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 directfriend 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 directfriends 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.

 

Note:

1. N is in range [1,200].
2. M[i][i] = 1 for all students.
3. If M[i][j] = 1, then M[j][i] = 1.

 

这个题的解法是有不唯一的。还有，这个题与那个统计油田的个数的BFS和DFS的模板题是不一样的。

参考链接：http://www.cnblogs.com/grandyang/p/6686983.html

第一种解法：

BFS

对于某个人，我们可以先遍历其好友（如果他有好友），然后再遍历好友的好友，一直继续下去，直到不能继续找到好友为止，可以找出一个朋友圈内的所有好友。接下来再对于还没有遍历的人，继续同样的方法，直到遍历完我所有人为止，此时，我们可以统计朋友圈的个数。注意，每当遍历到人时，要标记已经遍历过的。

 

C++代码：

class Solution {
public:
    int findCircleNum(vector<vector<int>>& M) {
        queue<int> q;
        int m = M.size();
        int res = 0;
        vector<bool> vis(m,false);
        for(int i = 0; i < m; i++){
            if(vis[i] == true) continue;  //如果已经遍历过，就略过，因为这个人所在的朋友圈已经遍历了。
            q.push(i);
            while(!q.empty()){
                int t = q.front();q.pop();
                vis[t] = true;
                for(int j = 0; j < m; j++){
                    if(M[t][j] && !vis[j]){
                        q.push(j);
                    }           
                }
            }
            res++;
        }
        return res;
    }
};

 

 

第二种解法：

DFS，思路和上面的BFS类似。

class Solution {
public:
    void dfs(vector<vector<int>> &M,int i,vector<bool> &vis){
        int m = M.size();
        vis[i] = true;  //把遍历到的人标记为已经遍历的。
        for(int j = 0; j < m; j++){
            if(M[i][j] == 1 && !vis[j]){
                dfs(M,j,vis);
            }
        }
    }
    int findCircleNum(vector<vector<int>>& M) {
        int m = M.size();
        int res = 0;
        vector<bool> vis(m,false);
        for(int i = 0; i < m; i++){
            if(vis[i] == true) continue;
            dfs(M,i,vis);
            res++; //一路“莽”到底后，递增res，指的是朋友圈的个数的增加。
        }
        return res;
    }
};