547. Friend Circles ------------M ~~~~~~~~~~~!!!

 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.

 

Example 2:

Input: 
[[1,1,0],
 [1,1,1],
 [0,1,1]]
Output: 1
Explanation:The 0th and 1st students are direct friends, the 1st and 2nd students are direct friends, 
so the 0th and 2nd students are indirect friends. All of them are in the same friend circle, so return 1.

 

Constraints:

• 1 <= N <= 200
• M[i][i] == 1
• M[i][j] == M[j][i]

A:

这个的transitive 比较巧妙，我一开始想成 dfs了.  其实下面用的是是BFS

class Solution {
public:
    int findCircleNum(vector<vector<int>>& M) {
        int N = M.size();
        int res = 0;
        for(int i =0; i<N;i++){
            if(M[i][i] != 1) // this number already been visited
                continue;
            res++;
            stack<int> row2check;
            row2check.push(i);
            while(not row2check.empty()){
                int row = row2check.top();
                M[row][row] = -1;
                row2check.pop();                
                for(int j = 0;j<N;j++){
                    if(M[row][j]==1){
                        M[row][j] = -1;
                        row2check.push(j);
                    }
                }
            }
        }
        return res;
    }
};

其实DFS也是可以的。就是找到一个新的数字以后，要dfs 其每一条line