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：
2 2 1 0 1
输入样例#2：
4 1 1 2 3 4 1 2 2 3 3 4 1 1
输出样例#2：
4
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#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\n",ans[i]); return 0;
}
