SDUT Ubiquitous Religions（並查集+哈希）
Ubiquitous Religions
Time Limit: 1000MS Memory limit: 65536K
題目描述
There are so many different religions in the world today that it is difficult to keep track of them all. You are interested in finding out how many different religions students in your university believe in.
You know that there are n students in your university (0 < n <= 50000). It is infeasible for you to ask every student their religious beliefs. Furthermore, many students are not comfortable expressing their beliefs. One
way to avoid these problems is to ask m (0 <= m <= n(n-1)/2) pairs of students and ask them whether they believe in the same religion (e.g. they may know if they both attend the same church). From this data, you may not know what each person believes in, but
you can get an idea of the upper bound of how many different religions can be possibly represented on campus. You may assume that each student subscribes to at most one religion.
輸入
The input consists of a number of cases. Each case starts with a line specifying the integers n and m. The next m lines each consists of two integers i and j, specifying that students i and j believe in the same religion. The students are numbered 1 to n.
The end of input is specified by a line in which n = m = 0.
輸出
For each test case, print on a single line the case number (starting with 1) followed by the maximum number of different religions that the students in the university believe in.
示例輸入
10 9
1 2
1 3
1 4
1 5
1 6
1 7
1 8
1 9
1 10
10 4
2 3
4 5
4 8
5 8
0 0
示例輸出
Case 1: 1
Case 2: 7
提示
給n個大學生，然後給出下面m次詢問，每次兩個數a，b代表編號為a的大學生和編號為b的大學生信仰同一個宗教，問最多有多少種不同的宗教。就是問有多少個不同的集合。
並完掃一遍+hash
#include
#include
#include
#include
#include
using namespace std;
int father[100010],num[100010];
void Make_set(int n)//初始化
{
for(int i=1;i<=n;i++)
father[i]=i;
}
int Find(int x)//查
{
if(x!=father[x])
father[x]=Find(father[x]);//路徑壓縮
return father[x];
}
void Union(int x,int y)//並
{
int fx=Find(x);
int fy=Find(y);
if(fx==fy)
return ;
father[fy]=fx;
}
int max(int x,int y)
{
return x>y?x:y;
}
int main()
{
int n,m,i,x,y,T=1;
while(~scanf("%d%d",&n,&m))
{
if(!n&&!m)break;
memset(num,0,sizeof(num));
Make_set(n);
while(m--)
{
scanf("%d%d",&x,&y);
Union(x,y);
}
for(i=1;i<=n;i++)
{
int f=Find(i);
num[f]++;
}
int cnt=0;
for(i=1;i<=n;i++)
if(num[i])
cnt++;
printf("Case %d: %d\n",T++,cnt);
}
return 0;
}
