LeetCode 547 Friend Circles（并查集）

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 i_th and j_th 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.

题目大意：给出一个N*N的矩阵M，如果M[i][j]==1，说明i和j有直接朋友关系，如果M[i][j]==1并且M[j][k]==1那么i和k就有间接朋友关系。求所有人之间的朋友圈子有几个。

解题思路：利用并查集，把有有关系的人放到一个集合里，然后计算集合的个数。

代码如下：

struct DisjointSet{
    int par;
    int rank;
};
#define maxn 222
struct DisjointSet ds[maxn];

void init()
{
    for(int i = 0;i < maxn;i++){
        ds[i].par = i;
        ds[i].rank = 1;
    }
}

int find(int x)
{
    if(x == ds[x].par) return  x;
    return ds[x].par = find(ds[x].par);
}

void _union(int x, int y)
{
    x = find(x);
    y = find(y);
    if(x == y) return ;
    if(ds[x].rank < ds[y].rank)
        ds[x].par = y;
    else{
        if(ds[x].rank == ds[y].rank)
            ds[x].rank++;
        ds[y].par = x;
    }
}

int same(int x, int y)
{
    return find(x) == find(y);
}

int findCircleNum(int** M, int MRowSize, int MColSize) {
    init();
    for(int i = 0;i < MRowSize;i++){
        for(int j = i + 1;j < MColSize;j++){
            if(M[i][j] && M[j][i]) _union(i,j);
        }
    }
    int count = 0;
    for(int i = 0;i < MRowSize;i++){
        if(ds[i].par == i) count++;
    }
    return count;
}