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 $c_{v} cv . The tree root is a vertex with number 1.$

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

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 $(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 $v_{j},k_{j} vj,kj $ (1<=v_{j}<=n; 1<=k_{j}<=10^{5}) $ .$

输出格式：

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

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=200005;
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=0;char x=getchar();
    while(x<'0'||x>'9') x=getchar();
    while(x>='0'&&x<='9') a=(a<<3)+(a<<1)+(x^48),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,1,n) a[i]=gi(),bl[i]=(i-1)/blo+1;
    For(i,2,n) u=gi(),v=gi(),Add(u,v),Add(v,u);
    dfs(1,0);
    For(i,1,m) u=gi(),v=gi(),t[i]=node{inc[u],ouc[u],v,i};
    sort(t+1,t+m+1);
    for(RE int i=1,l=1,r=0;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,1,m) printf("%d
",ans[i]);
    return 0;
}