,v,lson);
    }
    else {
        Insert(p,v,rson);
    }
    pushup1(rt);
    pushup2(rt);
}
void updata(int p,int v,int l,int r,int rt)
{
    if(l == r){
        tree1[rt] = v;
        tree2[rt] = v;
        return ;
    }
    int m = (l+r)/2;
    if(m >= p){
        updata(p,v,lson);
    }
    else {
        updata(p,v,rson);
    }
    pushup1(rt);
    pushup2(rt);
}
int query1(int L,int R,int l,int r,int rt)
{
    if(L<=l && r<=R){
        return tree1[rt];
    }
    int m = (l+r)/2;
    int ans = -INF;
    if(m >= L){
        ans = max(ans,query1(L,R,lson));
    }
    if(m < R){
        ans = max(ans,query1(L,R,rson));
    }
    return ans;
}
int query2(int L,int R,int l,int r,int rt)
{
    if(L<=l && r<=R){
        return tree2[rt];
    }
    int m = (l+r)/2;
    int ans = 0;
    if(m >= L){
        ans += query2(L,R,lson);
    }
    if(m < R){
        ans += query2(L,R,rson);
    }
    return ans;
}
int Getmax(int x,int y)
{
    int ans = -INF;
    while(top[x] != top[y])
    {
        if(deq[top[x]] < deq[top[y]]){
            swap(x,y);
        }
        ans = max(ans,query1(w[top[x]],w[x],1,fn,1));
        x = fa[top[x]];
    }
    if(deq[x] < deq[y]){
        swap(x,y);
    }
    ans = max(ans,query1(w[y],w[x],1,fn,1));
    return ans;
}
int Getsum(int x,int y)
{
    int ans = 0;
    while(top[x] != top[y])
    {
        if(deq[top[x]] < deq[top[y]]){
            swap(x,y);
        }
        ans += query2(w[top[x]],w[x],1,fn,1);
        x = fa[top[x]];
    }
    if(deq[x] < deq[y]){
        swap(x,y);
    }
    ans += query2(w[y],w[x],1,fn,1);
    return ans;
}
int main()
{
    int i;
    while(~scanf("%d",&n))
    {
        ind = 1;
        memset(pre,-1,sizeof(pre));
        memset(tree1,0,sizeof(tree1));
        memset(tree2,0,sizeof(tree2));
        for(i=1; i<n; i++){
            int x,y;
            scanf("%d%d",&x,&y);
            add(x,y);
            add(y,x);
        }
        for(i=1; i<=n; i++){
            scanf("%d",&val[i]);
        }
        dfs1(1,1,1);

        fn = 0;
        dfs2(1,1);

        for(i=1; i<=n; i++){
            Insert(w[i],val[i],1,fn,1);
        }
        int q;
        char s[10];
        scanf("%d",&q);
        while(q--)
        {
            scanf("%s",s);
            int x,y;
            scanf("%d%d",&x,&y);
            if(s[0] == ‘C‘){
                updata(w[x],y,1,fn,1);
            }
            else if(s[1] == ‘S‘){
                printf("%d\n",Getsum(x,y));
            }
            else {
                printf("%d\n",Getmax(x,y));
            }
        }
    }
}