题目
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.
分析
比较经典的问题，并查集中的合并操作
如果这两个人是朋友就合并
最后我们判断有多少个根即可和上次我说的联盟是一种问题
代码
class Solution {
public:
int uf[200];
int find( int u ){
return uf[u] == u ? u : uf[u] = find( uf[u] ) ;
}
void Union( int u , int v ){
int fu = find( u ) ;
int fv = find( v ) ;
if( fu == fv )
return ;
uf[fv] = fu ;
}
int findCircleNum(vector<vector<int>>& M) {
for( int i=0 ; i<M.size() ; i++){
uf[i] = i ;
}
for( int i=0 ; i<M.size() ; i++){
for( int j =0 ; j<M.size() ; j++){
if( M[i][j] == 1 ){
Union( i , j ) ;
}
}
}
int ans =0 ;
for( int i=0 ; i<M.size() ; i++){
if( uf[i] == i )
ans++;
}
return ans ;
}
};