spoj COT2 - Count on a tree II

COT2 - Count on a tree II

http://www.spoj.com/problems/COT2/

#tree

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 tov.

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

Submit solution!

题意：问树上两点间有多少不同的权值

树上莫队

开始狂T，发现自己竟是按节点编号划分的块！！

dfs分块。。

#include<cmath>

#include<cstdio>

#include<algorithm>

using namespace std;

#define N 40001

#define M 100001

int n,m,siz,tmp;

int hash[N],key[N];

int front[N],to[N*],nxt[N*],tot;

int fa[N],deep[N],id[N],son[N],bl[N],block[N];

bool vis[N];

int sum[N],ans[M];

struct node

{

    int l,r,id;

    bool operator < (node p) const

    {

        if(block[l]!=block[p.l]) return block[l]<block[p.l];

        return block[r]<block[p.r];

    }

}e[M];

int read(int &x)

{

    x=; char c=getchar();

    while(c<''||c>'') c=getchar();

    while(c>=''&&c<='') { x=x*+c-''; c=getchar(); }

}

void add(int x,int y)

{

    to[++tot]=y; nxt[tot]=front[x]; front[x]=tot;

    to[++tot]=x; nxt[tot]=front[y]; front[y]=tot;

}

void dfs(int x)

{

    son[x]++;

    for(int i=front[x];i;i=nxt[i])

    {

        if(to[i]==fa[x]) continue;

        deep[to[i]]=deep[x]+;

        fa[to[i]]=x;

        dfs(to[i]);

        son[x]+=son[to[i]];

    }

}

void dfs2(int x,int top)

{

    id[x]=++tot;

    bl[x]=top;

    block[x]=(tot-)/siz+;

    int y=;

    for(int i=front[x];i;i=nxt[i])

    {

        if(to[i]==fa[x]) continue;

        if(son[to[i]]>son[y]) y=to[i];

    }

    if(!y) return;

    dfs2(y,top);

    for(int i=front[x];i;i=nxt[i])

    {

        if(to[i]==fa[x]||to[i]==y) continue;

        dfs2(to[i],to[i]);

    }

}

void point(int u)

{

    if(vis[u]) tmp-=(!--sum[hash[u]]);

    else tmp+=(++sum[hash[u]]==);

    vis[u]^=;

}

void path(int u,int v)

{

    while(u!=v)

    {

        if(deep[u]>deep[v]) point(u),u=fa[u];

        else point(v),v=fa[v];

    }

}

int get_lca(int u,int v)

{

    while(bl[u]!=bl[v])

    {

        if(deep[bl[u]]<deep[bl[v]]) swap(u,v);

        u=fa[bl[u]];

    }

    return deep[u]<deep[v] ? u : v;

}

int main()

{

    read(n);read(m);  siz=sqrt(n);

    for(int i=;i<=n;i++) read(key[i]),hash[i]=key[i];

    sort(key+,key+n+);

    key[]=unique(key+,key+n+)-(key+);

    for(int i=;i<=n;i++) hash[i]=lower_bound(key+,key+key[]+,hash[i])-key;

    int x,y;

    for(int i=;i<n;i++)

    {

        read(x); read(y);

        add(x,y);

    }

    tot=;

    dfs();

    dfs2(,);

    for(int i=;i<=m;i++)

    {

        read(e[i].l); read(e[i].r);

        e[i].id=i;

    }

    sort(e+,e+m+);

    int L=,R=,lca;

    for(int i=;i<=m;i++)

    {

        if(id[e[i].l]>id[e[i].r]) swap(e[i].l,e[i].r);

        path(L,e[i].l);

        path(R,e[i].r);

        lca=get_lca(e[i].l,e[i].r);

        point(lca);

        ans[e[i].id]=tmp;

        point(lca);

        L=e[i].l; R=e[i].r;

    }

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

}