D. Tree and Queries time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output
You have a rooted tree consisting of 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. Then we represent the color of vertex v as cv. The tree root is a vertex with number 1.
In this problem you need to answer to m queries. Each query is described by two integers vj, kj. The answer to query vj, kj is the number of such colors of vertices x, that the subtree of vertex vj contains at least kj vertices of color x.
You can find the definition of a rooted tree by the following link: http://en.wikipedia.org/wiki/Tree_(graph_theory).
Input
The first line contains two integers n and m (2 ≤ n ≤ 105; 1 ≤ m ≤ 105). The next line contains a sequence of integers c1, c2, ..., cn(1 ≤ ci ≤ 105). The next n - 1 lines contain the edges of the tree. The i-th line contains the numbers ai, bi (1 ≤ ai, bi ≤ n; ai ≠ bi) — the vertices connected by an edge of the tree.
Next m lines contain the queries. The j-th line contains two integers vj, kj (1 ≤ vj ≤ n; 1 ≤ kj ≤ 105).
Output
Print m integers — the answers to the queries in the order the queries appear in the input.
Examples input
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
output
2
2
1
0
1
input
4 1
1 2 3 4
1 2
2 3
3 4
1 1
output
4
Note
A subtree of vertex v in a rooted tree with root r is a set of vertices {u : dist(r, v) + dist(v, u) = dist(r, u)}. Where dist(x, y) is the length (in edges) of the shortest path between vertices x and y.
题意：给一颗树和M个询问，每个节点有一种颜色，每个询问输出以该节点为根的子树节点数量大于等于K的颜色个数。
思路：先dfs序重编号，然后莫队分块处理即可，维护颜色i的数量f(i)和大于等于k节点的颜色数量sum(k)。
# include <bits/stdc++.h>
using namespace std;
const int maxn = 1e5+30;
int len, cnt=0, cnt2=0, in[maxn], out[maxn], Next[maxn];
int c[maxn], nc[maxn], f[maxn]={0}, sum[maxn]={0}, ans[maxn];
struct node
{
int v, next;
}edge[maxn<<1];
struct query
{
int l, r, block, id, k;
query(){}
query(int l, int r, int k, int id):l(l),r(r),k(k), id(id){block = (l-1)/len+1;}
bool operator <(const query b)const
{
if(block == b.block) return r < b.r;
return block < b.block;
}
}q[maxn];
void add_edge(int u, int v)
{
edge[cnt] = {v, Next[u]};
Next[u] = cnt++;
edge[cnt] = {u, Next[v]};
Next[v] = cnt++;
}
void dfs(int pre, int cur)
{
in[cur] = ++cnt2;
nc[cnt2] = c[cur];
for(int i=Next[cur]; i!=-1; i=edge[i].next)
{
int v = edge[i].v;
if(v == pre) continue;
dfs(cur, v);
}
out[cur] = cnt2;
}
int main()
{
int n, m;
memset(Next, -1, sizeof(Next));
scanf("%d%d",&n,&m);
len = sqrt(n*1.0);
for(int i=1; i<=n; ++i) scanf("%d",&c[i]);
for(int i=1; i<n; ++i)
{
int s, e;
scanf("%d%d",&s,&e);
add_edge(s, e);
}
dfs(0, 1);
for(int i=1; i<=m; ++i)
{
int s, e;
scanf("%d%d",&s,&e);
q[i] = query(in[s], out[s], e, i);
}
sort(q+1, q+1+m);
int L=1, R=0;
for(int i=1; i<=m; ++i)
{
query &t = q[i];
while(R < t.r)
{
++f[nc[++R]];
++sum[f[nc[R]]];
}
while(L > t.l)
{
++f[nc[--L]];
++sum[f[nc[L]]];
}
while(R > t.r)
{
--sum[f[nc[R]]];
--f[nc[R--]];
}
while(L < t.l)
{
--sum[f[nc[L]]];
--f[nc[L++]];
}
ans[t.id] = sum[t.k];
}
for(int i=1; i<=m; ++i)
printf("%d\n",ans[i]);
return 0;
}