HDU 5326 work 【并查集】
Work
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 1353 Accepted Submission(s): 813
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.
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
7 2 1 2 1 3 2 4 2 5 3 6 3 7
2
题目大意就是求子集节点数是K的有几个
然后做着做着就出来了
/*
#include <iostream>
#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;
int has[2][105][105];
int son[105];
int n,m,sum;
int fin(int x)
{
int num=0;
for(int j=1; j<=n; j++)
{
if(has[1][x][j]==1&&has[0][x][j]==0)
{
num++;
has[0][x][j]=1;
son[x]+=fin(j);
}
}
return son[x];
}
int main ()
{
while(~scanf("%d%d",&n,&m))
{
memset(has,0,sizeof(has));
memset(son,0,sizeof(son));
int a,aa;
sum=0;
for(int i=1; i<n; i++)
{
scanf("%d%d",&a,&aa);
has[1][a][aa]=1;
son[a]++;
}
for(int i=1; i<=n; i++)
{
for(int j=1; j<=n; j++)
{
if(has[1][i][j]==1&&has[0][i][j]==0)
{
has[0][i][j]=1;
son[i]+=fin(j);
}
}
}
/* for(int i=1; i<=n; i++)
{
printf("**%d\n",son[i]);
}
for(int i=1; i<=n; i++)
{
if(m==son[i])
sum++;
}
printf("%d\n",sum);
}
}*/
#include<stdio.h>
#include<string.h>
using namespace std;
int f[105];
int jihe[105];
void find2(int a)
{
int r=a;
while(f[r]!=r)
{
r=f[r];//r是一个上司。
jihe[r]++;//这个上司有一个员工。
}
}
int main()
{
int n,m;
while(~scanf("%d%d",&n,&m))
{
for(int i=1;i<=n;i++)
{
f[i]=i;
jihe[i]=0;
}
for(int i=0;i<n-1;i++)
{
int x,y;
scanf("%d%d",&x,&y);
f[y]=x;//让并查集数组指向上司
}
int output=0;
for(int i=1;i<=n;i++)
{
find2(i);//枚举每一个人，让他找他的上司
}
for(int i=1;i<=n;i++)
{
if(jihe[i]==m)
output++;//符合条件的输出
}
printf("%d\n",output);
}
}
//附几组数据
/*
7 6
1 2
1 3
2 4
2 5
3 6
3 7
7 2
1 2
1 3
2 4
2 5
3 6
3 7
7 2
7 6
7 5
6 4
6 3
5 2
5 1
7 1
1 2
2 3
3 4
4 5
5 6
6 7
*/