Work

J - Work

Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u

Submit  Status

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

</pre><pre name="code" class="cpp">//1:(并查集）  
#include<stdio.h>  
#include<string.h>  
#include<algorithm>  
using namespace std;  
const int maxm=205;  
int p[maxm];  
int num[maxm];  
int n,k;  
void Init()  
{  
    for(int i=1;i<=n;i++)  
    {  
        p[i]=i;  
        num[i]=0;  
    }  
    return;  
}  
void find(int x)  
{  
    if(p[x]!=x)  
    {  
        num[p[x]]++;  
        find(p[x]);  
        return;  
    }  
}  
int main()  
{  
    while(scanf("%d%d",&n,&k)!=EOF)  
    {  
        Init();  
        int cnt=0;  
        for(int i=1; i<=n-1; i++)  
        {  
            int x,y;  
            scanf("%d%d",&x,&y);  
            p[y]=x;  
        }  
        for(int i=1;i<=n;i++)  
        {  
            find(i);  
        }  
        for(int i=1; i<=n; i++)  
        {  
           if(num[i]==k)  
           {  
               cnt++;  
           }  
        }  
        printf("%d\n",cnt);  
    }  
    return 0;  
}  
//2:(看网上代码，不晓得这是啥方法，好理解）  
#include<stdio.h>  
#include<string.h>  
#include<algorithm>  
using namespace std;  
const int maxm=105;  
int vis[maxm][maxm];  
int num[maxm];  
int main()  
{  
    int n,r;  
    while(scanf("%d%d",&n,&r)!=EOF)  
    {  
        int x,y;  
        int cnt=0;  
        memset(vis,0,sizeof(vis));  
        memset(num,0,sizeof(num));  
        for(int i=1;i<=n-1;i++)  
        {  
            scanf("%d%d",&x,&y);  
            vis[x][y]=1;  
            num[x]++;  
        }  
        for(int i=1;i<=n;i++)  
        {  
            for(int j=1;j<=n;j++)  
            {  
                if(vis[i][j])  
                {  
                    for(int k=1;k<=n;k++)  
                    {  
                        if(vis[j][k])  
                        {  
                            vis[i][k]=1;  
                        }  
                    }  
                }  
            }  
        }  
        for(int i=1;i<=n;i++)  
        {  
            for(int j=1;j<=n;j++)  
            {  
                if(vis[i][j])  
                {  
                    num[i]+=num[j];  
                }  
            }  
        }  
        for(int i=1;i<=n;i++)  
        {  
            if(num[i]==r)  
            {  
                cnt++;  
            }  
        }  
        printf("%d\n",cnt);  
    }  
    return 0;  
}