hdu 5326 work 搜索 2015多校联合训练赛-优快云博客

本博客探讨了在给定公司组织结构的情况下，如何计算特定层级下管理一定人数的员工数量。通过构建关系图并进行深度优先搜索，实现算法解决实际问题。

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Work

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 557 Accepted Submission(s): 376

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

Sample Output

Source

2015 Multi-University Training Contest 3

求子树中结点树恰好为k（不包括子树的根）的结点个数

建图，直接深搜即可

#include<iostream>
#include<cstdio>
#include<cstring>
#include<vector>
using namespace std;

vector<int>haha[200];
int check[200];
int num[200];
int dfs(int u){
    if(check[u]) return num[u]+1;
    check[u] = 1;
    for(int i = 0; i < haha[u].size();i++)
        num[u] += dfs(haha[u][i]);
    return num[u]+1;
}


int main(){
    int n,k;
    while(scanf("%d%d",&n,&k)!=EOF){
        for(int i = 0;i <= n; i++)
            haha[i].clear();
        int u,v;
        for(int i =1;i < n; i++){
            scanf("%d%d",&u,&v);
            haha[u].push_back(v);
        }
        memset(check,0,sizeof(check));
        memset(num,0,sizeof(num));
        int ans = 0;
        for(int i = 1;i <= n; i++){

            if(check[i]) continue;
            dfs(i);
        }
        for(int i =1;i <= n; i++)
            if(num[i]==k) ans++;
        cout<<ans<<endl;
    }
    return 0;
}