C - Kefa and Park

CodeForces - 580C

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 consecutive vertices 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 aieither 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.

Example
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 2
1 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:

题意：已知你住在根节点处，你想到叶子节点处的饭店吃饭，你很怕猫，不能忍受路径上连续m个结点都有猫，问可以到几个饭店吃饭

解法：dfs，有猫标记为1，没猫为0，已知根节点为1，从根节点向下深搜，遇到超过m个结点有猫，则返回，否则走到叶子节点返回并ans加一

代码：

#include<iostream>
#include<vector>
using namespace std;
#define maxn 200005
bool cat[maxn];
bool viz[maxn];
vector<int> v[maxn];
int n,m,ans=0;
void dfs(int x,int cnt)
{
    viz[x]=1;
    if(cnt>m)
        return;
    if(v[x].size()==1&&viz[v[x][0]]){
        ans++;
        return;
    }
    for(int i=0;i<v[x].size();i++){
        int t=v[x][i];
        if(!viz[t]){
            if(cat[t])
                dfs(t,cnt+1);
            else
                dfs(t,0);
        }
    }
}
int main()
{
    cin>>n>>m;
    for(int i=1;i<=n;i++)
        cin>>cat[i];
    for(int i=1;i<n;i++){
        int a,b;
        cin>>a>>b;
        v[a].push_back(b);
        v[b].push_back(a);
    }
    dfs(1,cat[1]);
    cout<<ans<<endl;
}