BZOJ 4817 [LCT][线段树][树链剖分]

Description

Bob有一棵n个点的有根树，其中1号点是根节点。Bob在每个点上涂了颜色，并且每个点上的颜色不同。
定义一条路径的权值是：这条路径上的点（包括起点和终点）共有多少种不同的颜色。
Bob可能会进行这几种操作：

• 1 x：把点x到根节点的路径上所有的点染上一种没有用过的新颜色。
• 2 x y：求x到y的路径的权值。
• 3 x：在以x为根的子树中选择一个点，使得这个点到根节点的路径权值最大，求最大权值。

Bob一共会进行m次操作。

Solution

后面两个操作维护dfs序之后直接线段树查询一下就好了。
第一个操作好像就是access吧。
虚边相连的两个点颜色不同，实边相连的两个点颜色相同。这样的话每次access的时候都在splay之后对相应节点+1,−1就好了。

#include <cstdio>#include <cstdlib>#include <iostream>using namespace std;const int N = 101010;inline char get(void) {    static char buf[100000], *S = buf, *T = buf;    if (S == T) {        T = (S = buf) + fread(buf, 1, 100000, stdin);        if (S == T) return EOF;    }    return *S++;}inline void read(int &x) {    static char c; x = 0;    for (c = get(); c < '0' || c > '9'; c = get());    for (; c >= '0' && c <= '9'; c = get()) x = x * 10 + c - '0';}template<typename T>inline T Max(T a, T b) {    return a < b ? b : a;}int n, m, Gcnt, x, y, z, clc, opt;struct edge {    int to, next;    edge (int t = 0, int n = 0):to(t), next(n) {}};edge G[N << 1];int head[N];namespace Tree {    int pre[N], post[N], son[N], size[N], dep[N], top[N], fa[N];    inline void dfs1(int u) {        int to; size[u] = 1;        for (int i = head[u]; i; i = G[i].next) {            to = G[i].to; if (to == fa[u]) continue;            dep[to] = dep[u] + 1; fa[to] = u;            dfs1(to); size[u] += size[to];            if (size[son[u]] < size[to]) son[u] = to;        }    }    inline void dfs2(int u, int t) {        top[u] = t; pre[u] = ++clc;        if (son[u]) dfs2(son[u], t);        for (int i = head[u]; i; i = G[i].next)            if (G[i].to != son[u] && G[i].to != fa[u])                dfs2(G[i].to, G[i].to);        post[u] = clc;     }    inline int LCA(int x, int y) {        while (top[x] != top[y]) {            if (dep[top[x]] > dep[top[y]]) swap(x, y);            y = fa[top[y]];        }        return dep[x] < dep[y] ? x : y;    }}namespace Seg {    int mx[N << 2], add[N << 2];    inline void PushDown(int o) {        if (add[o]) {            mx[o << 1] += add[o]; mx[o << 1 | 1] += add[o];            add[o << 1] += add[o]; add[o << 1 | 1] += add[o];            add[o] = 0;        }    }    inline void Add(int o, int l, int r, int L, int R, int x) {        if (l >= L && r <= R) {            add[o] += x; mx[o] += x;            return;        }        int mid = (l + r) >> 1;        PushDown(o);        if (L <= mid) Add(o << 1, l, mid, L, R, x);        if (R > mid) Add(o << 1 | 1, mid + 1, r, L, R, x);        mx[o] = Max(mx[o << 1], mx[o << 1 | 1]);    }    inline int Query(int o, int l, int r, int L, int R) {        if (l >= L && r <= R) return mx[o];        int mid = (l + r) >> 1, res = 0;        PushDown(o);        if (L <= mid) res = Max(res, Query(o << 1, l, mid, L, R));        if (R > mid) res = Max(res, Query(o << 1 | 1, mid + 1, r, L, R));        return res;    }    inline void Add(int u, int x) {        Add(1, 1, n, Tree::pre[u], Tree::post[u], x);    }    inline int Query(int u) {        return Query(1, 1, n, Tree::pre[u], Tree::post[u]);    }    inline int Dist(int u) {        return Query(1, 1, n, Tree::pre[u], Tree::pre[u]);    }}namespace LCT {    struct node {        node *ch[2];        node *fa;        int id;    };    node *null;    node mem[N];    node *T[N];    inline void Init(int n) {        for (int i = 0; i <= n; i++) T[i] = mem + i;        null = T[0];        null->ch[0] = null->ch[1] = null;        null->fa = null;        for (int i = 1; i <= n; i++) {            T[i]->ch[0] = T[i]->ch[1] = null;            T[i]->fa = T[Tree::fa[i]]; T[i]->id = i;        }    }    inline bool IsRoot(node *x) {        return x->fa == null || (x->fa->ch[0] != x && x->fa->ch[1] != x);    }    inline void Rotate(node* &x) {        node *y = x->fa, *z = y->fa;        int l = (y->ch[0] != x), r = l ^ 1;        if (!IsRoot(y) && z != null) {            if (z->ch[0] == y) z->ch[0] = x;            else z->ch[1] = x;        }        x->fa = z; y->fa = x; x->ch[r]->fa = y;        y->ch[l] = x->ch[r]; x->ch[r] = y;    }    inline void Splay(node* &x) {        while (!IsRoot(x)) {            node *y = x->fa, *z = y->fa;            if (!IsRoot(y)) {                if (y->ch[0] == x ^ z->ch[0] == y) Rotate(x);                else Rotate(y);            }            Rotate(x);        }    }    inline void Access(node* x) {        node *t, *y;        for (y = null; x != null; x = x->fa) {            Splay(x);            for (t = x->ch[1]; t->ch[0] != null; t = t->ch[0]);            if (t != null) Seg::Add(t->id, 1);            for (t = y; t->ch[0] != null; t = t->ch[0]);            if (t != null) Seg::Add(t->id, -1);            x->ch[1] = y; y = x;        }    }}inline void AddEdge(int from, int to) {    G[++Gcnt] = edge(to, head[from]); head[from] = Gcnt;    G[++Gcnt] = edge(from, head[to]); head[to] = Gcnt;}int main(void) {    read(n); read(m);    for (int i = 1; i < n; i++) {        read(x); read(y);        AddEdge(x, y);    }    Tree::dfs1(1); Tree::dfs2(1, 1);    LCT::Init(n);    for (int i = 1; i <= n; i++)        Seg::Add(Tree::pre[i], 1);    while (m--) {        read(opt); read(x);        if (opt == 1) {            LCT::Access(LCT::T[x]);        } else if (opt == 2) {            read(y); z = Tree::LCA(x, y);            printf("%d\n", Seg::Dist(x) + Seg::Dist(y) - 2 * Seg::Dist(z) + 1);        } else {            printf("%d\n", Seg::Query(x));        }    }    return 0;}