HDOJ-----5326

Work

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1709    Accepted Submission(s): 1026

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 21 21 32 42 53 63 7

 

Sample Output

2

给出n-1组数x和y，x是y的上级，上级可以直接或间接管理下属，并查集，然后给一个数k，问正好管理k个人的领导有几个，这道题只需要把并查集计算到一半

#include<cstdio>#include<cstring>#define maxn 110000int pre[maxn], vis[maxn];void find(int a){int r = a;while(r != vis[r]){r = vis[r];pre[r]++;}}int main(){    int t, m, n, x, y;    while(~scanf("%d%d", &m, &n)){    for(int i = 0; i <= m; i++){    pre[i] = 0;    vis[i] = i;}for(int i = 1; i < m; i++){scanf("%d%d", &x, &y);vis[y] = x;//标记直接的上下级关系}for(int i = 1; i <= m; i++){find(i);//计算完整的上下级关系，题目也说明了所有人的直接关系都已给出，不用考虑其他情况}int ans = 0;for(int i = 1; i <= m; i++){if(pre[i] == n){ans++;}}printf("%d\n", ans);}    return 0;}