CodeForces 580C 트리 + dfs 검색

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.
제목의 뜻 해석

나무 하나를 건설하여 뿌리에서 잎 노드까지 하나의 경로

로 만든다.

길에는 m개 이상의'연속consecutive'고양이가 있는 노드가 있을 수 없다

수 에 일치하는 경로가 몇 개 있음

첫 번째 ac 코드 없음

#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){//            
    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
",road);
    return 0;
}

오류 원인

트리 작성 시 기본 입력 데이터는 루트 - 노드이며 단방향 그림 작성 ##수정 ##

양방향 그림 만들기

새로운 버그: 단방향 도면을 볼 때 잎 노드의 판단은 끝이 없는 것이 바로 잎 노드이다. - {전형적인 수정 후에 이전에 문제를 풀었던 사고방식과 어떻게 출구가 있는지 기억하지 못하고 출력 디버깅을 해서 오랫동안 위치를 정했다}

수정: 걸을 수 없는 연결 노드가 바로 잎 노드

목록 내용

최종 수정 ac 코드

#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){//            
     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
",road);
    return 0;
}