CodeForces 580C 樹+dfs搜索

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 ≤ 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.

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.

題意解析

• 建成一顆樹從根出發到葉節點爲一條路徑
• 路上不能有大於m個“連續consecutive”有貓的節點
• 數有幾條符合的路徑

第一次未ac代碼

#include<iostream>
#include<cstdio>
#include<string>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<vector>
#include<queue>

using namespace std;

int n,m;
bool cat[100005];
bool visited[100005];
vector<int> tree[100005];
int road=0;

void dfs(int v,int mao){//到當前結點有連續幾隻貓 
    if(mao>m)return;
    int l=tree[v].size();
    if(!l){//leave

        road++;
        return ;
    }
    int w;
    for(int i=0;i<l;i++){
        w=tree[v][i];
        if(cat[w]){
            dfs(w,mao+1);
        }
        else{
            dfs(w,0);
        }
    }
}
int main(){
    scanf("%d%d",&n,&m);
    int d;
    for(int i=1;i<=n;i++){
        scanf("%d",&d);
        cat[i]=d;
    }
    int u,v;
    for(int i=1;i<n;i++){
        scanf("%d%d",&u,&v);
        tree[u].push_back(v);
    }
    if(cat[1]) dfs(1,1);
    else
    dfs(1,0);
    printf("%d\n",road);
    return 0;
}

錯誤原因

• 建樹時默認輸入數據爲根到節點，建立單向圖
  ## 修改 ##
• 建立雙向圖
• 新bug：單向圖時葉節點的判斷採用無出邊即爲葉節點
  -{典型修改之後不記得跟之前做題的思路哪裏有出路，又輸出調試找了好久定位bug}

• 修改：無可走的連通節點即爲葉節點

• 列表內容

最終修改ac代碼

#include<iostream>
#include<cstdio>
#include<string>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<vector>
#include<queue>

using namespace std;

int n,m;
bool cat[100005];
bool visited[100005];
vector<int> tree[100005];
int road=0;

void dfs(int v,int mao){//到當前結點有連續幾隻貓 
     visited[v]=1;
   // cout<<"dfs"<<endl;
    if(mao>m)return;
    int l=tree[v].size();
    /*if(!l){//leave

    }
    */
    bool flag=true;
    int w;
    for(int i=0;i<l;i++){
        w=tree[v][i];
        if(!visited[w])
        {   //cout<<"w: "<<w<<endl; 
             flag=false;
            if(cat[w]){
                dfs(w,mao+1);
            }
            else{
                dfs(w,0);
            }
            visited[w]=1;
        }
    }
    if(flag){//leave
    //  cout<<"resturant:"<<v<<endl;
        road++;
        return ;
    }
}
int main(){
    scanf("%d%d",&n,&m);
    int d;
    for(int i=1;i<=n;i++){
        scanf("%d",&d);
        cat[i]=d;
    }
    int u,v;
    for(int i=1;i<n;i++){
        scanf("%d%d",&u,&v);
        tree[u].push_back(v);
        tree[v].push_back(u);
    }

    if(cat[1]) dfs(1,1);
    else
    dfs(1,0);
    printf("%d\n",road);
    return 0;
}