題解

老年選手的程式碼康復計劃QAQ
這題又沒一遍A，難受

每個節點維護這個節點子樹內聯通塊的大小

維護所有節點輕兒子的\(g[u][0]\)表示所有輕兒子白色的聯通塊總數
\(g[u][1]\)表示所有輕兒子黑色聯通塊總數

更新一個點為新顏色的時候，是\(g[u][c[u] ^ 1] + 1\)再加上重兒子價值（如果修改後顏色相同

然後我們向上更新，修改到和父親節點顏色相同的最淺的節點

如果改後顏色和父親顏色不同，那麼這些點都減去原來的值
否則都加上新值
然後在經過輕邊的時候順帶維護一下\(g\)即可

對於找顏色相同的最淺節點，用線段樹維護一個區間顏色是否相同即可

程式碼

#include <bits/stdc++.h>
#define fi first
#define se second
#define pii pair<int,int>
#define pdi pair<db,int>
#define mp make_pair
#define pb push_back
#define enter putchar('\n')
#define space putchar(' ')
#define eps 1e-8
#define mo 974711
#define MAXN 100005
//#define ivorysi
using namespace std;
typedef long long int64;
typedef double db;
template<class T>
void read(T &res) {
    res = 0;char c = getchar();T f = 1;
    while(c < '0' || c > '9') {
    if(c == '-') f = -1;
    c = getchar();
    }
    while(c >= '0' && c <= '9') {
    res = res * 10 + c - '0';
    c = getchar();
    }
    res *= f;
}
template<class T>
void out(T x) {
    if(x < 0) {x = -x;putchar('-');}
    if(x >= 10) {
    out(x / 10);
    }
    putchar('0' + x % 10);
}
int N,M;
struct node {
    int to,next;
}E[MAXN * 2];
int sumE,head[MAXN],c[MAXN];
int dfn[MAXN],son[MAXN],siz[MAXN],top[MAXN],dep[MAXN],fa[MAXN],idx,seq[MAXN];
int g[MAXN][2];
void add(int u,int v) {
    E[++sumE].to = v;
    E[sumE].next = head[u];
    head[u] = sumE;
}
void dfs1(int u) {
    dep[u] = dep[fa[u]] + 1;
    siz[u] = 1;
    for(int i = head[u] ; i ; i = E[i].next) {
    int v = E[i].to;
    if(v != fa[u]) {
        fa[v] = u;
        dfs1(v);
        siz[u] += siz[v];
        if(siz[v] > siz[son[u]]) son[u] = v;
    }
    }
    for(int i = head[u] ; i ; i = E[i].next) {
    int v = E[i].to;
    if(v != fa[u] && v != son[u]) {
        g[u][1] += siz[v];
    }
    }
}
void dfs2(int u) {
    dfn[u] = ++idx;seq[idx] = u;
    if(!top[u]) top[u] = u;
    if(son[u]) {top[son[u]] = top[u];dfs2(son[u]);}
    for(int i = head[u] ; i ; i = E[i].next) {
    int v = E[i].to;
    if(v != fa[u] && v != son[u]) dfs2(v);
    }
}
namespace seg_tr {
    struct node {
    int L,R;
    int val,lz;
    bool sc,col;
    }tr[MAXN * 4];
#define lc(u) u << 1
#define rc(u) u << 1 | 1
#define val(u) tr[u].val
#define lz(u) tr[u].lz
#define sc(u) tr[u].sc
#define col(u) tr[u].col
    void addlz(int u,int v) {
    val(u) += v;lz(u) += v;
    }
    void push_down(int u) {
    if(tr[u].lz) {
        addlz(lc(u),tr[u].lz);
        addlz(rc(u),tr[u].lz);
    }
    tr[u].lz = 0;
    } 
    void update(int u) {
    if(sc(lc(u)) && sc(rc(u)) && !(col(lc(u)) ^ col(rc(u)) ) ) {
        sc(u) = 1;col(u) = col(lc(u));
    }
    else sc(u) = 0;
    }
    void build(int u,int l,int r) {
    tr[u].L = l;tr[u].R = r;
    if(l == r) {
        tr[u].sc = 1;tr[u].col = c[seq[l]];
        tr[u].val = siz[seq[l]];tr[u].lz = 0;
        return;
    }
    int mid = (l + r) >> 1;
    build(u << 1,l,mid);
    build(u << 1 | 1,mid + 1,r);
    update(u);
    }
    int query(int u,int p) {
    if(tr[u].L == tr[u].R) return tr[u].val;
    push_down(u);
    int mid = (tr[u].L + tr[u].R) >> 1;
