2015 Multi-University Training Contest 3 hdu 5326 Work

Work

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 306    Accepted Submission(s): 217

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

 

Source

2015 Multi-University Training Contest 3

 

解题：一顿乱搞

 

 #include <bits/stdc++.h>
 using namespace std;
 const int maxn = ;
 struct arc {
     int to,next;
     arc(int x = ,int y = -) {
         to = x;
         next = y;
     }
 } e[maxn];
 int head[maxn],tot;
 void add(int u,int v) {
     e[tot] = arc(v,head[u]);
     head[u] = tot++;
     e[tot] = arc(u,head[v]);
     head[v] = tot++;
 }
 int ind[maxn],cnt[maxn],ret,n,k;
 void dfs(int u,int fa) {
     cnt[u] = ;
     for(int i = head[u]; ~i; i = e[i].next) {
         if(e[i].to == fa) continue;
         dfs(e[i].to,u);
         cnt[u] += cnt[e[i].to];
     }
     if(k +  == cnt[u]) ++ret;
 }
 int main() {
     int u,v;
     while(~scanf("%d%d",&n,&k)) {
         memset(head,-,sizeof head);
         memset(ind,,sizeof ind);
         ret = tot = ;
         for(int i = ; i < n; ++i) {
             scanf("%d%d",&u,&v);
             add(u,v);
             ++ind[v];
         }
         for(int i = ; i <= n; ++i)
             if(!ind[i]) {
                 dfs(i,-);
                 break;
             }
         printf("%d\n",ret);
     }
     return ;
 }