Problem Description
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 = 105;
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, -1, sizeof(head));
cnt = 0;
}
int n, k;
int fa[maxN], p[maxN];
void input()
{
initEdge();
memset(p, -1, sizeof(p));
int u, v;
for (int i = 1; i < n; ++i)
{
scanf("%d%d", &u, &v);
addEdge(u, v);
}
}
void dfs(int now)
{
p[now] = 0;
int to;
for (int i = head[now]; i != -1; i = edge[i].next)
{
to = edge[i].to;
if (p[to] == -1)
dfs(to);
p[now] += p[to]+1;
}
}
void work()
{
int ans = 0;
for (int i = 1; i <= n; ++i)
{
if (p[i] != -1)
{
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 0;
}