1107 Social Clusters (30 分)(並查集)

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. A social cluster is 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 integer N (≤1000), the total number of people in a social network. Hence the people are numbered from 1 to N. Then N lines follow, each gives the hobby list of a person in the format:

Ki: hi[1] hi[2] … hi[Ki]

where K**i (>0) is the number of hobbies, and h**i[j] is the index of the j-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

注意這裏考察點：求出每個並查集的個數。我們將它分成兩個問題：

• 如何求並查集的根節點
• 如何在根節點記錄此並查集的個數

以下是代碼：

#include <iostream>
#include <stdio.h>
#include <math.h>
#include <vector>
#include <algorithm>

using namespace std;

const int maxn = 1e3+10;

int k;
int n;
int Rank[maxn];
int parent[maxn];
int cnt[maxn];
vector<int> hobbies[maxn];

int findParent(int x)
{
    int x_root = x;
    while(parent[x_root] != -1)//注意是它的parent不爲空
    {
        x_root = parent[x_root];
    }
    return x_root;
}

int Union(int x, int y)
{
    int x_root = findParent(x);
    int y_root = findParent(y);
    if(x_root != y_root)
    {
        if(Rank[x_root] > Rank[y_root]){
            parent[y_root] = x_root;
            cnt[x_root] += cnt[y_root];
        }
        else if(Rank[x_root] < Rank[y_root]){
            parent[x_root] = y_root;
            cnt[y_root] += cnt[x_root];
        }
        else{
            parent[x_root] = y_root;
            cnt[y_root] += cnt[x_root];
            Rank[y_root]++;
        }
        return 1;
    }
    return 0;
}

int main()
{
    /*1118 Birds in Forest (25 分)*/
    #ifndef ONLINE_JUDGE
        freopen("test.txt", "r", stdin);
    #endif // ONLINE_JUDGE
    scanf("%d", &n);
    fill(Rank, Rank+maxn, 0);
    fill(cnt, cnt+maxn, 1);/*cnt初始化爲1，每個合併的時候，加到根節點上*/
    fill(parent, parent+maxn, -1);
    for(int i = 1; i <= n; i++)
    {
        scanf("%d: ", &k);
        for(int j = 0; j < k; j++)
        {
            int hobby;
            scanf("%d", &hobby);
            hobbies[hobby].push_back(i);
        }
    }
    for(int i = 1; i <= 1000; i++)
    {
        if(hobbies[i].size() > 1)
        {
            int x = hobbies[i][0];
            for(int j = 1; j < hobbies[i].size(); j++)
                Union(x, hobbies[i][j]);
        }
    }
    vector<int> ans;
    for(int i = 1; i <= n; i++)
    {
        if(findParent(i) == i)
        {/*根節點的parent是它自己*/
            ans.push_back(cnt[i]);
        }
    }
    sort(ans.begin(), ans.end());
    cout << ans.size() << endl;
    for(int i = ans.size()-1; i >= 0; i--)
    {
        if(i == 0)
            cout << ans[i];
        else
            cout <<ans[i] << " ";
    }
}

以上代碼可以在parent數組保存內自定義結構體，其中包括其索引和cnt，也就是上述代碼中cnt和parent數組合並。