边刷leetcode边学编程-547Friend Circles

547. Friend Circles

题目描述

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.

想到的是用图来做。

图的遍历

设无向图G，其中所有顶点标记为没有访问过。
首先选择G的一个顶点$\nu$ 为已经访问过。然后递归地搜索。如果还有没有访问过的，则再选一次没访问过的顶点开始访问，重复直至都访问一次为止。

深度优先

• 伪代码
  DFSTraverse(G){
  count = 1;
  for(all $\nu \in V$)
  mark v”new”;
  while(if $\nu \in V$ mark new)
  search(v);
  }
  void search(v){
  dfn[v] = count;
  count++;
  mark v “old”;
  for(each $\omega \in L[\nu]$)
  if($\omega$ is “new”){
  search(w);
  }
  }

---

或者
当到达某一顶点$\nu$时，依次考查所有相关的没有被访问过的顶点，然后再继续访问下一个顶点。

广度优先

• 伪代码
  BFSTraverse(G){
  count = 1;
  for(all $\nu \in V$)
  mark $\nu$ “new;
  while($\nu \in V$marked “new”)
  BSearch($\nu$);
  }
  void BSearch($\nu$){
  MakeNull(Q);
  bfn[v] = count;
  count = count + 1;
  mark v”old”;
  EnQueue($\nu$,Q);
  while(! Empty(Q)){
  $\nu$=Front(Q);
  DeQueue(Q);
  for(each $\omega \in L[\nu ]$)
  if($\omega$ is “new){
  bfn[$\omega$] =count;
  count++;
  mark $\omega$”old”;
  EnQueue($\omega$,Q);
  }
  }
  }

---

leetcode457 Friend Circles

leetcode457的问题就是问需要遍历多少次图；

class Solution {
public:
    int findCircleNum(vector<vector<int>>& M) {
        if(M.empty())return 0;
        vector<int> visited(M.size(), 0);
        int ans = 0;
        for(int i = 0; i != M.size(); ++i){
            if(visited[i] == 0){
                ++ans;
                dfVisit(M, visited, i);
            }
        }
        return ans;
    }
    void dfVisit(vector<vector<int>>& M, vector<int>& visited, int i){
        visited[i] = 1;
        for(int j = 0; j != M.size(); ++j){
            if(visited[j])continue;
            if(M[i][j] == 1){
                dfVisit(M, visited, j);
            }
        }
    }

};

结果