洛谷题单指南-图论之树-P3178 [HAOI2015] 树上操作

原题链接：https://www.luogu.com.cn/problem/P3178

题意解读：对树上节点进行操作：1、单点修改，权值增加一个数 2、子树修改，子树所有节点权值增加一个数 3、路径权值和查询

解题思路：就是一个典型的树链剖分+线段树的应用，直接上代码。

100分代码：

#include <bits/stdc++.h>   
using namespace std;

typedef long long LL;
const int N = 100005;

struct TreeNode
{
    int v, w;
};
vector<TreeNode> g[N]; //邻接表

struct SegmentTree
{
    int l, r;
    LL sum, tag;
} tr[N << 2]; //线段树

int a[N]; //节点初始权值
int son[N], top[N], siz[N], depth[N], fa[N], dfn[N], rk[N], idx; //树链剖分相关
int n, m;

void dfs1(int u, int p, int d)
{
    siz[u] = 1;
    depth[u] = d;
    fa[u] = p;
    for(auto item : g[u])
    {
        int v = item.v;
        if(v == p) continue;
        dfs1(v, u, d + 1);
        siz[u] += siz[v];
        if(siz[v] > siz[son[u]]) son[u] = v;
    }
}

void dfs2(int u, int t)
{
    top[u] = t;
    dfn[u] = ++idx;
    rk[idx] = u;
    if(son[u]) dfs2(son[u], t);
    for(auto item : g[u])
    {
        int v = item.v;
        if(v != son[u] && v != fa[u]) dfs2(v, v);
    }
}

void pushup(int u)
{
    tr[u].sum = tr[u << 1].sum + tr[u << 1 | 1].sum;
}

void addtag(int u, LL add)
{
    tr[u].tag += add;
    tr[u].sum += add * (tr[u].r - tr[u].l + 1);
}

void pushdown(int u)
{
    if(tr[u].tag)
    {
        addtag(u << 1, tr[u].tag);
        addtag(u << 1 | 1, tr[u].tag);
        tr[u].tag = 0;
    }
}

void build(int u, int l, int r)
{
    tr[u] = {l, r};
    if(l == r)
    {
        tr[u].sum = a[rk[l]];
        return;
    }
    int mid = l + r >> 1;
    build(u << 1, l, mid);
    build(u << 1 | 1, mid + 1, r);
    pushup(u);
}

void update(int u, int x, LL add)
{
    if(tr[u].l == tr[u].r)
    {
        addtag(u, add);
        return;
    }
    pushdown(u);
    int mid = tr[u].l + tr[u].r >> 1;
    if(x <= mid) update(u << 1, x, add);
    else update(u << 1 | 1, x, add);
    pushup(u);
}

void updateRange(int u, int l, int r, LL add)
{
    if(l <= tr[u].l && tr[u].r <= r)
    {
        addtag(u, add);
        return;
    }
    else if(tr[u].l > r || tr[u].r < l) return;
    pushdown(u);
    int mid = tr[u].l + tr[u].r >> 1;
    if(l <= mid) updateRange(u << 1, l, r, add);
    if(r > mid) updateRange(u << 1 | 1, l, r, add);
    pushup(u);
}

LL query(int u, int l, int r)
{
    if(l <= tr[u].l && tr[u].r <= r) return tr[u].sum;
    else if(tr[u].l > r || tr[u].r < l) return 0;
    pushdown(u);
    int mid = tr[u].l + tr[u].r >> 1;
    LL sum = 0;
    if(l <= mid) sum += query(u << 1, l, r);
    if(r > mid) sum += query(u << 1 | 1, l, r);
    return sum;
}

void updateSubTree(int x, int add)
{
    updateRange(1, dfn[x], dfn[x] + siz[x] - 1, add);
}

LL queryPath(int x, int y)
{
    LL res = 0;
    while(top[x] != top[y])
    {
        if(depth[top[x]] < depth[top[y]]) swap(x, y);
        res += query(1, dfn[top[x]], dfn[x]);
        x = fa[top[x]];
    }
    if(depth[x] > depth[y]) swap(x, y);
    res += query(1, dfn[x], dfn[y]);
    return res;
}

int main()
{
    cin >> n >> m;
    for(int i = 1; i <= n; i++) cin >> a[i];
    for(int i = 1; i < n; i++)
    {
        int u, v;
        cin >> u >> v;
        g[u].push_back({v, 0});
        g[v].push_back({u, 0});
    }
    dfs1(1, 0, 1);
    dfs2(1, 1);
    build(1, 1, n);
    int op, x, y;
    while(m--)
    {
        cin >> op;
        if(op == 1)
        {
            cin >> x >> y;
            update(1, dfn[x], y);
        }
        else if(op == 2)
        {
            cin >> x >> y;
            updateSubTree(x, y);
        }
        else
        {
            cin >> x;
            cout << queryPath(1, x) << endl;
        }
    }
    return 0;
}