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
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
解题:一顿乱搞
1 #include <bits/stdc++.h> 2 using namespace std; 3 const int maxn = 100010; 4 struct arc { 5 int to,next; 6 arc(int x = 0,int y = -1) { 7 to = x; 8 next = y; 9 } 10 } e[maxn]; 11 int head[maxn],tot; 12 void add(int u,int v) { 13 e[tot] = arc(v,head[u]); 14 head[u] = tot++; 15 e[tot] = arc(u,head[v]); 16 head[v] = tot++; 17 } 18 int ind[maxn],cnt[maxn],ret,n,k; 19 void dfs(int u,int fa) { 20 cnt[u] = 1; 21 for(int i = head[u]; ~i; i = e[i].next) { 22 if(e[i].to == fa) continue; 23 dfs(e[i].to,u); 24 cnt[u] += cnt[e[i].to]; 25 } 26 if(k + 1 == cnt[u]) ++ret; 27 } 28 int main() { 29 int u,v; 30 while(~scanf("%d%d",&n,&k)) { 31 memset(head,-1,sizeof head); 32 memset(ind,0,sizeof ind); 33 ret = tot = 0; 34 for(int i = 1; i < n; ++i) { 35 scanf("%d%d",&u,&v); 36 add(u,v); 37 ++ind[v]; 38 } 39 for(int i = 1; i <= n; ++i) 40 if(!ind[i]) { 41 dfs(i,-1); 42 break; 43 } 44 printf("%d ",ret); 45 } 46 return 0; 47 }