HDU Problem 5326 Work 【并查集】

Work

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

2015 Multi-University Training Contest 3

 

Recommend

wange2014   |   We have carefully selected several similar problems for you:  5780 5779 5778 5777 5776 

 

将树建好，从子节点依次往上加。顺便给 图片点一个赞。

#include <bitsstdc++.h>
#define MAX_N 105
using 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
", ans);
    }
    return 0;
}