SPOJ COT2 Count on a tree II（树上莫队）

题目链接：http://www.spoj.com/problems/COT2/

You are given a tree with N nodes.The tree nodes are numbered from 1 to N.Each node has an integer weight.

We will ask you to perfrom the following operation:

• u v : ask for how many different integers that represent the weight of nodes there are on the path from u to v.

Input

In the first line there are two integers N and M.(N<=40000,M<=100000)

In the second line there are N integers.The ith integer denotes the weight of the ith node.

In the next N-1 lines,each line contains two integers u v,which describes an edge (u,v).

In the next M lines,each line contains two integers u v,which means an operation asking for how many different integers that represent the weight of nodes there are on the path from u to v.

Output

For each operation,print its result.

题目大意：给一棵树，每个点有一个权值。多次询问路径(a, b)上有多少个权值不同的点。

思路：参考VFK WC 2013 糖果公园 park 题解（此题比COT2要难。）

http://vfleaking.blog.163.com/blog/static/174807634201311011201627/

代码（2.37S）：

 #include <bits/stdc++.h>
 using namespace std;

 const int MAXV = ;
 const int MAXE = MAXV << ;
 const int MAXQ = ;
 const int MLOG = ;

 namespace Bilibili {

 int head[MAXV], val[MAXV], ecnt;
 int to[MAXE], next[MAXE];
 int n, m;

 int stk[MAXV], top;
 int block[MAXV], bcnt, bsize;

 struct Query {
     int u, v, id;
     void read(int i) {
         id = i;
         scanf("%d%d", &u, &v);
     }
     void adjust() {
         if(block[u] > block[v]) swap(u, v);
     }
     bool operator < (const Query &rhs) const {
         if(block[u] != block[rhs.u]) return block[u] < block[rhs.u];
         return block[v] < block[rhs.v];
     }
 } ask[MAXQ];
 int ans[MAXQ];
 /// Graph
 void init() {
     memset(head + , -, n * sizeof(int));
     ecnt = ;
 }

 void add_edge(int u, int v) {
     to[ecnt] = v; next[ecnt] = head[u]; head[u] = ecnt++;
     to[ecnt] = u; next[ecnt] = head[v]; head[v] = ecnt++;
 }

 void gethash(int a[], int n) {
     static int tmp[MAXV];
     int cnt = ;
     for(int i = ; i <= n; ++i) tmp[cnt++] = a[i];
     sort(tmp, tmp + cnt);
     cnt = unique(tmp, tmp + cnt) - tmp;
     for(int i = ; i <= n; ++i)
         a[i] = lower_bound(tmp, tmp + cnt, a[i]) - tmp + ;
 }

 void read() {
     scanf("%d%d", &n, &m);
     for(int i = ; i <= n; ++i) scanf("%d", &val[i]);
     gethash(val, n);
     init();
     for(int i = , u, v; i < n; ++i) {
         scanf("%d%d", &u, &v);
         add_edge(u, v);
     }
     for(int i = ; i < m; ++i) ask[i].read(i);
 }
 /// find_block
 void add_block(int &cnt) {
     while(cnt--) block[stk[--top]] = bcnt;
     bcnt++;
     cnt = ;
 }

 void rest_block() {
     while(top) block[stk[--top]] = bcnt - ;
 }

 int dfs_block(int u, int f) {
     int size = ;
     for(int p = head[u]; ~p; p = next[p]) {
         int v = to[p];
         if(v == f) continue;
         size += dfs_block(v, u);
         if(size >= bsize) add_block(size);
     }
     stk[top++] = u;
     size++;
     if(size >= bsize) add_block(size);
     return size;
 }

 void init_block() {
     bsize = max(, (int)sqrt(n));
     dfs_block(, );
     rest_block();
 }
 /// ask_rmq
 int fa[MLOG][MAXV];
 int dep[MAXV];

 void dfs_lca(int u, int f, int depth) {
     dep[u] = depth;
     fa[][u] = f;
     for(int p = head[u]; ~p; p = next[p]) {
         int v = to[p];
         if(v != f) dfs_lca(v, u, depth + );
     }
 }

 void init_lca() {
     dfs_lca(, -, );
     for(int k = ; k +  < MLOG; ++k) {
         for(int u = ; u <= n; ++u) {
             if(fa[k][u] == -) fa[k + ][u] = -;
             else fa[k + ][u] = fa[k][fa[k][u]];
         }
     }
 }

 int ask_lca(int u, int v) {
     if(dep[u] < dep[v]) swap(u, v);
     for(int k = ; k < MLOG; ++k) {
         if((dep[u] - dep[v]) & ( << k)) u = fa[k][u];
     }
     if(u == v) return u;
     for(int k = MLOG - ; k >= ; --k) {
         if(fa[k][u] != fa[k][v])
             u = fa[k][u], v = fa[k][v];
     }
     return fa[][u];
 }
 /// modui
 bool vis[MAXV];
 int diff, cnt[MAXV];

 void xorNode(int u) {
     if(vis[u]) vis[u] = false, diff -= (--cnt[val[u]] == );
     else vis[u] = true, diff += (++cnt[val[u]] == );
 }

 void xorPathWithoutLca(int u, int v) {
     if(dep[u] < dep[v]) swap(u, v);
     while(dep[u] != dep[v])
         xorNode(u), u = fa[][u];
     while(u != v)
         xorNode(u), u = fa[][u],
         xorNode(v), v = fa[][v];
 }

 void moveNode(int u, int v, int taru, int tarv) {
     xorPathWithoutLca(u, taru);
     xorPathWithoutLca(v, tarv);
     //printf("debug %d %d\n", ask_lca(u, v), ask_lca(taru, tarv));
     xorNode(ask_lca(u, v));
     xorNode(ask_lca(taru, tarv));
 }

 void make_ans() {
     for(int i = ; i < m; ++i) ask[i].adjust();
     sort(ask, ask + m);
     int nowu = , nowv = ; xorNode();
     for(int i = ; i < m; ++i) {
         moveNode(nowu, nowv, ask[i].u, ask[i].v);
         ans[ask[i].id] = diff;
         nowu = ask[i].u, nowv = ask[i].v;
     }
 }

 void print_ans() {
     for(int i = ; i < m; ++i)
         printf("%d\n", ans[i]);
 }

 void solve() {
     read();
     init_block();
     init_lca();
     make_ans();
     print_ans();
 }

 };

 int main() {
     Bilibili::solve();
 }