HDU 5326 Work

WorkTime limit:2000/1000 MS (java/others) Memory limit:32768/32768 K (java/others)
Total submission (s): 549 Accepted Submission (s): 375


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 have a title, everyone except the boss have a direct leader, and all the Relati Onship forms a tree. If a ' s title is higher than B (A was the direct or indirect leader of B), we call it A manages B.
Now, give you the relation for a company, can-calculate how many people manage k people.

Inputthere is multiple test cases.
Each test case is begins with a integers n and k, n indicates the number of stuff of the company.
Each of the following n-1 lines have both integers a and B, means a is the direct leader of B.

1 <= N <=, 0 <= k < n
1 <= A, B <= N

Outputfor each test case, output the answer as described above.
Sample Input

7 21 21 32 42 53 63 7

Sample Output

2

Source2015 multi-university Training Contest 3

#include <cstdio> #include <cstring> #include <algorithm> #include <iostream>using namespace    std;struct node{int u;    int V; int next;}    Edge[100000];int cnt;int head[1000],vis[1000],sum[1000];void Add (int u,int v) {edge[cnt].u=u;    Edge[cnt].v=v;    Edge[cnt].next=head[u]; head[u]=cnt++;}    void Dfs (int u) {vis[u]=1;        for (int i=head[u];i!=-1;i=edge[i].next) {int v=edge[i].v;            if (vis[v]==0) {sum[u]++;            DFS (v);        SUM[U]+=SUM[V];    }}}int Main () {int n,k,i,a,b,v;         while (scanf ("%d%d", &n,&k)!=eof) {cnt=0;        memset (head,-1,sizeof (head));          for (i=1;i<=n-1;i++) {scanf ("%d%d", &a,&b);        Add (A, b);         } memset (Sum,0,sizeof (sum));        memset (vis,0,sizeof (VIS));            for (i=1;i<=n;i++) {if (vis[i]==0) {DFS (i);        }} v=0; for (i=1;i<=n;i++)       {if (sum[i]==k) {v++;    }} printf ("%d\n", V); } return 0;}

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HDU 5326 Work