Dylans loves tree

\(trick\)

如果有3个点，权值都为0，尽管异或为0，但0是出现了奇数次，应该输出0....

\(code\)

树剖模板
const int N = 100005;

int n, q, tot, cnt;
int head[N], pa[N], dep[N], sz[N], son[N], top[N], id[N], rk[N];

struct node { int to, next; } e[N * 2];

void Inite() {
    cnt = tot = 0;
    mem(head, -1);
}

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

void DFS1(int u, int p, int deep) {
    pa[u] = p;
    sz[u] = 1;
    dep[u] = deep;
    for (int i = head[u]; ~i; i = e[i].next) if (e[i].to != p) {
        int v = e[i].to;
        DFS1(v, u, deep + 1);
        sz[u] += sz[v];
        if (son[u] == -1 || sz[v] > sz[son[u]]) son[u] = v;
    }
}

void DFS2(int u, int st) {
    top[u] = st;
    id[u] = ++cnt;
    rk[cnt] = u;
    if (son[u] == -1) return;
    DFS2(son[u], st);
    for (int i = head[u]; ~i; i = e[i].next) if (e[i].to != son[u] && e[i].to != pa[u]) DFS2(e[i].to, e[i].to);
}

int a[N], sum[4 * N];

void Pushup(int root) {
    sum[root] = sum[lson] ^ sum[rson];
}

void Build(int l, int r, int root) {
    if (l == r) {
        sum[root] = a[rk[++cnt]] + 1;
        return;
    }
    int mid = (l + r) >> 1;
    Build(l, mid, lson);
    Build(mid + 1, r, rson);
    Pushup(root);
}

void Update(int l, int r, int root, int pos, int x) {
    if (l == r) {
        sum[root] = x;
        return;
    }
    int mid = (l + r) >> 1;
    if (pos <= mid) Update(l, mid, lson, pos, x);
    else Update(mid + 1, r, rson, pos, x);
    Pushup(root);
}

int Query(int l, int r, int root, int L, int R) {
    if (l > R || r < L) return 0;
    if (L <= l && r <= R) return sum[root];
    int mid = (l + r) >> 1;
    int ans = 0;
    ans ^= Query(l, mid, lson, L, R);
    ans ^= Query(mid + 1, r, rson, L, R);
    return ans;
}

int check(int u, int v) {
    int ans = 0;
    int pu = top[u], pv = top[v];

    while(pu != pv) {
        if (dep[pu] > dep[pv]) {
            ans ^= Query(1, n, 1, id[pu], id[u]);
            u = pa[pu];
        }
        else {
            ans ^= Query(1, n, 1, id[pv], id[v]);
            v = pa[pv];
        }
        pu = top[u], pv = top[v];
    }

    if (id[u] <= id[v]) ans ^= Query(1, n, 1, id[u], id[v]);
    else ans ^= Query(1, n, 1, id[v], id[u]);

    return ans;
}

int main()
{
    BEGIN() {
        sc(n), sc(q);

        Inite();
        Rep(i, 2, n) {
            int u, v;
            sc(u), sc(v);
            addedge(u, v);
            addedge(v, u);
        }
        Rep(i, 1, n) sc(a[i]);

        mem(son, -1);
        DFS1(1, 1, 0);
        DFS2(1, 1);

        cnt = 0;
        mem(sum, 0);
        Build(1, n, 1);

        while(q--) {
            int op;
            sc(op);
            if (op == 0) {
                int x, y;
                sc(x), sc(y);
                Update(1, n, 1, id[x], y + 1);
            }
            else {
                int u, v;
                sc(u), sc(v);
                pr(check(u, v) - 1);
            }
        }
    }
    return 0;
}

原文地址：https://www.cnblogs.com/zgglj-com/p/9822969.html