Work
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 230 Accepted Submission(s): 171
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
Source
Recommend
1 #include <stdio.h> 2 #include <string.h> 3 #include <algorithm> 4 using namespace std; 5 6 int n,k,s; 7 int a[105][105]; 8 9 int dfs(int g) 10 { 11 for(int i=1;i<=n;i++) 12 { 13 if(a[g][i]==1) 14 { 15 s++; 16 dfs(i); 17 } 18 } 19 return 0; 20 } 21 22 int main() 23 { 24 int i,j,x,y; 25 while(scanf("%d %d",&n,&k)!=EOF) 26 { 27 memset(a,0,sizeof(a)); 28 for(i=1;i<=n-1;i++) 29 { 30 scanf("%d %d",&x,&y); 31 a[x][y]=1; 32 } 33 int ans=0; 34 for(i=1;i<=n;i++) 35 { 36 s=0; 37 dfs(i); 38 if(s==k) 39 ans++; 40 } 41 printf("%d ",ans); 42 } 43 }