Description
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.
Input
Output
Sample Input
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
Sample Output
Case 1: 1 Case 2: 7
Hint
很簡單的一道題,採用並查集很快即可求解。
代碼如下:
#include<iostream>
#include<cstdio>
#include<algorithm>
using namespace std;
const int MAX_N = 50010;
int set[MAX_N];
int height[MAX_N];
void init(int n)
{
for (int i = 0; i < n;i++)
{
set[i] = i;
height[i] = 1;
}
}
int cha(int x)
{
int r = x;
while (set[r] != r)
r = set[r];
int i = x;
while (set[i] != r)
{
i = set[i];
set[i] = r;
}
return r;
}
void unite(int x, int y)
{
x = cha(x);
y = cha(y);
if (x == y)
{
return;
}
if (height[x] < height[y])
{
set[x] = y;
}
else
{
set[y] = x;
if (height[x] == height[y])
height[x] ++;
}
}
int main()
{
#ifdef ONLINE_JUDGE
#else
freopen("D:\\in.txt", "r", stdin);
freopen("D:\\out.txt", "w", stdout);
#endif
int n(0), m(0);
int a(0), b(0);
int cnt(0);
while (scanf("%d%d",&n,&m)!=EOF)
{
if (n == 0 && m == 0)
break;
init(n+1);
cnt++;
int ans(0);
for (int i = 0; i < m; i++)
{
scanf("%d%d", &a, &b);
unite(a, b);
}
for (int i = 1; i <=n; i++)
{
if (set[i] == i)
ans++;
}
printf("Case %d: %d\n", cnt, ans);
}
return 0;
}