Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 1665    Accepted Submission(s): 995

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

 

 

Author

ZSTU

 

 

Source

 

 

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将树建好，从子节点依次往上加。顺便给 图片点一个赞。

#include 
      
       #define MAX_N 105using namespace std;const int INF = 1e9;const double ESP = 1e-5;int par[MAX_N], num[MAX_N];int  n, k;void init() {    for (int i = 0; i < 104; i++) {        par[i] = i; num[i] = 0;    }}void unite(int x, int y) {    par[y] = x;}void solve(int x) {    int t = x;    while (t != par[t]) {        num[par[t]]++;        t = par[t];    }}int main() {    int a, b;    while (scanf("%d%d", &n, &k) != EOF) {        init(); int ans = 0;        for (int i = 0; i < n - 1; i++) {            scanf("%d%d", &a, &b);            unite(a, b);        }        for (int i = 1; i <= n; i++) {           solve(i);        }        for (int i = 1; i <= n; i++) {            if (num[i] == k) ans++;        }        printf("%d\n", ans);    }    return 0;}