hdu5326 树形dp

C - Work

Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u

Submit Status Practice HDU 5326

Appoint description: System Crawler  (Aug 24, 2016 8:38:50 AM)

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 21 21 32 42 53 63 7

Sample Output

2

题意：

在一个公司里，A - B B-C 表示A是B的直接上级，B是C的直接上级，A是C的间接上级，不管是直接上级环视间接上级

都属于A的管辖人员，有n个员工n-1条关系（树形关系网），问管辖K个人的人有多少个；

树形dp嘛，问几个节点有K个孩子，但是有一个坑，有可能有多个树根，有可能 A-B C-B ，这种关系，A C 关系，但是

A C 都管辖B，所以要在每个树根处都要进行dfs，每棵树上的管辖k个人的节点都要计算入结果

#include <iostream>#include <vector>#include <stdio.h>#include <algorithm>#include <string.h>using namespace std;const int maxn=1000;vector<int>edge[maxn];int ru[maxn];int leaf[maxn];void dfs(int u,int pre){    if(edge[u].size()==0)    {        return;    }    for(int i=0; i<edge[u].size(); i++)    {        int v=edge[u][i];        if(v==pre)            continue;        dfs(v,u);        leaf[u]+=leaf[v];    }}int main(){    int n,k;    int u,v;    while(scanf("%d%d",&n,&k)!=EOF)    {        memset(ru,0,sizeof(ru));        for(int i=1; i<=n; i++)        {            edge[i].clear();            leaf[i]=1;        }        for(int i=1; i<n; i++)        {            scanf("%d%d",&u,&v);            {                ru[v]++;                edge[u].push_back(v);               // edge[v].push_back(u);            }        }        int ans=0;        int root;        for(int i=1; i<=n; i++)        {            if(ru[i]==0)            {                for(int i=1;i<=n;i++)                    leaf[i]=1;                dfs(i,0);                for(int i=1;i<=n;i++)                {                      //cout<<i<<" "<<leaf[i]-1<<" ";                    if(leaf[i]-1==k)                        ans++;                }            }        }        printf("%d\n",ans);    }    return 0;}