Work
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 992 Accepted Submission(s): 613
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.
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
題解:比如1領導2,再去看2能領導幾個人,這裏就可以想到用遞歸做。用個計數器記錄1能領導(包括間接)的人數,記錄下所有人能領導的人數,遍歷一遍就能求出答案。
參考代碼:
#include<stdio.h>
#include<string.h>
struct node
{
int num,s[105];
}res[105];
int ans;
void judge(int i)
{
if(res[i].num==0)
return ;
ans+=res[i].num;
for(int j=0;j<res[i].num;j++)
judge(res[i].s[j]);
}
int main()
{
int n,k,a,b,cnt;
while(~scanf("%d%d",&n,&k))
{
memset(res,0,sizeof(res));
for(int i=0;i<n-1;i++)
{
scanf("%d%d",&a,&b);
res[a].s[res[a].num]=b;
res[a].num++;
}
cnt=0;
for(int i=1;i<=n;i++)
{
ans=res[i].num;
for(int j=0;j<res[i].num;j++)
judge(res[i].s[j]);
if(ans==k)
cnt++;
}
printf("%d\n",cnt);
}
return 0;
}