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
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
Recommend
#include <bits\stdc++.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\n", ans); } return 0; }