树链剖分模板

洛谷P3384【模板】树链剖分·第一个树剖祭

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题目描述

如题，已知一棵包含N个结点的树（连通且无环），每个节点上包含一个数值，需要支持以下操作：

操作1： 格式： 1 x y z 表示将树从x到y结点最短路径上所有节点的值都加上z

操作2： 格式： 2 x y 表示求树从x到y结点最短路径上所有节点的值之和

操作3： 格式： 3 x z 表示将以x为根节点的子树内所有节点值都加上z

操作4： 格式： 4 x 表示求以x为根节点的子树内所有节点值之和

输入输出格式

输入格式：

第一行包含4个正整数N、M、R、P，分别表示树的结点个数、操作个数、根节点序号和取模数（即所有的输出结果均对此取模）。

接下来一行包含N个非负整数，分别依次表示各个节点上初始的数值。

接下来N-1行每行包含两个整数x、y，表示点x和点y之间连有一条边（保证无环且连通）

接下来M行每行包含若干个正整数，每行表示一个操作，格式如下：

操作1： 1 x y z

操作2： 2 x y

操作3： 3 x z

操作4： 4 x

输出格式：

输出包含若干行，分别依次表示每个操作2或操作4所得的结果（对P取模）

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代码：（感觉自己写的有点长```>_<```······）

#include<bits/stdc++.h>
using namespace std;
const int MAX = 100005;
typedef struct Node{
    long long val, tag;
    int l, r;
    Node *rson, *lson;
    int len() {return r - l;}
    int mid() {return (l + r) >> 1;}
    Node(): rson(NULL), lson(NULL), l(0), r(0), val(0), tag(0){}
}node, *pointer;

int mod;

class Segment_Tree
{
    public:
        pointer rt = new Node();
        long long a[MAX];
    
    void PushUp(pointer rt)
        {
            rt->val = rt->rson->val + rt->lson->val;
        }
    
    void PushDown(pointer rt)
        {
            if(rt->tag)
                {
                    rt->lson->tag += rt->tag;
                    rt->rson->tag += rt->tag;
                    rt->lson->val += rt->tag * rt->lson->len();
                    rt->rson->val += rt->tag * rt->rson->len();
                    rt->tag = 0;
                }
        }
    
    void Build(pointer rt,int l, int r)
        {
            rt->l = l;
            rt->r = r;
            if(l + 1 == r)
                {
                    rt->val = a[l];
                    //rt->pos = l;
                    rt->rson = NULL;
                    rt->lson = NULL;
                    return ;
                }
            rt->rson = new Node();
            rt->lson = new Node();
            Build(rt->lson, l, rt->mid());
            Build(rt->rson, rt->mid(), r);
            PushUp(rt); 
        }
        
    void UpData(pointer rt, int L, int R, int k)
        {
            if(L <= rt->l && rt->r <= R)
                {
                    rt->val += k * rt->len();
                    rt->tag += k;
                }
            else
                {
                    if(rt->tag != 0)    PushDown(rt);
                    if(L < rt->mid())    UpData(rt->lson, L, R, k);
                    if(R > rt->mid())     UpData(rt->rson, L, R, k);
                    PushUp(rt);
                }
        }
        
    long long Query(pointer rt, int L, int R)
        {
            if(L <= rt->l && rt->r <= R)    return rt->val;
            else
                {
                    if(rt->tag != 0)    PushDown(rt);
                    long long ret = 0;
                    if(L < rt->mid())    ret += Query(rt->lson, L, R);
                    if(R > rt->mid()) ret += Query(rt->rson, L, R);
                    return ret;
                }
        }
};

class Tree_Chain_Division
{
private:
    int Head[MAX], Next[MAX << 1], To[MAX << 1], edgenum = 0;
    int size[MAX], Hson[MAX], Father[MAX], Deep[MAX], pos[MAX];
    int tid[MAX], ti = 0, top[MAX];
    Segment_Tree tree;

    void First_Dfs(int u, int fa)
        {
            size[u] = 1, Father[u] = fa;
            for(int i = Head[u]; i != -1; i = Next[i])
                {
                    int v = To[i];
                    if(v == fa) continue;
                    Deep[v] = Deep[u] + 1;
                    First_Dfs(v, u);
                    size[u] += size[v];
                    if(size[v] > size[Hson[u]] || Hson[u] == 0)    Hson[u] = v;
                }
            return;
        }

