模板 - 数据结构 - 重链剖分

一般都是重链剖分的多，重链剖分在维护路径信息时比较优秀。

也可以使用长链剖分，使用深度最大的儿子作为“重儿子”即可，看了一下用于某种与深度有关的启发式合并比重链剖分好，等以后再研究吧。

#include<bits/stdc++.h>
#define lc (o<<1)
#define rc (o<<1|1)
typedef long long ll;
using namespace std;

const int MAXN = 100000 + 5;
int dep[MAXN], siz[MAXN],  son[MAXN], fa[MAXN], top[MAXN], tid[MAXN], rnk[MAXN], cnt;

int n, m, r, mod;
int a[MAXN];

int head[MAXN], etop;

struct Edge {
    int v, next;
} e[MAXN * 2];

inline void init(int n) {
    etop = 0;
    memset(head, -1, sizeof(head[0]) * (n + 1));
}

inline void addedge(int u, int v) {
    e[++etop].v = v;
    e[etop].next = head[u];
    head[u] = etop;
    e[++etop].v = u;
    e[etop].next = head[v];
    head[v] = etop;
}

struct SegmentTree {
    int sum[MAXN * 4], lz[MAXN * 4];
    void pushup(int o) {
        sum[o] = (sum[lc] + sum[rc]) % mod;
    }
    void pushdown(int o, int l, int r) {
        if(lz[o]) {
            lz[lc] = (lz[lc] + lz[o]) % mod;
            lz[rc] = (lz[rc] + lz[o]) % mod;
            int m = l + r >> 1;
            sum[lc] = (1ll * lz[o] * (m - l + 1) + sum[lc]) % mod;
            sum[rc] = (1ll * lz[o] * (r - m) + sum[rc]) % mod;
            lz[o] = 0;
        }
    }

    void build(int o, int l, int r) {
        if (l == r)
            sum[o] = a[rnk[l]] % mod;
        else {
            int m = (l + r) >> 1;
            build(lc, l, m);
            build(rc, m + 1, r);
            pushup(o);
        }
        lz[o] = 0;
    }

    void update(int o, int l, int r, int ql, int qr, int v) {
        if (ql <= l && r <= qr) {
            lz[o] = (lz[o] + v) % mod;
            sum[o] = (sum[o] + 1ll * v * (r - l + 1)) % mod;
        } else {
            pushdown(o, l, r);
            int m = (l + r) >> 1;
            if (ql <= m)
                update(lc, l, m, ql, qr, v);
            if (qr >= m + 1)
                update(rc, m + 1, r, ql, qr, v);
            pushup(o);
        }
    }

    int query(int o, int l, int r, int ql, int qr) {
        if (ql <= l && r <= qr) {
            return sum[o];
        } else {
            pushdown(o, l, r);
            int m = (l + r) >> 1;
            int res = 0;
            if (ql <= m)
                res += query(lc, l, m, ql, qr);
            if (qr >= m + 1)
                res += query(rc, m + 1, r, ql, qr);
            return res % mod;
        }
    }
} st;

void init1() {
    dep[r] = 1;
}

void dfs1(int u, int t) {
    siz[u] = 1, son[u] = -1, fa[u] = t;
    for (int i = head[u]; i != -1; i = e[i].next) {
        int v = e[i].v;
        if(v == t)
            continue;
        dep[v] = dep[u] + 1;
        dfs1(v, u);
        siz[u] += siz[v];
        if (son[u] == -1 || siz[v] > siz[son[u]])
            son[u] = v;
    }
}

void init2() {
    cnt = 0;
}

void dfs2(int u, int t) {
    top[u] = t;
    tid[u] = ++cnt;
    rnk[cnt] = u;
    if (son[u] == -1)
        return;
    dfs2(son[u], t);
    for (int i = head[u]; i != -1; i = e[i].next) {
        int v = e[i].v;
        if(v==fa[u]||v==son[u])
            continue;
        dfs2(v, v);
    }
}

int query1(int u, int v) {
    ll ret = 0;
    int tu = top[u], tv = top[v];
    while (tu != tv) {
        if (dep[tu] >= dep[tv]) {
            ret += st.query(1, 1, n, tid[tu], tid[u]);
            u = fa[tu];
            tu = top[u];
        } else {
            ret += st.query(1, 1, n, tid[tv], tid[v]);
            v = fa[tv];
            tv = top[v];
        }
    }
    if(tid[u] <= tid[v])
        ret += st.query(1, 1, n, tid[u], tid[v]);
    else
        ret += st.query(1, 1, n, tid[v], tid[u]);
    return ret % mod;
}

inline int query2(int u) {
    return st.query(1, 1, n, tid[u], tid[u] + siz[u] - 1);
}

inline void update1(int u, int v, int val) {
    val %= mod;
    int tu = top[u], tv = top[v];
    while (tu != tv) {
        if (dep[tu] >= dep[tv]) {
            st.update(1, 1, n, tid[tu], tid[u], val);
            u = fa[tu];
            tu = top[u];
        } else {
            st.update(1, 1, n, tid[tv], tid[v], val);
            v = fa[tv];
            tv = top[v];
        }
    }
    if(tid[u] <= tid[v])
        st.update(1, 1, n, tid[u], tid[v], val);
    else
        st.update(1, 1, n, tid[v], tid[u], val);
}

inline void update2(int u, int val) {
    val %= mod;
    st.update(1, 1, n, tid[u], tid[u] + siz[u] - 1, val);
}

void op1() {
    int u, v, val;
    scanf("%d%d%d", &u, &v, &val);
    update1(u, v, val);
}

void op2() {
    int u, v;
    scanf("%d%d", &u, &v);
    printf("%d\n", query1(u, v) % mod);
}

void op3() {
    int u, val;
    scanf("%d%d", &u, &val);
    update2(u, val);
}

void op4() {
    int u;
    scanf("%d", &u);
    printf("%d\n", query2(u) % mod);
}

int main() {
#ifdef Yinku
    freopen("Yinku.in", "r", stdin);
#endif // Yinku
    scanf("%d%d%d%d", &n, &m, &r, &mod);
    for(int i = 1; i <= n; ++i) {
        scanf("%d", &a[i]);
    }
    init(n);
    for(int i = 1, u, v; i <= n - 1; ++i) {
        scanf("%d%d", &u, &v);
        addedge(u, v);
    }
    init1();
    dfs1(r, -1);
    init2();
    dfs2(r, r);
    st.build(1, 1, n);
    for(int i = 1, op; i <= m; ++i) {
        scanf("%d", &op);
        switch(op) {
        case 1:
            op1();
            break;
        case 2:
            op2();
            break;
        case 3:
            op3();
            break;
        case 4:
            op4();
            break;
        }
    }
    return 0;
}

原文地址：https://www.cnblogs.com/KisekiPurin2019/p/12231730.html