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.
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
For each test case, output the answer as described above.
1 #include <iostream>
2 #include <cstdio>
3 #include <cstring>
4 #include <
string>
5 #include <cmath>
6 #include <algorithm>
7 #include <vector>
8 #include <queue>
9 #include <
set>
10 #include <map>
11 #include <stack>
12 #include <limits.h>
13 using namespace std;
14 typedef
long long LL;
15 #define y1 y234
16 #define MAXN 1000010
// 1e617 vector <
int> edge[MAXN];
18 int n, k;
19 int sum[MAXN];
20 void DFS(
int u) {
21 int res =
0;
22 int len = edge[u].size();
23 for(
int i =
0; i < len; i++) {
24 int v = edge[u][i];
25 if(!sum[v]) {
26 DFS(v);
27 }
28 res++;
29 res += sum[v];
30 }
31 sum[u] = res;
32 }
33 int main() {
34 while(~scanf(
"%d%d", &n, &k)) {
35 int ans =
0;
36 memset(sum,
0,
sizeof(sum));
37 for(
int i =
0; i <= n; i++) edge[i].clear();
38 for(
int i =
0; i < n -
1; i++) {
39 int a, b;
40 scanf(
"%d%d",&a, &b);
41 edge[a].push_back(b);
42 }
43 for(
int i =
1; i <= n; i++) {
44 if(sum[i])
continue;
45 DFS(i);
46 }
47 for(
int i =
1; i <= n; i++) {
48 if(sum[i] == k) ans++;
49 }
50 printf(
"%d\n", ans);
51 }
52 return 0;
