HDU 5326 Work 并查集 (2015 Multi-University Training Contest 3 2015多校联合)

-_- 好吧我不得不承认这道题是签到题。

这道题方法很多，可以用dfs等等，我是用并查集做的。思路很简单，就是求管理k个人的人数有多少。

就是先用一个数组pre来表示该节点的父节点，初始化都为自己，输入的时候直接录入，然后查找，直到找到父节点是自己的结点为止，这样就可以求出每个结点下边的子孙结点有多少个，直接和题目给的数字比较一下然后累加就可以了。

看着代码自己感悟吧。~

Work

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 784    Accepted Submission(s): 504

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

题目链接：HDU5326 Work

#include <iostream>#include <cstdio>#include <cmath>#include <cstring>#include <string>using namespace std;int  pre[105];int t[105];void find(int x){    if(x!=pre[x])    {        t[pre[x]]++;        find(pre[x]);    }    else        return ;}int main(){    int n,k,x,y,ans,i;    while(~scanf("%d%d",&n,&k))    {        ans=0;        for(i=0;i<=n;i++)            pre[i]=i;        memset(t,0,sizeof(t));        for(i=1;i<=n-1;i++)        {            scanf("%d%d",&x,&y);            pre[y]=x;        }        for(i=1;i<=n;i++)        {            if(pre[i]==i)                continue;            else                find(i);        }        for(i=1;i<=n;i++)        {            if(t[i]==k)                ans++;        }        printf("%d\n",ans);    }    return 0;}