Codeforces 375D:Tree and Queries 点化成区间 莫队

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 queryvj, 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 ≤ nai ≠ 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.

Sample test(s)
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)}. Wheredist(x, y) is the length (in edges) of the shortest path between vertices x and y.


题目是给出了一个树,树上的每一个节点都有一个颜色。然后给出了m个查询(v,k),每次查询以这个点v的子树里面,颜色数量至少为k的有多少种颜色。

深搜一遍,将每一个节点的子树查询转换为包含的区间查询,记录各个颜色的数量,然后因为查询的k至少是1,所以用一个数组统计至少为k的数量。

代码:

#pragma comment(linker, "/STACK:102400000,102400000")
#pragma warning(disable:4996)  
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
using namespace std;
#define INF 0x3fffffff
typedef long long ll;

const int mod = 1e9 + 7;
const int maxn = 1e5 + 5;

int n, m, num, bk;
vectoredg[maxn];
int cor[maxn], color[maxn], cor_num[maxn], L[maxn], R[maxn], cnt[maxn], res[maxn];

struct no
{
	int le;
	int ri;
	int id;
	int k;
}qu[maxn];

void input()
{
	int i, u, v;
	scanf("%d%d", &n, &m);

	for (i = 1; i <= n; i++)
	{
		scanf("%d", &cor[i]);
	}
	for (i = 1; i <= n - 1; i++)
	{
		scanf("%d%d", &u, &v);
		edg[u].push_back(v);
		edg[v].push_back(u);
	}
}

void dfs(int x, int google)
{
	color[++num] = cor[x];
	L[x] = num;

	int i;
	int sz = edg[x].size();

	for (i = 0; i < sz; i++)
	{
		if (edg[x][i] != google)
		{
			dfs(edg[x][i], x);
		}
	}
	R[x] = num;
}

bool cmp(no a, no b)
{
	if (a.le / bk == b.le / bk)
	{
		return a.ri < b.ri;
	}
	else
	{
		return a.le / bk < b.le / bk;
	}
}

void solve()
{
	num = 0;
	dfs(1, -1);

	int i, j, c, k;
	bk = sqrt(1.0*n);
	for (i = 1; i <= m; i++)
	{
		scanf("%d%d", &c, &k);
		qu[i].id = i;
		qu[i].k = k;
		qu[i].le = L[c];
		qu[i].ri = R[c];
	}
	sort(qu + 1, qu + m + 1, cmp);

	int pl = 1, pr = 0;
	int id;
	for (i = 1; i <= m; i++)
	{
		id = qu[i].id;
		if (qu[i].le == qu[i].ri)
		{
			res[id] = qu[i].k > 1 ? 0 : 1;
			continue;
		}
		if (pr < qu[i].ri)
		{
			for (j = pr + 1; j <= qu[i].ri; j++)
			{
				cor_num[++cnt[color[j]]]++;
			}
		}
		else
		{
			for (j = pr; j > qu[i].ri; j--)
			{
				cor_num[cnt[color[j]]--]--;
			}
		}
		pr = qu[i].ri;

		if (pl < qu[i].le)
		{
			for (j = pl; j < qu[i].le; j++)
			{
				cor_num[cnt[color[j]]--]--;
			}
		}
		else
		{
			for (j = pl - 1; j >= qu[i].le; j--)
			{
				cor_num[++cnt[color[j]]]++;
			}
		}
		pl = qu[i].le;
		res[id] = cor_num[qu[i].k];
	}
	for (i = 1; i <= m; i++)
	{
		printf("%d\n", res[i]);
	}
}

int main() 
{
	//freopen("i.txt","r",stdin);
	//freopen("o.txt","w",stdout);
	
	input();
	solve();

	return 0;
}

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