CF375D Tree and Queries

题意翻译

给出一棵 n 个结点的树，每个结点有一个颜色 c i 。询问 q 次，每次询问以 v 结点为根的子树中，出现次数 ≥k 的颜色有多少种。树的根节点是1。

感谢@elijahqi 提供的翻译

题目描述

You have a rooted tree consisting of n n n vertices. Each vertex of the tree has some color. We will assume that the tree vertices are numbered by integers from 1 to n n n . Then we represent the color of vertex v v v as cv c_{v} cv . The tree root is a vertex with number 1.

In this problem you need to answer to m m m queries. Each query is described by two integers vj,kj v_{j},k_{j} vj,kj . The answer to query vj,kj v_{j},k_{j} vj,kj is the number of such colors of vertices x x x , that the subtree of vertex vj v_{j} vj contains at least kj k_{j} kj vertices of color x x x .

You can find the definition of a rooted tree by the following link: http://en.wikipedia.org/wiki/Tree\_(graph\_theory).

输入输出格式

输入格式：

The first line contains two integers n n n and m m m $ (2<=n<=10^{5}; 1<=m<=10^{5}) $ . The next line contains a sequence of integers c1,c2,...,cn c_{1},c_{2},...,c_{n} c1,c2,...,cn (1<=ci<=105) (1<=c_{i}<=10^{5}) (1<=ci<=105) . The next n−1 n-1 n−1 lines contain the edges of the tree. The i i i -th line contains the numbers ai,bi a_{i},b_{i} ai,bi $ (1<=a_{i},b_{i}<=n; a_{i}≠b_{i}) $ — the vertices connected by an edge of the tree.

Next m m m lines contain the queries. The j j j -th line contains two integers vj,kj v_{j},k_{j} vj,kj $ (1<=v_{j}<=n; 1<=k_{j}<=10^{5}) $ .

输出格式：

Print m m m integers — the answers to the queries in the order the queries appear in the input.

输入输出样例

输入样例#1：

8 5

1 2 2 3 3 2 3 3

1 2

1 5

2 3

2 4

5 6

5 7

5 8

1 2

1 3

1 4

2 3

5 3

输出样例#1：

输入样例#2：

输出样例#2：

说明

A subtree of vertex v v v in a rooted tree with root r r r is a set of vertices $ {u :dist(r,v)+dist(v,u)=dist(r,u)} $ . Where dist(x,y) dist(x,y) dist(x,y) is the length (in edges) of the shortest path between vertices x x x and y y y .

Solution：

　　本题树上莫队。

　　求子树颜色个数，可以直接弄出dfs序，统计每个子树的入栈时间$inc$和$ouc$，然后对于询问变为dfs序上的区间颜色数查询，直接莫队就好了。

代码：

/*Code by 520 -- 10.19*/

#include<bits/stdc++.h>

#define il inline

#define ll long long

#define RE register

#define For(i,a,b) for(RE int (i)=(a);(i)<=(b);(i)++)

#define Bor(i,a,b) for(RE int (i)=(b);(i)>=(a);(i)--)

using namespace std;

const int N=;

int n,m,to[N],net[N],h[N],cnt,a[N],bl[N],c[N];

int rc,dfn[N],inc[N],ouc[N],sum[N],ans[N];

struct node{

    int l,r,k,id;

    bool operator < (const node &a) const {return bl[l]==bl[a.l]?r<a.r:l<a.l;}

}t[N];

int gi(){

    int a=;char x=getchar();

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

    while(x>=''&&x<='') a=(a<<)+(a<<)+(x^),x=getchar();

    return a;

}

il void Add(int u,int v){to[++cnt]=v,net[cnt]=h[u],h[u]=cnt;}

void dfs(int u,int lst){

    dfn[++rc]=u,inc[u]=rc;

    for(RE int i=h[u];i;i=net[i])

        if(to[i]!=lst) dfs(to[i],u);

    ouc[u]=rc;

}

il void add(int x){sum[++c[a[x]]]++;}

il void del(int x){sum[c[a[x]]--]--;}

int main(){

    n=gi(),m=gi(); int blo=sqrt(n),u,v;

    For(i,,n) a[i]=gi(),bl[i]=(i-)/blo+;

    For(i,,n) u=gi(),v=gi(),Add(u,v),Add(v,u);

    dfs(,);

    For(i,,m) u=gi(),v=gi(),t[i]=node{inc[u],ouc[u],v,i};

    sort(t+,t+m+);

    for(RE int i=,l=,r=;i<=m;i++){

        while(l<t[i].l) del(dfn[l]),l++;

        while(l>t[i].l) --l,add(dfn[l]);

        while(r<t[i].r) ++r,add(dfn[r]);

        while(r>t[i].r) del(dfn[r]),r--;

        ans[t[i].id]=sum[t[i].k];

    }

    For(i,,m) printf("%d\n",ans[i]);

    return ;

}