ACM学习历程—HDU 5326 Work(树形递推)

It’s an interesting experience to move from ICPC to work, end my college life and start a brand new journey in company.
As
is known to all, every stuff in a company has a title, everyone except
the boss has a direct leader, and all the relationship forms a tree. If
A’s title is higher than B(A is the direct or indirect leader of B), we
call it A manages B.
Now, give you the relation of a company, can you calculate how many people manage k people.
Each test case begins with two integers n and k, n indicates the number of stuff of the company.
Each of the following n-1 lines has two integers A and B, means A is the direct leader of B.
1 <= n <= 100 , 0 <= k < n
1 <= A, B <= n
这个题目是个递推,不过由于是树形的,需要dfs来完成递推的过程。
关键在于p[now] += p[to]+1;如果now能manage to的话。
此处采用链式前向星来保存关系图。
代码:
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <set>
#include <map>
#include <queue>
#include <string>
#include <algorithm>
#define LL long long using namespace std; const int maxN = ; struct Edge
{
int to, next;
}edge[maxN]; int head[maxN], cnt; void addEdge(int u, int v)
{
edge[cnt].to = v;
edge[cnt].next = head[u];
head[u] = cnt;
cnt++;
} void initEdge()
{
memset(head, -, sizeof(head));
cnt = ;
} int n, k;
int fa[maxN], p[maxN]; void input()
{
initEdge();
memset(p, -, sizeof(p));
int u, v;
for (int i = ; i < n; ++i)
{
scanf("%d%d", &u, &v);
addEdge(u, v);
}
} void dfs(int now)
{
p[now] = ;
int to;
for (int i = head[now]; i != -; i = edge[i].next)
{
to = edge[i].to;
if (p[to] == -)
dfs(to);
p[now] += p[to]+;
}
} void work()
{
int ans = ;
for (int i = ; i <= n; ++i)
{
if (p[i] != -)
{
if (p[i] == k)
ans++;
}
else
{
dfs(i);
if (p[i] == k)
ans++;
}
}
printf("%d\n", ans);
} int main()
{
//freopen("test.in", "r", stdin);
while (scanf("%d%d", &n, &k) != EOF)
{
input();
work();
}
return ;
}
ACM学习历程—HDU 5326 Work(树形递推)的更多相关文章
- ACM学习历程—HDU1041 Computer Transformation(递推 && 大数)
Description A sequence consisting of one digit, the number 1 is initially written into a computer. A ...
- ACM学习历程—ZOJ3777 Problem Arrangement(递推 && 状压)
Description The 11th Zhejiang Provincial Collegiate Programming Contest is coming! As a problem sett ...
- ACM学习历程——HDU4472 Count(数学递推) (12年长春区域赛)
Description Prof. Tigris is the head of an archaeological team who is currently in charge of an exca ...
- ACM学习历程—HDU 5451 Best Solver(Fibonacci数列 && 快速幂)(2015沈阳网赛1002题)
Problem Description The so-called best problem solver can easily solve this problem, with his/her ch ...
- ACM学习历程—HDU 5459 Jesus Is Here(递推)(2015沈阳网赛1010题)
Sample Input 9 5 6 7 8 113 1205 199312 199401 201314 Sample Output Case #1: 5 Case #2: 16 Case #3: 8 ...
- ACM学习历程—HDU 5512 Pagodas(数学)
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=5512 学习菊苣的博客,只粘链接,不粘题目描述了. 题目大意就是给了初始的集合{a, b},然后取集合里 ...
- ACM学习历程—HDU 3915 Game(Nim博弈 && xor高斯消元)
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=3915 题目大意是给了n个堆,然后去掉一些堆,使得先手变成必败局势. 首先这是个Nim博弈,必败局势是所 ...
- ACM学习历程—HDU 5536 Chip Factory(xor && 字典树)
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=5536 题目大意是给了一个序列,求(si+sj)^sk的最大值. 首先n有1000,暴力理论上是不行的. ...
- ACM学习历程—HDU 5534 Partial Tree(动态规划)
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=5534 题目大意是给了n个结点,让后让构成一个树,假设每个节点的度为r1, r2, ...rn,求f(x ...
随机推荐
- MySQL技术内幕InnoDB存储引擎(表&索引算法和锁)
表 4.1.innodb存储引擎表类型 innodb表类似oracle的IOT表(索引聚集表-indexorganized table),在innodb表中每张表都会有一个主键,如果在创建表时没有显示 ...
- 认识 service worker
离线缓存可以提升用户体验,可以节省网络资源,但是,浏览器对资源缓存和自定义网络请求的控制一直不够完善,service worker 的出现就是为了解决这些问题 它可以解决目前离线应用的问题,同时也可以 ...
- Android-Android进程间通讯之messenger
转自‘https://www.cnblogs.com/makaruila/p/4869912.html 平时一说进程间通讯,大家都会想到AIDL,其实messenger和AIDL作用一样,都可以进行进 ...
- Android 红色小圆球提示气泡 BadgeView
今天给大家分享两个实用有简单的一个小圆球提示气泡: BadgeView 参考地址: https://github.com/qstumn/BadgeView; 个人地址:http://git ...
- <转载> pycharm快捷键及一些常用设置
1.编辑(Editing ) Ctrl + Space 基本的代码完成(类.方法.属性)Ctrl + Alt + Space 快速导入任意类Ctrl + Shift + Enter 语句完成Ctrl ...
- Angular入门(一) 环境配置
angular/cli 安装 ♦ npm uninstall -g angular-cli /cnpm install -g angular-cli ※采用npm安装失败: Missing write ...
- python 搜集参数的共有项和所有项
搜集共性项和所有项 ###搜集共有参数值 def intersect(*args): res=[] for x in args[0]: for other in args[1:]: if x not ...
- Redis之java增删改查
jedis是java的redis客户端实现,要使用jedis须要加入jedis的maven依赖: <dependency> <groupId>redis.clients< ...
- MongoDB入门学习(三):MongoDB的增删查改
对于我们这样的菜鸟来说,最重要的不是数据库的管理,也不是数据库的性能,更不是数据库的扩展,而是怎么用好这款数据库,也就是一个数据库提供的最核心的功能,增删查改. 由于M ...
- long_query_time 设置不生效问题
由于原来的慢查询日志太大了,有1G多,并且其中包含上一次查询优化前的慢sql,所以想收集最近两天的慢查询语句,故 mysql> show global variables like 'slow% ...