# (BFS DFS 并查集) leetcode 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 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.

BFS

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;
}
};

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;
}
};

C++代码：

class Solution {
public:
int getRoot(vector<int> &root,int i){
while(i != root[i]){
root[i] = root[root[i]];
i = root[i];
}
return i;
}
int findCircleNum(vector<vector<int>>& M) {
int m = M.size();
int res = m;
vector<int> root(m);
for(int i = 0; i < m; i++) root[i] = i;
for(int i = 0; i < m; i++){
for(int j = i + 1; j < m; j++){
if(M[i][j] == 1){
int a = getRoot(root, i);
int b = getRoot(root, j);
if(a != b){
root[a] = b;
res--;
}
}
}
}
return res;
}
};

posted @ 2019-05-17 16:27  PJCK  阅读(94)  评论(0编辑  收藏