    void Second_Dfs(int u, int anc)
        {
            top[u] = anc, tid[u] = ++ti, pos[ti] = u;
            if(Hson[u] == 0)    return;
            Second_Dfs(Hson[u], anc);
            for(int i = Head[u]; i != -1; i = Next[i])
                {
                    int v = To[i];
                    if(Father[u] == v || Hson[u] == v)  continue;
                    Second_Dfs(v, v);
                }
            return ;
        }

public:
    int a[MAX];
    
    void init()
        {
            memset(Head, -1, sizeof(Head));
        }

    void Add_edge(int from, int to)
        {
            Next[++ edgenum] = Head[from];
            To[edgenum] = to;
            Head[from] = edgenum;
        }

    void Chain_Division(int root, int n)
        {
            Deep[root] = 1;
            First_Dfs(root, 0);
            Second_Dfs(root, root);
            for(int i = 1; i <= n; ++ i)    tree.a[i] = a[pos[i]];
            tree.Build(tree.rt, 1, n + 1);
            return;
        }

    int Get_LCA(int x, int y)
        {
            int Father_x = top[x], Father_y = top[y];
            for(;Father_x != Father_y;)
                {
                    if(Deep[Father_x] < Deep[Father_y])
                        swap(Father_y, Father_x), swap(x, y);
                    x = Father[Father_x], Father_x = top[x];
                }
            return Deep[x] < Deep[y] ? x : y;
        }

    void Chai_Add(int x, int y, int w) //x y z 表示将树从x到y结点最短路径上所有节点的值都加上w
        {
            int Father_x = top[x], Father_y = top[y];
            for(;Father_x != Father_y;)
                {
                    if(Deep[Father_x] < Deep[Father_y])
                        swap(Father_y, Father_x), swap(x, y);
                    tree.UpData(tree.rt, tid[Father_x], tid[x] + 1, w);
                    x = Father[Father_x], Father_x = top[x];
                }
             if(Deep[x] < Deep[y]) swap(x, y);
             tree.UpData(tree.rt, tid[y], tid[x] + 1, w);
             return ;
        }

    long long Chai_Query(int x, int y)  //x y 表示求树从x到y结点最短路径上所有节点的值之和
        {
            long long ans = 0;
            int Father_x = top[x], Father_y = top[y];
            for(;Father_x != Father_y;)
                {
                    if(Deep[Father_x] < Deep[Father_y])
                        swap(Father_y, Father_x), swap(x, y);
                    ans = (ans + tree.Query(tree.rt, tid[Father_x], tid[x] + 1)) % mod;
                    x = Father[Father_x], Father_x = top[x];
                }
            if(Deep[x] < Deep[y])  swap(x, y);
            ans = (ans + tree.Query(tree.rt, tid[y], tid[x] + 1)) % mod;
            return ans;
        }

    void Tree_Add(int x, int w) //x w 表示将以x为根节点的子树内所有节点值都加上w
        {
            tree.UpData(tree.rt, tid[x], tid[x] + size[x], w);
            return;
        }

    long long Tree_Query(int x)  //x 表示求以x为根节点的子树内所有节点值之和
        {
            return tree.Query(tree.rt, tid[x], tid[x] + size[x]) % mod; 
        }
};

inline int Qread()
    {
        int x = 0, w = 0; char ch = getchar();
        for(;!isdigit(ch);w |= (ch == '-'), ch = getchar());
        for(;isdigit(ch); x = (x << 3) + (x << 1) + (ch ^ 48), ch = getchar());
        return w ? -x : x;
    }

Tree_Chain_Division division;
int main()
{
//    freopen("sp.in", "r", stdin);
//    freopen("sp.out", "w", stdout);
    
    int x, y, opt, z, n = Qread(), m = Qread(), root = Qread(); mod = Qread();
    for(int i = 1; i <= n; ++ i) division.a[i] = Qread();
    division.init();
    for(int i = 1; i <  n; ++ i) 
        {
            x = Qread(), y = Qread();
            division.Add_edge(x, y), division.Add_edge(y, x);
        }
    division.Chain_Division(root, n);
    for(int i = 1; i <= m; ++ i)
        {
            opt = Qread();
            switch(opt)
            {
                case 1: x = Qread(), y = Qread(), z = Qread(); division.Chai_Add(x, y, z); break;
                case 2: x = Qread(), y = Qread(); printf("%lld
", division.Chai_Query(x, y)); break;
                case 3: x = Qread(), y = Qread(); division.Tree_Add(x, y); break;
                case 4: x = Qread(); printf("%lld
", division.Tree_Query(x)); break;
            }
        }
    return 0;
}