链接：http://acm.hdu.edu.cn/showproblem.php?pid=5326
Work
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1147 Accepted Submission(s): 695
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
直接暴力就OK
#include <iostream>
#include <cstdio>
#include <cstring>
#include <stack>
#include <queue>
#include <map>
#include <set>
#include <vector>
#include <cmath>
#include <algorithm>
using namespace std;
const double eps = 1e-6;
const double pi = acos(-1.0);
const int INF = 0x3f3f3f3f;
const int MOD = 1000000007;
int s[105][105],f[105];
int n,k;
int cnt,ans;
void solve(int m)
{
int t = m;
if (t>=n||f[t]==0)
return ;
for (int i=1; i<=n; i++)
{
if (s[t][i])
{
cnt+=f[i];
solve(i);
}
}
}
int main()
{
int a,b;
while (scanf ("%d%d",&n,&k)==2)
{
memset(s, 0, sizeof(s));
memset(f, 0, sizeof(f));
for (int j=1; j<n; j++)
{
scanf ("%d%d",&a,&b);
s[a][b] = 1;
f[a]++;
}
cnt=0,ans=0;
for (int i=1; i<=n; i++)
{
cnt = f[i];
solve(i);
if (cnt == k)
ans++;
}
printf ("%d\n",ans);
}
return 0;
}
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