547. Friend Circles

There are N students in a class. Some of them are friends, while some are not. Their friendship is transitive in nature. For example, if A is a direct friend of B, and B is a direct friend of C, then A is an indirect friend of C. And we defined a friend circle is a group of students who are direct or indirect friends.

Given a N*N matrix M representing the friend relationship between students in the class. If M[i][j] = 1, then the ith and jth students are direct friends with each other, otherwise not. And you have to output the total number of friend circles among all the students.

Example 1:

Input: 
[[1,1,0],
 [1,1,0],
 [0,0,1]]
Output: 2
Explanation:The 0th and 1st students are direct friends, so they are in a friend circle. 
The 2nd student himself is in a friend circle. So return 2.

统计朋友圈个数。

解决思路：用Union-Find做。初始化res等于总人数，index数组储存所属朋友圈，初始化每个人属于自己的朋友圈。如果两个人是朋友，首先Find，如果属于不同的朋友圈，就Union，且res减1。

class Solution {
public:
    int Find(vector<int>& index, int i) {
        while(index[i]!=i)
            i = index[i];
        return i;
    }
    void Union(vector<int>& index, int i1, int i2) {
        for(int i=0; i<index.size(); ++i)
            if(index[i]==i1)
                index[i] = i2;
    }
    int findCircleNum(vector<vector<int>>& M) {
        int len = M.size();
        int res = len;
        vector<int> index;
        for(int l=0; l<len; ++l)
            index.push_back(l);
        int i,j;
        for(i=0; i<len; ++i)
            for(j=i+1; j<len; ++j)
                if(M[i][j] && Find(index, i)!=Find(index, j)) {
                    --res;
                    Union(index, Find(index, i), Find(index, j));
                }
       return res;     
    }
};