Friend Circles

本此的題目是關於深度優先算法(DFS)的應用，Friend Circles，題目如下:

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 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: 2Explanation: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: 1Explanation: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.

主要思路是找尋朋友圈數，透過鄰接矩陣我們可以得出所有人之間的關係，M[i][j] = 1代表i和j是朋友，朋友的朋友也包括在該朋友圈內，由上述給出的例子可以理解這點。

int findCircleNum(vector<vector<int> >& m)
{
int count = 0;
int size = m.size();
vector<bool> visited(size, false);
    if(m.empty())
    {
    return count;
}
for(int i = 0; i < size; i++)
{
if(!visited[i])
{
count++;
dfs(i, visited, m);
}
}
return count;
} 

我的主要想法為透過在 findCircleNum函數內調用dfs函數的次數可作為朋友圈的數量，也可以理解為有多少個連通部件，在基本概念上還是多採用dfs的寫法。以下是本此題目所寫的dfs函數:

void dfs(int i, vector<bool> &visited, vector<vector<int> >& m)
{
visited[i] = true;
for(int j = 0; j < visited.size(); j++)
{
if(i!=j && (m[i][j] == 1))
{
dfs(j, visited, m);
}
}
}

其實目的只要單純遍歷所有節點，所以在每個節點的訪問與利用上部是主要的問題，不必多寫複雜的代碼。