題意；

　　一棵N個節點的樹，有點權。M次詢問，每次詢問點（u，v）路徑上有多少個權值不同的點。

題解：

　　樹上開莫隊，分塊方法可以參照BZOJ1086題的方式。按照詢問點（u，v）所在塊將詢問進行排序。更新路徑時用vis數組標記路徑上的點是否訪問過。

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn = 4e4+10;
int n, m;
int u, v;
int a[maxn], w[maxn], c[maxn];
int depth[maxn];
int num[maxn];
 
int f[maxn], fa[maxn], vis[maxn], lca_[maxn*3];
int blk, tot, top, s[maxn], bel[maxn];
int no, now_u, now_v, la_u, la_v, la_lca;
int ans[maxn*3];
vector<int> g[maxn];
struct node {
    int to, id;
    node(int a, int b) {
        to = a; id = b;
    }
};
vector<node> q[maxn];
void add_edge(int
 
 u, int v) {
    g[u].push_back(v);
    g[v].push_back(u);
}
int find(int x) {
    return x==f[x]?x:f[x] = find(f[x]);
}
struct ask {
    int l, r, id;
    ask(int a, int b, int c) {
        l = a; r = b; id = c;
    }
    ask() {}
    bool operator < (const ask &a)const {
        if(bel[l]==bel[a.l]) return
 
 bel[r] < bel[a.r];
        return bel[l] < bel[a.l];
    } 
};
ask as[maxn*3];
void dfs(int u, int pre, int d) {
    fa[u] = pre;
    depth[u] = d;
    int tmp = top;
    int len = g[u].size();
    for(int i = 0; i < len; i++) {
        if(g[u][i]==pre) continue;
        dfs(g[u][i], u, d+1);
            if(top-tmp >= blk) {
              tot++;
            while (top != tmp) bel[s[top--]] = tot;
        }
    }
    s[++top] = u;
}
void tarjan(int u, int pre) {
    int len = g[u].size();
    for(int i = 0; i < len; i++) {
        int v = g[u][i];
        if(v==pre) continue;
        tarjan(v, u);
        f[v] = u;
    }
    vis[u] = 1;
    int lenn = q[u].size();
    for(int i = 0; i < lenn; i++) {
        int v = q[u][i].to;
        if(vis[v]) lca_[q[u][i].id] = find(v);
    } 
}
void xornode(int x) {
    if(vis[x]) {
        vis[x]--;
        num[a[x]]--;
        if(!num[a[x]]) no--;
    }
    else {
        vis[x]++;
        if(!num[a[x]]) no++;
        num[a[x]]++;
    }
}
void xorpath(int u, int u_to) {
    if(depth[u] < depth[u_to]) swap(u, u_to);
    while(depth[u] > depth[u_to]) {
        xornode(u);
        u = fa[u];
    }
    while(u != u_to) {
        xornode(u);
        xornode(u_to);
        u = fa[u];
        u_to = fa[u_to];
    }
}
int main() {
    scanf("%d%d", &n, &m);
    for(int i = 1; i <= n; i++) {
        f[i] = i;
        scanf("%d", &w[i]);
        c[i] = w[i];
    }
    sort(c+1, c+n+1);
    int num = unique(c+1, c+n+1)-c;
    for(int i = 1; i <= n; i++) a[i] = lower_bound(c+1, c+num, w[i])-c;
    for(int i = 1; i < n; i++) {
        scanf("%d%d", &u, &v);
        add_edge(u, v);
    }
    blk = sqrt(n);
    dfs(1, 0, 0);
    for(int i = 1; i <= m; i++) {
        scanf("%d%d", &u, &v);
        if(bel[u] > bel[v]) swap(u, v);
        q[u].push_back(node(v, i));
        q[v].push_back(node(u, i));
        as[i] = ask(u, v, i);
    }
    sort(as+1, as+m+1);
    tarjan(1, 0);
    memset(vis, 0, sizeof(vis));
    la_u = la_v = la_lca = 1; xornode(1);
     for(int i = 1; i <= m; i++) {
        xorpath(as[i].l, la_u);
        xorpath(as[i].r, la_v);
        xornode(la_lca);
        xornode(lca_[as[i].id]);
        la_u = as[i].l; la_v = as[i].r; la_lca = lca_[as[i].id];
        ans[as[i].id] = no;
    }
    for(int i = 1; i <= m; i++) printf("%d\n", ans[i]);
}

SPOJ - COT2 Count on a tree II