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.
Input
There are multiple test cases.
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
Output
For each test case, output the answer as described above.
Sample Input
7 2
1 2
1 3
2 4
2 5
3 6
3 7
Sample Output
2
这是这场比赛里面最简单的一道题，个人觉的这道题不错，就把它拿来写博客了。
这个问题模拟公司里的职位关系，一个职工它在公司里可能会有下属，也可能会有上司。现在给你n个人，输入n-1有向边，请问拥有下属的个数等于k的有多少人。
分析：很明显这样的关系组出来就是一棵树，每个员工都只会有一个上司，换句话说就是，每一个人都只会有一个父亲节点，除了根节点的父亲是他自己外。
根据n-1条有向边建立节点间的父子关系。每次在新添加一条父子关系事，将当前父亲节点往上回溯到根节点，这些节点都是这个儿子节点的“父亲”或称祖先。
代码：
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <iostream>
#include <string>
#include <vector>
#include <queue>
#include <math.h>
#define eps 1e-8
#include <algorithm>
using namespace std;
int son[200];
int fa[200];
int main()
{
int n, k;
int i, j;
int u, v;
while(scanf("%d %d", &n, &k)!=EOF){
memset(son, 0, sizeof(son));//出事哈每个人都有0个儿子
for(i=1; i<=n; i++){
fa[i]=i; //每个人的父亲是它自己
}
for(i=1; i<n; i++){
scanf("%d %d", &u, &v);
fa[v]=u; //将v的父亲确定为u
son[u]++; //u的儿子数量+1
while( fa[u]!=u ){//如果u的父亲不是他自己 说明u不是根节点
u=fa[u]; //往上面回溯 直到根节点跳出
son[u]++; //经过的这些节点的儿子数全部+1
}
}
int cnt=0;
for(i=1; i<=n; i++)
if(son[i]==k)
cnt++;
printf("%d\n", cnt ); //遍历一下 输出
}
return 0;
}