    if(p <= mid) return query(lc(u),p);
    else if(p > mid) return query(rc(u),p);
    }
    void add(int u,int l,int r,int v) {
    if(tr[u].L == l && tr[u].R == r) {addlz(u,v);return;}
    push_down(u);
    int mid = (tr[u].L + tr[u].R) >> 1;
    if(r <= mid) add(lc(u),l,r,v);
    else if(l > mid) add(rc(u),l,r,v);
    else {
        add(lc(u),l,mid,v);
        add(rc(u),mid + 1,r,v);
    }
    }
    void change(int u,int p,int v,bool on) {
    if(tr[u].L == tr[u].R) {
        tr[u].val = v;
        tr[u].col = on;
        return;
    }
    push_down(u);
    int mid = (tr[u].L + tr[u].R) >> 1;
    if(p <= mid) change(lc(u),p,v,on);
    else change(rc(u),p,v,on);
    update(u);
    }
    int query_sc(int u,int l,int r) {
    if(tr[u].L == l && tr[u].R == r) return tr[u].sc;
    push_down(u);
    int mid = (tr[u].L + tr[u].R) >> 1;
    if(r <= mid) return query_sc(lc(u),l,r);
    else if(l > mid) return query_sc(rc(u),l,r);
    else {
        return query_sc(lc(u),l,mid) && query_sc(rc(u),mid + 1,r) && c[seq[mid]] == c[seq[mid + 1]];
    }
    }
    int Query(int p) {
    return query(1,dfn[p]);
    }
    void Add(int l,int r,int v) {
    add(1,l,r,v);
    }
    void Change(int p,int v,bool on) {
    change(1,dfn[p],v,on);
    }
    bool Query_sc(int l,int r) {
    return query_sc(1,l,r);
    }
}
int Find_same_color(int u) {
    while(1) {
    if(seg_tr::Query_sc(dfn[top[u]],dfn[u])) {
        if(top[u] == 1) return 1;
        if(c[fa[top[u]]] != c[u]) return top[u];
        u = fa[top[u]];
    }
    else {
        int L = dfn[top[u]] + 1,R = dfn[u];
        while(L < R) {
        int mid = (L + R) >> 1;
        if(seg_tr::Query_sc(mid,dfn[u])) R = mid;
        else L = mid + 1;
        }
        return seq[L];
    }
    }
}
void Change_Path(int u,int v,int d) {
    while(top[u] != top[v]) {
    seg_tr::Add(dfn[top[u]],dfn[u],d);
    g[fa[top[u]]][c[top[u]]] += d;
    u = fa[top[u]];
    }
    seg_tr::Add(dfn[v],dfn[u],d);
    if(v == top[u]) g[fa[v]][c[v]] += d;
}
void Init() {
    read(N);
    int u,v;
    for(int i = 1 ; i <= N ; ++i) c[i] = 1;
    for(int i = 1 ; i < N ; ++i) {
    read(u);read(v);
    add(u,v);add(v,u);
    }
    dfs1(1);dfs2(1);
    seg_tr::build(1,1,N);
}
void Solve() {
    read(M);
    int op,u;
    for(int i = 1 ; i <= M ; ++i) {
    read(op);read(u);
    if(!op) {
        out(seg_tr::Query(Find_same_color(u)));enter;
    }
    else {
        int nv = 1 + g[u][c[u] ^ 1] + (son[u] && c[son[u]] == (c[u] ^ 1) ? seg_tr::Query(son[u]) : 0);
        int ov = seg_tr::Query(u);
        if(u == 1) {
        seg_tr::Change(u,nv,c[u] ^ 1);c[u] ^= 1;
        }
        else {
        if(u == top[u]) {
            g[fa[u]][c[u]] -= ov;
            g[fa[u]][c[u] ^ 1] += nv;
        }
        int t = Find_same_color(fa[u]);
        
        if(c[fa[u]] == c[u]) Change_Path(fa[u],t,-ov);
        else Change_Path(fa[u],t,nv);
        seg_tr::Change(u,nv,c[u] ^ 1);c[u] ^= 1;
        }
    }
    }
}
int main() {
#ifdef ivorysi
    freopen("f1.in","r",stdin);
#endif
    Init();
    Solve();
    return 0;
}