codeforces-580C-Kefa and Park【dfs】

codeforces-580C-Kefa and Park

                    time limit per test2 seconds        memory limit per test256 megabytes

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 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.

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: 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 vertex 2.

Note to the second sample test: The restaurants are located at vertices 4, 5, 6, 7. Kefa can’t go to restaurants 6, 7.

题目链接：cf-580C

题目大意：A的家在1，其余没有子节点的点都是餐厅，途中不能经过m次以上的红点(值为1)问他能去几个餐厅？

题目思路：深搜。需要注意的是他是无向的。

以下是代码：

#include <vector>
#include <map>
#include <set>
#include <algorithm>
#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <string>
#include <cstring>
using namespace std;
vector <int> v[100010];
int a[100010];
int vis[100010];
int ret = 0;
int n,m;
int dfs(int cur,int ret)   //层数 ， 当前连续有猫数
{   
    if (a[cur]) ret++;    //如果有猫，当前连续值+1
    else ret = 0;        //没猫，连续值置0
    if (ret > m || vis[cur]) return 0;     //如果连续有猫大于m或者该点被访问过返回0
    vis[cur] = 1;   
    int ans = 0;
    if (cur > 1 && v[cur].size() == 1) return 1;    //如果访问到餐厅(没有子节点)
    for (int i = 0; i < v[cur].size(); i++)
    {
        ans += dfs(v[cur][i],ret);     
    }
    return ans;
}

int main(){
    cin >> n >> m;
    for (int i = 1; i <= n; i++)
    {
        cin >> a[i];
    }
    for (int i = 0; i < n - 1; i++)
    {
        int n1,n2;
        cin >> n1 >> n2;
        v[n1].push_back(n2);
        v[n2].push_back(n1);
    }
    cout << dfs(1,0) << endl;
    return 0;
}