HDU 5326 Work

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Work

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 601    Accepted Submission(s): 401

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

 

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DFS 回溯的时候更新 具体看代码

/* ***********************************************
Author        :
Created Time  :2015/7/29 18:45:22
File Name     :1.cpp
************************************************ */
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <stdio.h>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <math.h>
#include <stdlib.h>
#include <iomanip>
#include <list>
#include <deque>
#include <stack>
#define ull unsigned long long
#define ll long long
#define mod 90001
#define INF 1<<30
#define maxn 10000+10
#define cle(a) memset(a,0,sizeof(a))
const ull inf = 1LL << 61;
const double eps=1e-5;
using namespace std;

bool cmp(int a,int b){
    return a>b;
}
int n,k;
vector<int>vec[maxn];
int sum[maxn];
int vis[maxn];
int fa[maxn];
int first;
void init(){
	for(int i=1;i<=n;i++)
		fa[i]=i;
}
int findfa(int x){
	if(x==fa[x])return x;
	else return fa[x]=findfa(fa[x]);
}

void dfs(int u){
	for(int i=0;i<vec[u].size();i++){
		int v=vec[u][i];
		if(!vis[v]){
			vis[v]=1;
			dfs(v);
			vis[v]=0;
			sum[u]+=sum[v];
			sum[u]++;
		}
	}
}
int main()
{
    #ifndef ONLINE_JUDGE
    freopen("in.txt","r",stdin);
    #endif
    //freopen("out.txt","w",stdout);
	int a,b;
	while(cin>>n>>k){
		cle(sum);
		cle(vis);
		for(int i=0;i<=n;i++)
			vec[i].clear();
		for(int i=1;i<n;i++){
			scanf("%d%d",&a,&b);
			vec[a].push_back(b);
			int x,y;
			x=findfa(a);
			y=findfa(b);
			if(x!=y)fa[x]=y;
		}
		for(int i=1;i<=n;i++)
			if(fa[i]!=i){
				first=i;break;
			}
		dfs(first);// 改成dfs(1)也可以  数据比较弱，并查集是为了找根节点
		int num=0;
		for(int i=1;i<=n;i++){
			if(sum[i]==k)num++;
		}
		printf("%d\n",num);
	}
    return 0;
}

    
    
    
    

     
     
     
     

    
    
    
    

moron