算法分析与设计——LeetCode Problem.547 Friend Circles

问题详情

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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.

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

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问题分析及思路

题目的意思是给出一个矩阵，若M[i][j]=1则说明i与j是朋友关系，而M[j][k]=1且M[i][k]!=1则说明i与k是间接朋友关系。将所有的有朋友关系的人作为一个子集，求出矩阵中所有元素一共划分为几个子集。
因此一个元素只能在一个子集中，我认为可以对所有的进行遍历，然后在遍历中采用深度优先算法，对当前遍历元素的朋友进行遍历。
同时需要一个数组来保存已经遍历过的元素，以防止重复遍历，减少算法复杂度。

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具体代码

class Solution {
public:
    int findCircleNum(vector<vector<int>>& M) {
        if (M.empty()) 
            return 0;
        vector<bool> visited;
        int n = M.size();
        for(int i=0;ifalse);
        int a = 0;
        for (int i = 0; i < n; i++) {
            a += DeepFirst(visited, M, i, n) > 0;
        }
        return a;

    }

    int DeepFirst(vector<bool>& visited, vector<vector<int>>& M, int i, int n) {
        if (visited[i]) {
            return 0;
        }
        int a = 1;
        visited[i] = true;
        for (int j = 0; j < n; j++) {
            if (i != j && M[i][j]) {
                a += DeepFirst(visited, M, j, n);
            }
        }
        return a;
    }
};