洛谷P3703 树点涂色

题意：

解：

发现我们每次染的都是不同的颜色，那么用lct维护的话一个颜色就会在一个splay里。染色是access。

维护每个节点到根路径上的虚边数量。

虚边的切换只会在access和link中出现。于是access的时候顺便修改那个子树。lct上啥也不用维护。

查询链可以用两端点减去lca。

#include <bits/stdc++.h>

const int N = 100010, INF = 0x3f3f3f3f;

struct Edge {
    int nex, v;
}edge[N << 1]; int tp;

int e[N], n, siz[N], pos[N], id[N], d[N], num, top[N], father[N], son[N]; /// tree
int fa[N], s[N][2], stk[N], Top; /// lct
bool rev[N];
int large[N << 2], tag[N << 2]; /// seg

inline void add(int x, int y) {
    tp++;
    edge[tp].v = y;
    edge[tp].nex = e[x];
    e[x] = tp;
    return;
}

void DFS_1(int x, int f) { /// father siz son d
    siz[x] = 1;
    fa[x] = father[x] = f;
    for(int i = e[x]; i; i = edge[i].nex) {
        int y = edge[i].v;
        if(y == f) continue;
        d[y] = d[x] + 1;
        DFS_1(y, x);
        siz[x] += siz[y];
        if(siz[y] > siz[son[x]]) {
            son[x] = y;
        }
    }
    return;
}

void DFS_2(int x, int f) { /// top id pos
    top[x] = f;
    pos[x] = ++num;
    id[num] = x;
    if(son[x]) DFS_2(son[x], f);
    for(int i = e[x]; i; i = edge[i].nex) {
        int y = edge[i].v;
        if(y == father[x] || y == son[x]) continue;
        DFS_2(y, y);
    }
    return;
}

inline void Pushdown(int o) {
    if(tag[o]) {
        int ls = o << 1, rs = o << 1 | 1;
        tag[ls] += tag[o];
        tag[rs] += tag[o];
        large[ls] += tag[o];
        large[rs] += tag[o];
        tag[o] = 0;
    }
    return;
}

void build(int l, int r, int o) {
    if(l == r) {
        large[o] = d[id[r]];
        return;
    }
    int mid = (l + r) >> 1;
    build(l, mid, o << 1);
    build(mid + 1, r, o << 1 | 1);
    large[o] = std::max(large[o << 1], large[o << 1 | 1]);
    return;
}

void add(int L, int R, int v, int l, int r, int o) {
    if(L <= l && r <= R) {
        large[o] += v;
        tag[o] += v;
        return;
    }
    Pushdown(o);
    int mid = (l + r) >> 1;
    if(L <= mid) add(L, R, v, l, mid, o << 1);
    if(mid < R) add(L, R, v, mid + 1, r, o << 1 | 1);
    large[o] = std::max(large[o << 1], large[o << 1 | 1]);
    return;
}

int getMax(int L, int R, int l, int r, int o) {
    if(L <= l && r <= R) return large[o];
    int mid = (l + r) >> 1, ans = -INF;
    Pushdown(o);
    if(L <= mid) ans = std::max(ans, getMax(L, R, l, mid, o << 1));
    if(mid < R) ans = std::max(ans, getMax(L, R, mid + 1, r, o << 1 | 1));
    return ans;
}

int ask(int p, int l, int r, int o) {
    if(l == r) return large[o];
    Pushdown(o);
    int mid = (l + r) >> 1;
    if(p <= mid) return ask(p, l, mid, o << 1);
    else return ask(p, mid + 1, r, o << 1 | 1);
}

inline void pushdown(int x) {
    if(rev[x]) {
        std::swap(s[x][0], s[x][1]);
        if(s[x][0]) rev[s[x][0]] ^= 1;
        if(s[x][1]) rev[s[x][1]] ^= 1;
        rev[x] = 0;
    }
    return;
}

inline void pushup(int x) {
    return;
}

inline bool no_root(int x) {
    return (s[fa[x]][0] == x) || (s[fa[x]][1] == x);
}

inline void rotate(int x) {
    int y = fa[x];
    int z = fa[y];
    bool f = (s[y][1] == x);

    fa[x] = z;
    if(no_root(y)) {
        s[z][s[z][1] == y] = x;
    }
    s[y][f] = s[x][!f];
    if(s[x][!f]) {
        fa[s[x][!f]] = y;
    }
    s[x][!f] = y;
    fa[y] = x;

    pushup(y);
    return;
}

inline void splay(int x) {
    int y = x;
    stk[++Top] = y;
    while(no_root(y)) {
        y = fa[y];
        stk[++Top] = y;
    }
    while(Top) {
        pushdown(stk[Top]);
        Top--;
    }

    y = fa[x];
    int z = fa[y];
    while(no_root(x)) {
        if(no_root(y)) {
            (s[z][1] == y) ^ (s[y][1] == x) ?
            rotate(x) : rotate(y);
        }
        rotate(x);
        y = fa[x];
        z = fa[y];
    }
    pushup(x);
    return;
}

inline void change(int x, int v) {
    pushdown(x);
    while(s[x][0]) {
        x = s[x][0];
        pushdown(x);
    }
    /// x
    add(pos[x], pos[x] + siz[x] - 1, v, 1, n, 1);
    return;
}

inline void access(int x) {
    int y = 0;
    while(x) {
        splay(x);
        if(s[x][1]) change(s[x][1], 1);
        if(y) change(y, -1);
        s[x][1] = y;
        pushup(x);
        y = x;
        x = fa[x];
    }
    return;
}

inline int lca(int x, int y) {
    while(top[x] != top[y]) {
        if(d[top[x]] < d[top[y]]) {
            y = father[top[y]];
        }
        else {
            x = father[top[x]];
        }
    }
    return d[x] < d[y] ? x : y;
}

inline int ask(int x, int y) {
    //printf("ask : %d %d lca %d \n", x, y, lca(x, y));
    //printf("ask : %d + %d - 2 * %d \n", ask(pos[x], 1, n, 1), ask(pos[y], 1, n, 1), ask(pos[lca(x, y)], 1, n, 1));
    return ask(pos[x], 1, n, 1) + ask(pos[y], 1, n, 1) - 2 * ask(pos[lca(x, y)], 1, n, 1);
}

inline int query(int x) {
    return getMax(pos[x], pos[x] + siz[x] - 1, 1, n, 1);
}

int main() {
    int m;
    scanf("%d%d", &n, &m);
    for(int i = 1, x, y; i < n; i++) {
        scanf("%d%d", &x, &y);
        add(x, y); add(y, x);
    }
    /// init
    DFS_1(1, 0);
    DFS_2(1, 1);
    build(1, n, 1);

    for(int i = 1, f, x, y; i <= m; i++) {
        scanf("%d%d", &f, &x);
        if(f == 1) { /// change
            access(x);
        }
        else if(f == 2) {
            scanf("%d", &y);
            printf("%d\n", ask(x, y) + 1);
        }
        else { /// f == 3
            printf("%d\n", query(x) + 1);
        }
    }
    return 0;
}