题目链接：hdu 4897 Little Devil I

题目大意：给定一棵树，每条边有黑白两种颜色，初始都是白色，现在有三种操作：

• 1 u v：u到v路径上的边都取成相反的颜色
• 2 u v：u到v路径上相邻的边都取成相反的颜色（相邻即仅有一个节点在路径上）
• 3 u v：查询u到v路径上有多少个黑色边

解题思路：树链剖分，用两个线段W和L维护，W对应的是每条的黑白情况，L表示的是每个节点的相邻边翻转情况

（对于轻链而言，重链直接在W上修改）

对于1操作，即为普通的树链剖分，直接在W上修改即可。

对于2操作，每次修改在L上，不过链的两端（即重链的部分）还需要在W上修改

对于3操作，每次查询，重链可以直接根据W中的值判断，轻链还要根据对应节点的L值来确定。

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;
const int maxn = 1e5+5;

#define lson(x) ((x)<<1)
#define rson(x) (((x)<<1)|1)

struct Tree {
    int lc[maxn << 2], rc[maxn << 2], fp[maxn << 2], s[maxn << 2];
    void splay(int u) {
        s[u] = rc[u] - lc[u] + 1 - s[u];
        fp[u] ^= 1;
    }
    void pushup(int u) {
        s[u] = s[lson(u)] + s[rson(u)];
    }
    void pushdown(int u) {
        if (fp[u]) {
            splay(lson(u));
            splay(rson(u));
            fp[u] = 0;
        }
    }
    void build (int u, int l, int r) {
        lc[u] = l;
        rc[u] = r;
        fp[u] = s[u] = 0;

        if (l == r)
            return;

        int mid = (l + r) / 2;
        build(lson(u), l, mid);
        build(rson(u), mid + 1, r);
        pushup(u);
    }
    void modify(int u, int l, int r) {
        if (l <= lc[u] && rc[u] <= r) {
            splay(u);
            return;
        }
        pushdown(u);
        int mid = (lc[u] + rc[u]) / 2;
        if (l <= mid)
            modify(lson(u), l, r);
        if (r > mid)
            modify(rson(u), l, r);
        pushup(u);
    }
    int query(int u, int l, int r) {
        if (l <= lc[u] && rc[u] <= r)
            return s[u];
        pushdown(u);
        int mid = (lc[u] + rc[u]) / 2, ret = 0;
        if (l <= mid)
            ret += query(lson(u), l, r);
        if (r > mid)
            ret += query(rson(u), l, r);
        pushup(u);
        return ret;
    }
}W, L;

int N, Q, E, first[maxn], jump[maxn * 2], link[maxn * 2];
int id, idx[maxn], top[maxn], far[maxn], son[maxn], dep[maxn], cnt[maxn];

inline void add_Edge(int u, int v) {
    link[E] = v;
    jump[E] = first[u];
    first[u] = E++;
}

void dfs (int u, int pre, int d) {
    far[u] = pre;
    son[u] = 0;
    cnt[u] = 1;
    dep[u] = d;

    for (int i = first[u]; i + 1; i = jump[i]) {
        int v = link[i];
        if (v == pre)
            continue;
        dfs(v, u, d + 1);
        cnt[u] += cnt[v];
        if (cnt[son[u]] < cnt[v])
            son[u] = v;
    }
}

void dfs (int u, int rot) {
    top[u] = rot;
    idx[u] = ++id;
    if (son[u])
        dfs(son[u], rot);
    for (int i = first[u]; i + 1; i = jump[i]) {
        int v = link[i];
        if (v == far[u] || v == son[u])
            continue;
        dfs(v, v);
    }
}

void init () {
    E = id = 0;
    memset(first, -1, sizeof(first));

    int u, v;
    scanf("%d", &N);
    for (int i = 1; i < N; i++) {
        scanf("%d%d", &u, &v);
        add_Edge(u, v);
        add_Edge(v, u);
    }
    dfs(1, 0, 0);
    dfs(1, 1);
    W.build(1, 1, N);
    L.build(1, 1, N);
}

void modify(int u, int v, int k) {
    int p = top[u], q = top[v];
    while (p != q) {
        if (dep[p] < dep[q]) {
            swap(p, q);
            swap(u, v);
        }

        if (k) {
            L.modify(1, idx[p], idx[u]);
            W.modify(1, idx[p], idx[p]);
            W.modify(1, idx[son[u]], idx[son[u]]);
        } else
            W.modify(1, idx[p], idx[u]);
        u = far[p];
        p = top[u];
    }

    if (dep[u] > dep[v])
        swap(u, v);

    if (k) {
        L.modify(1, idx[u], idx[v]);
        W.modify(1, idx[u], idx[u]);
        W.modify(1, idx[son[v]], idx[son[v]]);
    } else {
        if (u == v)
            return;
        W.modify(1, idx[son[u]], idx[v]);
    }
}

int query(int u, int v) {
    int p = top[u], q = top[v], ret = 0;
    while (p != q) {
        if (dep[p] < dep[q]) {
            swap(p, q);
            swap(u, v);
        }
        if (u != p)
            ret += W.query(1, idx[son[p]], idx[u]);
        ret += (W.query(1, idx[p], idx[p]) ^ L.query(1, idx[far[p]], idx[far[p]]));
        u = far[p];
        p = top[u];
    }

    if (u == v) return ret;

    if (dep[u] > dep[v])
        swap(u, v);
    ret += W.query(1, idx[son[u]], idx[v]);
    return ret;
}

int main () {
    int cas;
    scanf("%d", &cas);
    while (cas--) {
        init();

        int k, u, v;
        scanf("%d", &Q);
        while (Q--) {
            scanf("%d%d%d", &k, &u, &v);
            if (k == 3)
                printf("%d\n", query(u, v));
            else
                modify(u, v, k - 1);
        }
    }
    return 0;
}

hdu 4897 Little Devil I(树链剖分+线段树)