树链剖分模板

#include<iostream>
#include<stdio.h>
#include<vector>
#include<algorithm>
#include<string.h>
using namespace std;
typedef long long LL;
const int MAXN = 100000 + 5;
LL aa[MAXN];//树上结点初始值
int siz[MAXN];//子树大小包括自己
int dep[MAXN];//深度
int top[MAXN];//重链起点
int fa[MAXN];//父亲
int son[MAXN];//重链上的儿子
int num[MAXN];//线段树标号
int pre[MAXN];//num的反函数
int vis[MAXN];//标记走过
vector<int>v[MAXN];//边
LL tre[MAXN << 2];//线段树所在点的区间和
LL  add[MAXN << 2];//线段树laz
int n;//总节点数
LL ans;//树上查询的答案
int tot;//dfs序所需
void update(int t, int l, int r, int x, int y, LL k);//线段树更新
void query(int t, int l, int r, int x, int y, LL ad);//线段树查询
void build(int t, int l, int r);//用初值建线段树
int dfs1(int last, int t);//第一次dfs得到fa dep siz son
void dfs(int t, int ttop);//第二次dfs得到num pre top
void update_on_tree(int a, int b, LL z);//树上路径更新，a-b都加z
void query_on_tree(int a, int b);//树上路径查询a-b
int dfs1(int last, int t)
{
    fa[t] = last;
    dep[t] = dep[last] + 1;
    siz[t] = 1;
    vis[t] = 1;
    int I = -1;
    int maxson = 0;
    for (int i = 0; i < v[t].size(); i++)
    {
        if (vis[v[t][i]] == 0)
        {
            siz[t] += dfs1(t, v[t][i]);
            if (siz[v[t][i]] > maxson)
            {
                I = i;
                maxson = siz[v[t][i]];
            }
        }
    }
    if (I >= 0)
        son[t] = v[t][I];
    else
        son[t] = -1;
    return siz[t];
}
void dfs(int t, int ttop)
{
    top[t] = ttop;
    vis[t] = 1;
    num[t] = tot;
    pre[tot] = t;
    tot++;
    if (son[t] > 0)
        dfs(son[t], ttop);
    for (int i = 0; i < v[t].size(); i++)
    {
        if (vis[v[t][i]] == 0)
        {
            if (v[t][i] == son[t])
                continue;
            dfs(v[t][i], v[t][i]);
        }
    }
}
void build(int t, int l, int r)
{
    if (l == r)
    {
        tre[t] = aa[pre[l]];
        return;
    }
    int mid = l + r >> 1;
    build(t << 1, l, mid);
    build(t << 1 | 1, mid + 1, r);
    tre[t] = tre[t << 1] + tre[t << 1 | 1];
}
void update(int t, int l, int r, int x, int y, LL k)
{
    tre[t] += (min(r, y) - max(l, x) + 1)*k;
    if (l >= x && r <= y)
    {
        add[t] += k;
        return;
    }
    int mid = l + r >> 1;
    if (y <= mid)
    {
        update(t << 1, l, mid, x, y, k);
    }
    else if (x > mid)
    {
        update(t << 1 | 1, mid + 1, r, x, y, k);
    }
    else
    {
        update(t << 1, l, mid, x, y, k);
        update(t << 1 | 1, mid + 1, r, x, y, k);
    }
}
void query(int t, int l, int r, int x, int y, LL ad)
{
    if (l >= x && r <= y)
    {
        ans += tre[t];
        ans += ad * (r - l + 1);
        return;
    }
    int mid = l + r >> 1;
    if (y <= mid)
    {
        query(t << 1, l, mid, x, y, ad + add[t]);
    }
    else if (x > mid)
    {
        query(t << 1 | 1, mid + 1, r, x, y, ad + add[t]);
    }
    else
    {
        query(t << 1, l, mid, x, y, ad + add[t]);
        query(t << 1 | 1, mid + 1, r, x, y, ad + add[t]);
    }
}
void update_on_tree(int a, int b, LL z)
{
    if (dep[top[a]] < dep[top[b]])
        swap(a, b);
    if (top[a] == top[b])
    {
        a = num[a];
        b = num[b];
        if (a > b)
            swap(a, b);
        update(1, 1, n, a, b, z);
        return;
    }
    update(1, 1, n, num[top[a]], num[a], z);
    update_on_tree(fa[top[a]], b, z);
}
void query_on_tree(int a, int b)
{
    if (dep[top[a]] < dep[top[b]])
        swap(a, b);
    if (top[a] == top[b])
    {
        a = num[a];
        b = num[b];
        if (a > b)
            swap(a, b);
        query(1, 1, n, a, b, 0);
        return;
    }
    query(1, 1, n, num[top[a]], num[a], 0);
    query_on_tree(fa[top[a]], b);
    return;
}
int main()
{
    int m;
    scanf("%d%d", &n,&m);//n个点m个操作
    for (int i = 1; i <= n; i++)//输入树结点的初始值
    {
        scanf("%lld", &aa[i]);
    }
    for (int i = 1; i < n; i++)//建边
    {
        int a, b;
        scanf("%d%d", &a, &b);
        v[a].push_back(b);
        v[b].push_back(a);
    }
    int root = 1;//根为1
    dep[root] = 1;
    tot = 1;
    dfs1(root, root);
    memset(vis, 0, sizeof(vis));
    dfs(root, root);
    for (int i = 0; i < m; i++)
    {
        int Q,x,y,z;
        ans = 0;
        scanf("%d", &Q);
        switch (Q)
        {
        case 1://x到y路径上的结点都加z
            scanf("%d%d%d", &x, &y, &z);
            update_on_tree(x, y, z);
            break;
        case 2://求x到y路径上的结点和
            scanf("%d%d", &x, &y);
            query_on_tree(x, y);
            printf("%lld\n", ans);
            break;
        case 3://以x为根的子树所有节点都加z
            scanf("%d%d",&x,&z);
            update(1, 1, n, num[x], num[x] + siz[x] - 1, z);
            break;
        case 4://求以x为根的子树所有结点之和
            scanf("%d", &x);
            query(1, 1, n, num[x], num[x] + siz[x] - 1, 0);
            printf("%lld\n", ans);
            break;
        default:
            break;
        }
    }
}