SPOJcot2 Count on a tree II （树上莫队）

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 perform 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 i-th integer denotes the weight of the i-th 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.

Example

Input:
8 2
105 2 9 3 8 5 7 7
1 2
1 3
1 4
3 5
3 6
3 7
4 8
2 5
7 8

Output:
4
4

题意：求两点间点权值的种类数量。

之前讲过了如何“皇室联邦分块”，由于此题没有要求在线，就先分块，然后莫队就行了。

每次转移的时候怕出错，就干脆vis标记即可。

此题还有很多在线的做法。强得不行。。。ORZ

#include<cmath>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
const int maxn=;
const int maxm=;
int n,m,group[maxn];

struct in
{
    int x; int y; int id;
    friend bool operator< (in a,in b)
    {
         return group[a.x]!=group[b.x]?group[a.x]<group[b.x]:group[a.y]<group[b.y];
    }
}q[maxm];

struct Solve
{
    int maxi;
    int Next[maxn<<],Laxt[maxn],To[maxn<<],cnt,delta;
    int num[maxn],dep[maxn],anc[maxn][],w[maxn],wx[maxn];
    int B,stc[maxn],top,tot,ans[maxm],vis[maxn];

    int swap(int &u,int &v) { u^=v;v^=u;u^=v;}
    int read()
    {
        int res=;bool t=false; char c=getchar();
        while(c>''||c<'') { if(c=='-') t=true; c=getchar();}
        while(c<=''&&c>='') { res=(res<<)+(res<<)+c-'';c=getchar();}
        if(t) return -res; return res;
    }

    void add(int u,int v)
    {
        Next[++cnt]=Laxt[u];
        Laxt[u]=cnt; To[cnt]=v;
    }

    void init()
    {
        int u,v; B=sqrt(n);
        for(int i=;(<<i)<=n;i++) maxi=i;
        for(int i=;i<=n;i++) scanf("%d",&w[i]);
        for(int i=;i<=n;i++) wx[i]=w[i];
        sort(wx+,wx+n+);
        int tt=unique(wx+,wx+n+)-(wx+);
        for(int i=;i<=n;i++)
            w[i]=lower_bound(wx+,wx+tt+,w[i])-wx;
        for(int i=;i<n;i++){
            scanf("%d%d",&u,&v);
            add(u,v); add(v,u);
        }
        dep[]=; dfs();
        while(top)  group[stc[top--]]=tot;//最后剩余部分莫忘liao。
        for(int i=;i<=m;i++) {
            q[i].id=i,  scanf("%d%d",&q[i].x,&q[i].y);
            if(group[q[i].x]<group[q[i].y]) swap(q[i].x,q[i].y);//稍微调整一下，可能会优化。
        }
        sort(q+,q+m+);   get_LCA();
    }

    void dfs(int u)
    {
        int Now=top;
        for(int i=Laxt[u];i;i=Next[i])
         if(!dep[To[i]]){
             anc[To[i]][]=u;  dep[To[i]]=dep[u]+;  dfs(To[i]);
             if(top-Now>=B) {
                 ++tot;  while(top!=Now) group[stc[top--]]=tot;
             }
         }
        stc[++top]=u;
    }

    void get_LCA()
    {
        for(int i=;i<=maxi;i++)
         for(int j=;j<=n;j++)
          anc[j][i]=anc[anc[j][i-]][i-];
    }

    int LCA(int u,int v)
    {
        if(dep[u]<dep[v]) swap(u,v);
        for(int i=maxi;i>=;i--)
           if(dep[anc[u][i]]>=dep[v]) u=anc[u][i];
        if(u==v) return u;
        for(int i=maxi;i>=;i--){
            if(anc[u][i]!=anc[v][i]){
                u=anc[u][i];
                v=anc[v][i];
            }
        } return anc[u][];
    }    

    void trans(int &u)
    {
         if(vis[u]) //已被记录，则本次去掉此点
         {
            if(--num[w[u]] == ) delta--;
         }
         else if(++num[w[u]] == ) delta++;
         vis[u] ^= ;
         u=anc[u][];
    }

     void solve()
    {
        int su=,sv=; delta=;
        for(int i = ; i <= m; i++)
        {
            int tu=q[i].x,tv=q[i].y;
            int lca=LCA(su, tu);//两点朝lca移动，处理路径上的点
            while(su!=lca) trans(su);
            while(tu!=lca) trans(tu);
            lca=LCA(sv,tv);
            while(sv!=lca) trans(sv);
            while(tv!=lca) trans(tv);
            su=q[i].x,sv=q[i].y;
            lca=LCA(sv, su);
            ans[q[i].id]=delta+(!num[w[lca]]);//对lca特殊处理
        }
        for(int i=;i<=m;i++) printf("%d\n",ans[i]);
    }
}Tree;
int main()
{
    scanf("%d%d",&n,&m);
    Tree.init();
    Tree.solve();
    return ;
}