Title description ( portal )
Suppose there are N users, some of whom are friends and some are not. A and B are friends, B and C are friends, so ABC is a circle of friends, please count the number of circles of a given friend relationship.
Given an N * N matrix M, it represents the friend relationship between users. If M[i][j] = 1, it means that the i-th and j-th persons are known to be friends with each other, otherwise it is not known. You must output the total number of known Moments in all users.
输入描述:
第一行输入为表示用户数N
第2到1+N行表示朋友之间关系，如果值为1表示有朋友关系
输出描述:
输出朋友圈数
Example 1
输入
3
1,1,0
1,1,0
0,0,1
输出
2
说明
第0个用户和第1个用户组成一个朋友圈，第2个用户组成一个朋友圈
Example 2
输入
3
1,1,0
1,1,1
0,1,1
输出
1
说明
第0，1，2个用户组成了一个朋友圈
Problem solving ideas & code implementation
I studied and checked the collection yesterday, and today I will contact you by hitting the iron while it is hot.
I won’t say more about the combined search. You can view: [High-level data structure] Detailed explanation of combined search
import java.util.Scanner;
/**
* @ClassName Test
* @Description ：TODO
* @Author Josvin
* @Date 2021/01/15/21:19
*/
public class Test {
public static int count ;
private static int[] id;
private static int[] size;
public static void main(String[] args) {
Scanner scanner = new Scanner(System.in);
String s = scanner.nextLine();
int num = Integer.parseInt(s);
int[][] M = new int[num][num];
//牛客上练习，对输入输出的处理要一定熟练，自己感觉处理的不是特别好，如果有更好的方法可以交流交流
for (int i = 0; i < num; i++) {
String[] str = scanner.nextLine().split(",");
for (int j = 0; j < num; j++) {
M[i][j] = Integer.parseInt(str[j]);
}
}
Union_Find(M); //调用并查集
System.out.println(count);
}
private static void Union_Find(int[][] M) {
int N = M.length;// M的长度也就是总共的人数，也就是上个博客提到的集合总数
id = new int[N];// id 定义一个集合，大小为 N，也就是总共有N个人
count = N;// 朋友圈的数目，刚开始每个人就是一个朋友圈，初始化为N
size = new int[N];// 记录每个朋友圈的大小，在合并时用
// 初始化id数组，每个人都是一个朋友圈，将置为不同的数即可
// size 刚开始每个人都是一个朋友圈，他们的大小也就是1
for (int i = 0; i < N; i++) {
id[i] = i;
size[i] = 1;
}
//合并，将M数组元素，为1的，合并两者i和j
for (int i = 0; i < N; i++) {
for (int j = 0; j < N; j++) {
if (M[i][j] == 1) {
Union(i,j);
}
}
}
}
// 合并
private static void Union(int i, int j) {
int root1 = Find(i);
int root2 = Find(j);
if (root1 == root2) {
return;
}
if (size[i] > size[j]) {
id[root1] = root2;
size[root2] += size[root1];
} else {
id[root2] = root1;
size[root1] += size[root2];
}
count--;
}
// 查找
private static int Find(int p) {
if (p != id[p]) {
id[p] = Find(id[p]);
}
return id[p];
}
}