#include<bits/stdc++.h>
#define max 100005
using namespace std;
int cats[max];
vector<int>tree[max];
int visit[max];
int ans=;
int n,m;
vector<int>::iterator it;
void init()
{
ans=;
memset(visit,,sizeof(visit));
for(int i=;i<n;i++)
tree[i].clear();
memset(cats,,sizeof(cats));
}
void dfs(int node,int cat)
{
//cout<<"zhuangtai"<<node<<" "<<cat<<tree[node].size()<<" "<<endl;
visit[node]=;
if(cat>m) return;
else
{
//cout<<"jiancha"<<node<<" "<<cat<<" "<<tree[node].size()<<endl; //if(tree[node].size()==1&&node!=1)
//{
//cout<<"shenegui"<<node<<" "<<cat<<endl;
//ans++;
//}
bool ok=;
for(int i=;i<tree[node].size();i++)
{
if(!visit[tree[node][i]])
{
ok=;
//visit[tree[node][i]]=1;
if(cats[tree[node][i]])
dfs(tree[node][i],cat+cats[tree[node][i]]);
else dfs(tree[node][i],);
}
//ans++;
}
//for(it=tree[node].begin();it!=tree[node].end();it++){
// if(!visit[*it]){
// if
//}
//}
ans+=ok;
}
}
/*4 1
1 1 0 0
1 2
1 3
1 4 7 1
1 0 1 1 0 0 0
1 2
1 3
2 4
2 5
3 6
3 7 3 2
1 1 1
1 2
2 3
*/
int main()
{ while(scanf("%d%d",&n,&m)!=EOF)
{
init();
for(int i=;i<=n;i++)
{
cin>>cats[i];
}
int a,b;
//cout<<"ok"<<endl;
for(int i=;i<n;i++)
{
scanf("%d%d",&a,&b);
tree[a].push_back(b);
tree[b].push_back(a);
}
visit[]=;
dfs(,cats[]);
if(ans==) cout<<<<endl;
else
cout<<ans<<endl;
}
return ;
}
B - Kefa and Park

Time Limit:2000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u

Description

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 mconsecutive 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 "xiyi" (without the quotes) (1 ≤ xi, yi ≤ nxi ≠ 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.

Sample Input

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

Hint

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.

Kefa and Park的更多相关文章

  1. Codeforces Round #321 (Div. 2) C. Kefa and Park dfs

    C. Kefa and Park Time Limit: 1 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/580/probl ...

  2. CF580C Kefa and Park dfs

    Kefa decided to celebrate his first big salary by going to the restaurant. He lives by an unusual pa ...

  3. codeforces 580C Kefa and Park(DFS)

    题目链接:http://codeforces.com/contest/580/problem/C #include<cstdio> #include<vector> #incl ...

  4. 【CF580C】Kefa and Park

    题目大意:给定一棵 N 个节点的有根树(其中根节点始终为 1 号节点),点有点权,点权只有 1 和 0 两种,求从根节点到叶子节点的路径中,有多少条路径满足:路径上最大连续点权为 1 的节点个数不超过 ...

  5. 「日常训练」Kefa and Park(Codeforces Round #321 Div. 2 C)

    题意与分析(CodeForces 580C) 给你一棵树,然后每个叶子节点会有一家餐馆:你讨厌猫(waht?怎么会有人讨厌猫),就不会走有连续超过m个节点有猫的路.然后问你最多去几家饭店. 这题我写的 ...

  6. CodeForces - 580C Kefa and Park 【BFS】

    题目链接 http://codeforces.com/problemset/problem/580/C 题意 根节点是 1 然后所有的叶子结点都是饭店 从根节点到叶子结点的路径上 如果存在 大于m 个 ...

  7. Codeforces Round #321 (Div. 2) C Kefa and Park(深搜)

    dfs一遍,维护当前连续遇到的喵的数量,然后剪枝,每个统计孩子数量判断是不是叶子结点. #include<bits/stdc++.h> using namespace std; ; int ...

  8. Codeforces Round #321 (Div. 2) Kefa and Park 深搜

    原题链接: 题意: 给你一棵有根树,某些节点的权值是1,其他的是0,问你从根到叶子节点的权值和不超过m的路径有多少条. 题解: 直接dfs一下就好了. 代码: #include<iostream ...

  9. chd校内选拔赛题目+题解

    题目链接   A. Currency System in Geraldion 有1时,所有大于等于1的数都可由1组成.没有1时,最小不幸的数就是1. #include<iostream> ...

随机推荐

  1. 给setTimeout和setIntreval函数添加回调参数

    setTimeout和setInterval是两个很常见的计时函数.在以前,他们只接收两个参数,我们无法直接向他们的回调函数中添加参数,如果需要实现添加多个参数,可以在外层多嵌一层来实现类似的功能.现 ...

  2. Call Paralution Solver from Fortran

    Abstract: Paralution is an open source library for sparse iterative methods with special focus on mu ...

  3. 用sql合并列,两句话合为一句

    合并bc两列 UPDATE `test` SET `a`=concat(`b`,`c`) 清空a列 UPDATE `test` SET `a` = NULL

  4. cocos基础教程(12)点击交互的三种处理

    1.概述 游戏也好,程序也好,只有能与用户交互才有意义.手机上的交互大致可以分为两部分:点击和输入.其中点击更为重要,几乎是游戏中全部的交互.在Cocos2d-x 3.0中,更改了dispatch机制 ...

  5. Android相机、相册获取图片显示并保存到SD卡

    Android相机.相册获取图片显示并保存到SD卡 [复制链接]   电梯直达 楼主    发表于 2013-3-13 19:51:43 | 只看该作者 |只看大图  本帖最后由 happy小妖同学 ...

  6. 我的grub.cfg配置文件

    路径:/boot/grub/grub.cfg 配置文件如下: # # DO NOT EDIT THIS FILE # # It is automatically generated by grub-m ...

  7. 对比WDCP面板与AMH面板的区别与选择

    转载: http://www.laozuo.org/2760.html | 老左博客 随着VPS主机的性价比提高(其实就是降价)我们很多站长会越来越多的选择使用VPS搭建网站或者运营一些项目,相比较而 ...

  8. [转]Spring的IOC原理[通俗解释一下]

    1. IoC理论的背景我们都知道,在采用面向对象方法设计的软件系统中,它的底层实现都是由N个对象组成的,所有的对象通过彼此的合作,最终实现系统的业务逻辑. 图1:软件系统中耦合的对象 如果我们打开机械 ...

  9. JavaScript String 对象方法

    String 对象方法 方法 描述 anchor() 创建 HTML 锚. big() 用大号字体显示字符串. blink() 显示闪动字符串. bold() 使用粗体显示字符串. charAt() ...

  10. iOS 图片拉伸的解释

    以前对于ios的图片拉伸参数一直不太理解,终于看到一篇好文章,转载一下,原文地址:http://blog.csdn.net/q199109106q/article/details/8615661 主要 ...