C. Kefa and Park
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题目传送门
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 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.
Examples
input
Copy
4 1 1 1 0 0 1 2 1 3 1 4
output
Copy
2
input
Copy
7 1 1 0 1 1 0 0 0 1 2 1 3 2 4 2 5 3 6 3 7
output
Copy
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.
妙啊！简直是妙啊！！做一道这么有意思的题、领悟巧妙的算法难道不比玩游戏、看小说强上100倍？
题目大意：
有n个节点，小明住在根节点，现在他想去位于叶子节点的餐馆吃饭。有的节点上有猫存在，小明不能连续碰到超过m个猫，问你小明可以去的餐馆有多少个？
注意：题目下方提醒了一个树应该有n-1个节点。
解题思路：
深度优先搜索，每条路都往下搜一遍，看是否满足条件就行。
#include
#include
#include
using namespace std;
const int maxn=1e5+10;
bool iscat[maxn];//存从1到n节点的值0或1，同时用于判断连续猫的数量
int isleaf[maxn];//存节点的度数，用于判断是否是叶子节点
vector G[maxn];//vector向量存图
bool vis[maxn];//标记点，是否已经搜索过
int ans=0;
int n,m;
void dfs(int x,int cat)//猫的数量会随着向下递归而变化，所以需要在参数里面
{
if(vis[x]) return;
else vis[x]=true;
if(iscat[x]) cat++;//用于判断连续的猫的数量
else cat=0;
if (cat>m) return;//如果猫的数量多于指定值的话，return
if (isleaf[x]<2&&x!=1) ans++;//满足该节点是叶子节点且不是节点1，答案++
for(int i=0;i>n>>m;
for(int i=1;i<=n;i++) cin>>iscat[i];//注意节点是从1开始的，易错
for(int i=0;i>u>>v;
isleaf[u]++;//度数++
isleaf[v]++;
G[u].push_back(v);
G[v].push_back(u);//为什么要双向呢，因为题目中只说了这两个点是连接着的，并没有说两个点的顺序是怎样的
}
dfs(1,0);
cout< 我很弱，但是我很坚强，绝不会让信任我的人失望。