[洛谷P3950] 部落冲突

题目链接：##

点我

题目分析：##

树剖。
考虑把边权下放到点上进行染色，连通是0，不连通是1，然后有两种处理思路：

• 维护区间颜色段总数（参考另一篇题解：染色），每次查询判断整段内是否只有一种颜色，若是，判断是否是全连通（可能是全不连通）
• 维护区间最大值，若是0说明是全连通，若是1说明有不连通边

和染色一样，注意跳过LCA

代码：##

#include<bits/stdc++.h>
#define int long long
#define N (1000000 + 5)
using namespace std;

inline int read() {
    int cnt = 0, f = 1;	char c;
    c = getchar();
    while (!isdigit(c)) {
        if (c == '-') f = -f;
        c = getchar();
    }
    while (isdigit(c)) {
        cnt = (cnt << 3) + (cnt << 1) + c - '0';
        c = getchar();
    }
    return cnt * f;
}

int nxt[N], first[N], to[N], tot;
void Add(int x, int y) {
    nxt[++tot] = first[x];
    first[x] = tot;
    to[tot] = y;
}
int father[N], dep[N], siz[N], num[N], son[N], top[N], idx;
void dfs1(int cur, int fa) {
    father[cur] = fa, siz[cur] = 1, dep[cur] = dep[fa] + 1;
    for (register int i = first[cur]; i; i = nxt[i]) {
        int v = to[i];
        if (v != fa) {
            dfs1(v, cur);
            siz[cur] += siz[v];
            if (siz[son[cur]] < siz[v]) son[cur] = v;
        }
    }
}
void dfs2(int cur, int tp) {
    num[cur] = ++idx, top[cur] = tp;
    if (son[cur]) dfs2(son[cur], tp);
    for (register int i = first[cur]; i; i = nxt[i]) {
        int v = to[i];
        if (!num[v]) dfs2(v, v);
    }
}

struct node {
    int l, r;
    int dat;
    int lc;
    int rc;
    #define l(p) tree[p].l
    #define r(p) tree[p].r
    #define dat(p) tree[p].dat
    #define lc(p) tree[p].lc
    #define rc(p) tree[p].rc
} tree[N * 4];

int Lc, Rc;

void push_up(int p);

void build_tree(int p, int l, int r) {
    l(p) = l, r(p) = r;
    if (l(p) == r(p)) {
        dat(p) = 1;
        lc(p) = rc(p) = 0;
        return;
    }
    int mid = (l + r) >> 1;
    build_tree(p << 1, l, mid);
    build_tree(p << 1 | 1, mid + 1, r);
    push_up(p);
}

void push_up(int p) {
    lc(p) = lc(p << 1), rc(p) = rc(p << 1 | 1);
    dat(p) = dat(p << 1) + dat(p << 1 | 1);
    if (rc(p << 1) == lc(p << 1 | 1)) --dat(p);
}

void modify(int p, int x) {
    if (l(p) == r(p)) {
        lc(p) = 1 - lc(p);
        rc(p) = lc(p);
        return;
    }
    int mid = (l(p) + r(p)) >> 1;
    if (x <= mid) modify(p << 1, x);
    if (x > mid) modify(p << 1 | 1, x);
    push_up(p);
}

int query(int p, int l, int r) {
    if (l > r) return 0;
    if (l(p) == l) Lc = lc(p);
    if (r(p) == r) Rc = rc(p);
    if (l <= l(p) && r >= r(p)) {
        return dat(p);
    }
    long long val = 0;
    int mid = (l(p) + r(p)) >> 1;
    if (l > mid) val += query(p << 1 | 1, l, r);
    else if (r <= mid) val += query(p << 1, l, r);
    else {	
        if (rc(p << 1) == lc(p << 1 | 1)) 
            val += query(p << 1, l, r) + query(p << 1 | 1, l, r) - 1;
         else val += query(p << 1, l, r) + query(p << 1 | 1, l, r);
    }
    return val;
}

bool chain_query(int u, int v) {
    int res = 0;
    while (top[u] != top[v]) {
        if (dep[top[u]] < dep[top[v]]) swap(u, v);
        res = query(1, num[top[u]], num[u]);
        if (res > 1 || (res == 1 && Lc == 1)) return false;
        u = father[top[u]];
    }
    if (dep[u] < dep[v]) swap(u, v);
    res = query(1, num[v] + 1, num[u]);
    if (res > 1 || (res == 1 && Lc == 1)) return false;
    return true;
}
int m, n, x, y;
int war[N], id;
char opr;
void solve() {
    n = read(), m = read();
    for (register int i = 1; i < n; i++) {
        x = read(), y = read();
        Add(x, y); Add(y, x);
    }
    build_tree(1, 1, n);
    dfs1(1, 0); dfs2(1, 1);
    dat(1) = 0;
    for (register int i = 1; i <= m; i++) {
        scanf("%s", &opr);
        if (opr == 'Q') {
            x = read(); y = read();
            if (chain_query(x, y)) printf("Yes\n");
            else printf("No\n");
        }
        if (opr == 'C') {
            x = read(); y = read();
            if (dep[x] < dep[y]) swap(x, y);
            war[++id] = x;
            modify(1, num[x]);
        }
        if (opr == 'U') {
            x = read();
            modify(1, num[war[x]]);
        }
    }
}

signed main() {
    solve();
    return 0;
}