codeforces580C. Kefa and Park【dfs】

C. Kefa and Park

time limit per test
2 seconds

memory limit per test
256 megabytes

input
standard input

output
standard output

Kefa decided to celebrate his first big salary by going to the restaurant.

He lives by an unusual park. The park is a rooted tree consisting of n vertices with the root at vertex 1.
Vertex 1 also contains Kefa's house. Unfortunaely for our hero, the park also contains cats. Kefa has already found out what are the vertices
with cats in them.

The leaf vertices of the park contain restaurants. Kefa wants to choose a restaurant where he will go, but unfortunately he is very afraid of cats, so there is no way he will go to the restaurant if the path from the restaurant to his house contains more than m consecutivevertices
with cats.

Your task is to help Kefa count the number of restaurants where he can go.

Input

The first line contains two integers, n and m (2 ≤ n ≤ 105, 1 ≤ m ≤ n)
— the number of vertices of the tree and the maximum number of consecutive vertices with cats that is still ok for Kefa.

The second line contains n integers a1, a2, ..., an,
where each ai either
equals to 0 (then vertex i has
no cat), or equals to 1 (then vertex i has
a cat).

Next n - 1 lines contains the edges of the tree in the format "xi yi"
(without the quotes) (1 ≤ xi, yi ≤ n, xi ≠ yi),
where xi and yi are
the vertices of the tree, connected by an edge.

It is guaranteed that the given set of edges specifies a tree.

Output

A single integer — the number of distinct leaves of a tree the path to which from Kefa's home contains at most m consecutive vertices
with cats.

Examples

input

4 1
1 1 0 0
1 2
1 3
1 4

output

2

input

7 1
1 0 1 1 0 0 0
1 21 3
2 4
2 5
3 6
3 7

output

2

Note

Let us remind you that a tree is a connected graph on n vertices and n - 1 edge.
A rooted tree is a tree with a special vertex called root. In a rooted tree among any two vertices connected by an edge, one vertex
is a parent (the one closer to the root), and the other one is a child. A vertex is called a leaf, if it has no children.

Note to the first sample test:

The vertices containing
cats are marked red. The restaurants are at vertices 2, 3, 4. Kefa can't go only to the restaurant located at vertex2.

Note to the second sample test:

The restaurants are located
at vertices 4, 5, 6, 7. Kefa can't go to restaurants 6, 7.

题意：给出一颗树数的叶子节点有餐厅一个人的家在根节点树上的某个节点有猫当从给节点都一个餐厅的路上连续的猫的数量大于m时该餐厅不能去问有多少餐厅能去

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<queue>
#include<list>
#include<vector>
using namespace std;
const int maxn=100010;
vector<int>vec[maxn];
int num[maxn],n,m,ans=0;
bool vis[maxn];
void dfs(int node,int cnt,int max1){
if(vis[node])return;
if(max1<=m&&vec[node].size()==1&&node!=1){
ans++;
}
for(int i=0;i<vec[node].size();++i){
vis[node]=true;
if(num[node]&&num[vec[node][i]]){
dfs(vec[node][i],cnt+1,max(cnt+1,max1));
}
else {
dfs(vec[node][i],num[vec[node][i]],max(max1,num[vec[node][i]]));
}
vis[node]=false;
}
}
int main()
{
int i,j,k;
scanf("%d%d",&n,&m);
for(i=1;i<=n;++i){
scanf("%d",&num[i]);
}
int a,b;
for(i=1;i<n;++i){
scanf("%d%d",&a,&b);
vec[a].push_back(b);
vec[b].push_back(a);
}dfs(1,num[1],num[1]);
printf("%d\n",ans);
return 0;
}