HDOJ 5326 Work

Work

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1707    Accepted Submission(s): 1025

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 21 21 32 42 53 63 7

 

Sample Output

2

 

Author

ZSTU

 

Source

2015 Multi-University Training Contest 3

 

题意：我们都知道，所有的员工都有个头衔，每个头衔为BOSS的人就是领导。所以，如果A比B的的头衔大，我们就可以说A管理B。现在给你一个公司的关系图，问你有多少人管理着K个人。

第一行n表示人数，k不用说。后面n-1行表示关系，前面的是A后面的是B。

题解：这个题用的也是模拟压缩路径！每一次压缩新的节点的时候，不断往上查找，让它的父节点都+1表示增加了一个子节点。把每个节点连接的子节点个数存在数组里，最后直接取用。

#include<stdio.h>#include<string.h>int per[111],num[111];int Find(int x){int r=x;while(r!=per[r])r=per[r];int i=x,j;while(i!=r){j=per[i];num[j]++;i=j;}return r;}int main(){int i,a,b,n,k;while(~scanf("%d%d",&n,&k)){memset(num,0,sizeof(num));for(i=1;i<=n;i++)per[i]=i;for(i=1;i<n;i++){scanf("%d%d",&a,&b);per[b]=a;}for(i=1;i<=n;i++)Find(i);int ans=0;for(i=1;i<=n;i++){if(num[i]==k)ans++;}printf("%d\n",ans);}return 0;}