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 = 100010;
struct arc {
    int to,next;
    arc(int x = 0,int y = -1) {
        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] = 1;
    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 + 1 == cnt[u]) ++ret;
}
int main() {
    int u,v;
    while(~scanf("%d%d",&n,&k)) {
        memset(head,-1,sizeof head);
        memset(ind,0,sizeof ind);
        ret = tot = 0;
        for(int i = 1; i < n; ++i) {
            scanf("%d%d",&u,&v);
            add(u,v);
            ++ind[v];
        }
        for(int i = 1; i <= n; ++i)
            if(!ind[i]) {
                dfs(i,-1);
                break;
            }
        printf("%d
",ret);
    }
    return 0;
}