Count on a tree II

Posted 2020-10-15 mxzf0213

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
分析：求树上任意两点路径上权值的不同的个数；
　　　树上莫队的一般步骤为：
　　　预处理出树的LCA，以及树上每个顶点进来的时间st和退出的时间ed，以及时间戳对应的顶点序号seq；
　　　询问u→v(设st[u]≤st[v])的路径，无非是以下三种情况： 
　　　当u==v时，路径上包含一个点；
     1.当u==lca(u,v)时，那么链上的点就是时间戳从st[u]~st[v]中的所有出现一次的点，即seq中下标从st[u]~st[v]且只出现一次的点。
　　　2.当u!=lca(u,v)时，那么链上的点就是时间戳从ed[u]~st[v]中所有出现一次的点+lca(u,v)

代码：

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#include <climits>
#include <cstring>
#include <string>
#include <set>
#include <bitset>
#include <map>
#include <queue>
#include <stack>
#include <vector>
#include <cassert>
#include <ctime>
#define rep(i,m,n) for(i=m;i<=(int)n;i++)
#define inf 0x3f3f3f3f
#define mod 1000000007
#define vi vector<int>
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define ll long long
#define pi acos(-1.0)
#define pii pair<int,int>
#define sys system("pause")
#define ls (rt<<1)
#define rs (rt<<1|1)
#define all(x) x.begin(),x.end()
const int maxn=1e5+10;
const int N=1e5;
using namespace std;
ll gcd(ll p,ll q){return q==0?p:gcd(q,p%q);}
ll qmul(ll p,ll q,ll mo){ll f=0;while(q){if(q&1)f=(f+p)%mo;p=(p+p)%mo;q>>=1;}return f;}
ll qpow(ll p,ll q){ll f=1;while(q){if(q&1)f=f*p%mod;p=p*p%mod;q>>=1;}return f;}
int n,m,k,t,dfn[maxn],st[maxn],ed[maxn],bl[maxn],a[maxn],d[maxn],tot,cnt[maxn],fa[20][maxn],ans[maxn],dep[maxn],cl,ret;
bool vis[maxn];
struct node
{
    int l,r,fa,id;
    bool operator<(const node&p)const
    {
        return bl[l]==bl[p.l]?r<p.r:bl[l]<bl[p.l];
    }
}qu[maxn];
vi e[maxn];
void dfs(int x,int y)
{
    int i;
    dfn[++cl]=x;
    st[x]=cl;
    dep[x]=dep[y]+1;
    fa[0][x]=y;
    for(i=1;fa[i-1][fa[i-1][x]];i++)fa[i][x]=fa[i-1][fa[i-1][x]];
    rep(i,0,e[x].size()-1)
    {
        int z=e[x][i];
        if(z==y)continue;
        dfs(z,x);
    }
    dfn[++cl]=x;
    ed[x]=cl;
}
int lca(int x,int y)
{
    int i;
    if(dep[x]<dep[y])swap(x,y);
    for(i=19;i>=0;i--)if(dep[fa[i][x]]>=dep[y])x=fa[i][x];
    if(x==y)return x;
    for(i=19;i>=0;i--)
    {
        if(fa[i][x]!=fa[i][y])
        {
            x=fa[i][x],
            y=fa[i][y];
        }
    }
    return fa[0][x];
}
void modify(int x)
{
    if(vis[dfn[x]]^=1)
    {
        if(++cnt[a[dfn[x]]]==1)++ret;
    }
    else
    {
        if(--cnt[a[dfn[x]]]==0)--ret;
    }
}
int main(){
    int i,j;
    scanf("%d%d",&n,&m);
    rep(i,1,n)scanf("%d",&a[i]),d[i]=a[i];
    rep(i,1,n-1)
    {
        int x,y;
        scanf("%d%d",&x,&y);
        e[x].pb(y),e[y].pb(x);
    }
    sort(d+1,d+n+1);
    tot=unique(d+1,d+n+1)-d-1;
    rep(i,1,n)a[i]=lower_bound(d+1,d+tot+1,a[i])-d;
    dfs(1,0);
    int sz=round(sqrt(2*n)+0.5);
    rep(i,1,2*n)bl[i]=(i-1)/sz+1;
    rep(i,1,m)
    {
        int x,y;
        scanf("%d%d",&x,&y);
        if(st[x]>st[y])swap(x,y);
        int fa=lca(x,y);
        if(fa==x)
        {
            qu[i]=node{st[x],st[y],fa,i};
        }else qu[i]=node{ed[x],st[y],fa,i};
    }
    sort(qu+1,qu+m+1);
    int l=1,r=0;
    rep(i,1,m)
    {
        while(r<qu[i].r)modify(++r);
        while(l>qu[i].l)modify(--l);
        while(r>qu[i].r)modify(r--);
        while(l<qu[i].l)modify(l++);
        if(qu[i].fa!=dfn[qu[i].l])modify(st[qu[i].fa]);
        ans[qu[i].id]=ret;
        if(qu[i].fa!=dfn[qu[i].l])modify(st[qu[i].fa]);
    }
    rep(i,1,m)printf("%d\n",ans[i]);
    return 0;
}