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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:
Note to the second sample test:
妙啊！簡直是妙啊！！做一道這麼有意思的題、領悟巧妙的算法難道不比玩遊戲、看小說強上100倍？
題目大意：
有n個節點，小明住在根節點，現在他想去位於葉子節點的餐館喫飯。有的節點上有貓存在，小明不能連續碰到超過m個貓，問你小明可以去的餐館有多少個？
注意：題目下方提醒了一個樹應該有n-1個節點。
解題思路：
深度優先搜索，每條路都往下搜一遍，看是否滿足條件就行。
#include <iostream>
#include <algorithm>
#include <vector>
using namespace std;
const int maxn=1e5+10;
bool iscat[maxn];//存從1到n節點的值0或1，同時用於判斷連續貓的數量
int isleaf[maxn];//存節點的度數，用於判斷是否是葉子節點
vector<int> G[maxn];//vector向量存圖
bool vis[maxn];//標記點，是否已經搜索過
int ans=0;
int n,m;
void dfs(int x,int cat)//貓的數量會隨着向下遞歸而變化，所以需要在參數裏面
{
if(vis[x]) return;
else vis[x]=true;
if(iscat[x]) cat++;//用於判斷連續的貓的數量
else cat=0;
if (cat>m) return;//如果貓的數量多於指定值的話，return
if (isleaf[x]<2&&x!=1) ans++;//滿足該節點是葉子節點且不是節點1，答案++
for(int i=0;i<G[x].size();i++) dfs(G[x][i],cat);//繼續搜索其他路徑
return;
}
int main()
{
cin>>n>>m;
for(int i=1;i<=n;i++) cin>>iscat[i];//注意節點是從1開始的，易錯
for(int i=0;i<n-1;i++)
{
int u,v;
cin>>u>>v;
isleaf[u]++;//度數++
isleaf[v]++;
G[u].push_back(v);
G[v].push_back(u);//爲什麼要雙向呢，因爲題目中只說了這兩個點是連接着的，並沒有說兩個點的順序是怎樣的
}
dfs(1,0);
cout<<ans;
return 0;
}
我很弱，但是我很堅強，絕不會讓信任我的人失望。