[SP10707]COT2 - Count on a tree II

题目大意：有一棵$n$个节点的树，第$i$个点有一个颜色$C_i$，$m$组询问，每次问$x->y$的路径上有多少种颜色

题解：树上莫队，把树按欧拉序展开成一条链，令第$i$个节点第一次出现在序列中为$in_i$，第二次为$out_i$，每一个询问就是看$in_x->in_y$中只出现一次的节点的颜色，但发现如果$x$不为$x,y$的$lca$的话$lca$不会被计入答案，特判一下就行

卡点：1$\sim$2.数组未开大

3.$tarjan$求$lca$时加询问加错

4.为先加入第一个点导致答案多一

C++ Code：

#include <cstdio>
#include <algorithm>
#define maxn 40010
#define maxm 100010
#define N (maxn << 1)
#define bl(x) ((x) >> 9)
int n, m;
int in[maxn], out[maxn], date[N], idx;
struct Query {
    int l, r, lca, id;
    bool addlca;
    inline bool operator < (const Query &rhs) const {
        return (bl(l) == bl(rhs.l)) ? r < rhs.r : l < rhs.l;
    }
} q[maxm];
namespace tree {
    int head[maxn], cnt = 0;
    struct Edge {
        int to, nxt;
    } e[maxn << 1];
    inline void add(int a, int b) {
        e[++cnt] = (Edge) {b, head[a]}; head[a] = cnt;
        e[++cnt] = (Edge) {a, head[b]}; head[b] = cnt;
    }
    int fa[maxn];

    void dfs(int u) {
        date[in[u] = ++idx] = u;
        for (int i = head[u]; i; i = e[i].nxt) {
            int v = e[i].to;
            if (v != fa[u]) {
                fa[v] = u;
                dfs(v);
            }
        }
        date[out[u] = ++idx] = u;
    }
}
namespace tarjan {
    int head[maxn], cnt = 0;
    struct QUERY {
        int v, nxt, id;
    } Q[maxm << 1];
    inline void add(int a, int b, int c) {
        Q[++cnt] = (QUERY) {b, head[a], c}; head[a] = cnt;
        Q[++cnt] = (QUERY) {a, head[b], c}; head[b] = cnt;
    }

    int f[maxn];
    inline void init(int n) {
        for (int i = 1; i <= n; i++) f[i] = i;
    }
    int find(int x) {return (x == f[x] ? x : (f[x] = find(f[x])));}

    bool vis[maxn];
    void dfs(int u) {
        for (int i = tree::head[u]; i; i = tree::e[i].nxt) {
            int v = tree::e[i].to;
            if (v != tree::fa[u]) {
                dfs(v);
                f[v] = u;
            }
        }
        for (int i = tarjan::head[u]; i; i = tarjan::Q[i].nxt) q[Q[i].id].lca = find(Q[i].v);
    }
}
#define ONLINE_JUDGE
#include <cctype>
namespace R {
    int x;
    #ifdef ONLINE_JUDGE
    char *ch, op[1 << 26];
    inline void init() {
        fread(ch = op, 1, 1 << 26, stdin);
    }
    inline int read() {
        while (isspace(*ch)) ch++;
        for (x = *ch & 15, ch++; isdigit(*ch); ch++) x = x * 10 + (*ch & 15);
        return x;
    }
    #else
    char ch;
    inline int read() {
        ch = getchar();
        while (isspace(ch)) ch = getchar();
        for (x = ch & 15, ch = getchar(); isdigit(ch); ch = getchar()) x = x * 10 + (ch & 15);
        return x;
    }
    #endif
}

inline void swap(int &a, int &b) {a ^= b ^= a ^= b;}
int num[maxn], W[maxn], w[maxn], ans[maxm];
bool vis[maxn];
int main() {
    #ifdef ONLINE_JUDGE
    R::init();
    #endif
    tarjan::init(n = R::read()); m = R::read();
    for (int i = 1; i <= n; i++) W[i] = w[i] = R::read();
    int tot = (std::sort(W + 1, W + n + 1), std::unique(W + 1, W + n + 1) - W - 1);
    for (int i = 1; i <= n; i++) w[i] = std::lower_bound(W + 1, W + tot + 1, w[i]) - W;
    for (int i = 1; i < n; i++) tree::add(R::read(), R::read());
    tree::dfs(1);
    for (int i = 1; i <= m; i++) tarjan::add(q[i].l = R::read(), q[i].r = R::read(), q[i].id = i);
    tarjan::dfs(1);
    for (int i = 1; i <= m; i++) {
        int &l = q[i].l, &r = q[i].r;
        if (in[l] > in[r]) swap(l, r);
        l = (q[i].addlca = (q[i].lca != l)) ? out[l] : in[l];
        r = in[r];
    }
    std::sort(q + 1, q + m + 1);
    int l, r, res; l = 1, r = 1, res = 1;
    vis[date[1]] = 1; num[w[date[1]]]++;
    for (int i = 1; i <= m; i++) {
        while (l > q[i].l) (vis[date[--l]] ^= 1) ? (res += num[w[date[l]]]++ == 0) : (res -= --num[w[date[l]]] == 0);
        while (r < q[i].r) (vis[date[++r]] ^= 1) ? (res += num[w[date[r]]]++ == 0) : (res -= --num[w[date[r]]] == 0);
        while (l < q[i].l) (vis[date[l]] ^= 1) ? (res += num[w[date[l++]]]++ == 0) : (res -= --num[w[date[l++]]] == 0);
        while (r > q[i].r) (vis[date[r]] ^= 1) ? (res += num[w[date[r--]]]++ == 0) : (res -= --num[w[date[r--]]] == 0);
        ans[q[i].id] = res + (q[i].addlca && !num[w[q[i].lca]]);
    }
    for (int i = 1; i <= m; i++) printf("%d\n", ans[i]);
    return 0;
}