PAT 1107 Social Clusters(30分)

1107 Social Clusters(30分)

When register on a social network, you are always asked to specify your hobbies in order to find some potential friends with the same hobbies. Asocial clusteris a set of people who have some of their hobbies in common. You are supposed to find all the clusters.

Input Specification:

Each input file contains one test case. For each test case, the first line contains a positive integerN(≤1000), the total number of people in a social network. Hence the people are numbered from 1 toN. ThenNlines follow, each gives the hobby list of a person in the format:

K​i​​:h​i​​[1]h​i​​[2] ...h​i​​[K​i​​]

whereK​i​​(>0) is the number of hobbies, andh​i​​[j]is the index of thej-th hobby, which is an integer in [1, 1000].

Output Specification:

For each case, print in one line the total number of clusters in the network. Then in the second line, print the numbers of people in the clusters in non-increasing order. The numbers must be separated by exactly one space, and there must be no extra space at the end of the line.

Sample Input:

8
3: 2 7 10
1: 4
2: 5 3
1: 4
1: 3
1: 4
4: 6 8 1 5
1: 4

Sample Output:

3
4 3 1

思路

• 使用course[h]记录喜欢活动h的人的编号，使用findFather(course[h])查找这个人所在集合的根结点。最后合并，i 与 findFather(course[h])合并。

代码

#include 
#include 

using namespace std;

const int maxn = 1010;

int father[maxn];
int isRoot[maxn];
int course[maxn];

int n;

bool cmp(int a, int b)
{
    return a > b;    
}
void init(int n)
{
    for (int i = 1; i <= n; i ++)
        father[i] = i;
}

int findFather(int x)
{
    while (x != father[x])
    {
        x = father[x];
    }
    return x;
}

void Union(int a, int b)
{
    int faA = findFather(a);
    int faB = findFather(b);
    if (faA != faB)
        father[faB] = faA;
}

int main()
{
    //freopen("test.txt", "r", stdin);

    scanf("%d", &n);
    init(n);
    int k, h;
    for (int i = 1; i <= n; i ++)
    {
        scanf("%d:", &k);
        for (int j = 1; j <= k; j ++)
        {
            scanf("%d", &h);
            if (course[h] == 0)
                course[h] = i;
            Union(i, findFather(course[h]));
        }
    }
    for (int i = 1; i <= n; i ++)
    {
        isRoot[findFather(i)] ++;
    }
    
    int sum = 0;
    for (int i = 1; i <= n; i ++)
    {
        if (isRoot[i] != 0)
            sum ++;
    }
    
    sort(isRoot, isRoot + n + 1, cmp);  //注意isRoot中包括下标n，因此排序的最后一个位置要加1
    printf("%d\n", sum);
    for (int i = 0; i < sum; i ++)
    {
        printf("%d", isRoot[i]);
        if (i < sum - 1)
            printf(" ");
    }
    
    return 0;
}

参考资料：《算法笔记》.