给出一棵树,问每一层各有多少叶子节点

dfs遍历树

#include<bits/stdc++.h>

using namespace std;
vector<int>p[];
int n,m;
int node ,k;
int vis[];
int maxn=-;
void dfs(int node,int step)
{
if(p[node].empty()){
vis[step]++;
maxn=max(step,maxn);
}
else{
for(int i=;i<p[node].size();i++){
dfs(p[node][i],step+);
}
}
}
int main()
{
ios::sync_with_stdio(false);
cin>>n>>m;
for(int i=;i<m;i++){
cin>>node>>k;
int t;
for(int j=;j<k;j++){
cin>>t;
p[node].push_back(t);
}
}
dfs(,);
cout<<vis[];
for(int i=;i<=maxn;i++){
cout<<" "<<vis[i];
}
cout<<endl;
return ;
}

bfs遍历求树

 #include<bits/stdc++.h>

 using namespace std;
struct NODE
{
int data;
int step;
NODE(){}
NODE(int data1,int step1):data(data1),step(step1){}
};
vector<int>p[];
int n,m;
int node ,k;
int vis[];
int maxn=-;
void bfs(int node,int step)
{
queue<NODE>Q;
Q.push(NODE(node,step));
while(!Q.empty()){
NODE u=Q.front();
Q.pop();
if(p[u.data].empty()){
maxn=max(u.step,maxn);
vis[u.step]++;
}
for(int i=;i<p[u.data].size();i++){
Q.push(NODE(p[u.data][i],u.step+));
}
}
}
int main()
{
ios::sync_with_stdio(false);
cin>>n>>m;
for(int i=;i<m;i++){
cin>>node>>k;
int t;
for(int j=;j<k;j++){
cin>>t;
p[node].push_back(t);
}
}
bfs(,);
cout<<vis[];
for(int i=;i<=maxn;i++){
cout<<" "<<vis[i];
}
cout<<endl;
return ;
}

1004 Counting Leaves (30 分)(树的遍历)的更多相关文章

  1. 1004 Counting Leaves (30分) DFS

    1004 Counting Leaves (30分)   A family hierarchy is usually presented by a pedigree tree. Your job is ...

  2. PAT 1004 Counting Leaves (30分)

    1004 Counting Leaves (30分) A family hierarchy is usually presented by a pedigree tree. Your job is t ...

  3. 【PAT甲级】1004 Counting Leaves (30 分)(BFS)

    题意:给出一棵树的点数N,输入M行,每行输入父亲节点An,儿子个数n,和a1,a2,...,an(儿子结点编号),从根节点层级向下依次输出当前层级叶子结点个数,用空格隔开.(0<N<100 ...

  4. 1004 Counting Leaves (30 分)

    A family hierarchy is usually presented by a pedigree tree. Your job is to count those family member ...

  5. PTA 1004 Counting Leaves (30)(30 分)(dfs或者bfs)

    1004 Counting Leaves (30)(30 分) A family hierarchy is usually presented by a pedigree tree. Your job ...

  6. 1004. Counting Leaves (30)

    1004. Counting Leaves (30)   A family hierarchy is usually presented by a pedigree tree. Your job is ...

  7. PAT 解题报告 1004. Counting Leaves (30)

    1004. Counting Leaves (30) A family hierarchy is usually presented by a pedigree tree. Your job is t ...

  8. PAT Advanced 1004 Counting Leaves (30) [BFS,DFS,树的层序遍历]

    题目 A family hierarchy is usually presented by a pedigree tree. Your job is to count those family mem ...

  9. PAT 1004. Counting Leaves (30)

    A family hierarchy is usually presented by a pedigree tree.  Your job is to count those family membe ...

  10. PAT A 1004. Counting Leaves (30)【vector+dfs】

    题目链接:https://www.patest.cn/contests/pat-a-practise/1004 大意:输出按层次输出每层无孩子结点的个数 思路:vector存储结点,dfs遍历 #in ...

随机推荐

  1. Java nio socket与as3 socket(粘包解码)连接的应用实例

    对Java nio socket与as3 socket连接的简单应用 <ignore_js_op>Java nio socket与as3 socket连接的应用实例.rar (9.61 K ...

  2. an exception occurred while initializing the database.

    对于手动删除本地的LocalDB数据库之后出现标题所示异常的,推荐下面的命令: sqllocaldb.exe stop v11.0 sqllocaldb.exe delete v11.0 在程序包管理 ...

  3. CUDA中多维数组以及多维纹理内存的使用

    纹理存储器(texture memory)是一种只读存储器,由GPU用于纹理渲染的图形专用单元发展而来,因此也提供了一些特殊功能.纹理存储器中的数据位于显存,但可以通过纹理缓存加速读取.在纹理存储器中 ...

  4. 5.Spring Cloud初相识-------Hystrix熔断器

    前言: 1.介绍Hystrix 在一个分布式系统里,许多依赖不可避免的会调用失败,比如超时.异常等,如何能够保证在一个依赖出问题的情况下,不会导致整体服务失败,这个就是Hystrix需要做的事情.Hy ...

  5. 微信小程序日期选择器

    /* JS代码部分 */ const date = new Date() const years = [] const months = [] const days = [] const hours ...

  6. javascript常用代码片段

    /** * * @desc 判断两个数组是否相等 * @param {Array} arr1 * @param {Array} arr2 * @return {Boolean} */ function ...

  7. Python3.6+pyinstaller+Django

    方案(一)Python3.6+pyinstaller+windows服务 一.Python3.6(64位)环境清单 Django==1.11.7 django-windows-tools==0.2 P ...

  8. python3爬取咪咕音乐榜信息(附源代码)

    参照上一篇爬虫小猪短租的思路https://www.cnblogs.com/aby321/p/9946831.html,继续熟悉基础爬虫方法,本次爬取的是咪咕音乐的排名 咪咕音乐榜首页http://m ...

  9. 裸机——210SD卡启动

    1.通过阅读iROM_Application_note可以获取关于启动的全部信息 2.记录下代码 制作SD卡启动的代码,即添加校验和的 #include <strings.h> #incl ...

  10. POJ-2251 三维迷宫

    题目大意:给一个三维图,可以前后左右上下6种走法,走一步1分钟,求最少时间(其实就是最短路) 分析:这里与二维迷宫是一样的,只是多了2个方向可走,BFS就行(注意到DFS的话复杂度为O(6^n)肯定会 ...