CodeForces 580C 树+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.

题意解析

• 建成一颗树从根出发到叶节点为一条路径
• 路上不能有大于m个“连续consecutive”有猫的节点
• 数有几条符合的路径

第一次未ac代码

#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){//到当前结点有连续几只猫     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;}

错误原因

• 建树时默认输入数据为根到节点，建立单向图
  ## 修改 ##
• 建立双向图
• 新bug：单向图时叶节点的判断采用无出边即为叶节点
  -{典型修改之后不记得跟之前做题的思路哪里有出路，又输出调试找了好久定位bug}

• 修改：无可走的连通节点即为叶节点

• 列表内容

最终修改ac代码

#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){//到当前结点有连续几只猫      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;}