문제 링크
1. 아이디어
무방향 그래프에서 연결 요소의 개수를 구하는 문제로 BFS나 DFS를 활용해서 각 노드에 대해 방문하지 않았으면 모든 연결된 노드를 방문 처리하고 개수를 하나씩 세도 되며, 유니온 파인드 알고리즘을 활용해서 서로소 집합을 만든 후 집합의 그룹장의 개수를 세어도 된다.
2. 복잡도
| 접근 | 시간 | 공간 |
|---|
| BFS | $O(N + E)$ | $O(N + E)$ |
| DFS | $O(N + E)$ | $O(N + E)$ |
| 유니온 파인드 | $O(N + E)$ | $O(N)$ |
($N$ = 입력값 n, $E$ = 입력값 m)
3. 코드
풀이 1: BFS [Java][C++]
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
| import java.io.*;
import java.util.*;
public class Main {
static List<Integer>[] adj;
public static void main(String[] args) throws IOException {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
StringTokenizer st = new StringTokenizer(br.readLine());
int n = Integer.parseInt(st.nextToken());
int m = Integer.parseInt(st.nextToken());
adj = new ArrayList[1 + n];
for (int i = 1; i <= n; i++) {
adj[i] = new ArrayList<>();
}
while (m-- > 0) {
st = new StringTokenizer(br.readLine());
int u = Integer.parseInt(st.nextToken());
int v = Integer.parseInt(st.nextToken());
adj[u].add(v);
adj[v].add(u);
}
boolean[] vis = new boolean[1 + n];
int cnt = 0;
for (int node = 1; node <= n; node++) {
if (vis[node]) continue;
bfs(node, vis);
cnt++;
}
System.out.println(cnt);
}
static void bfs(int start, boolean[] vis) {
Queue<Integer> q = new ArrayDeque<>();
q.offer(start);
vis[start] = true;
while (!q.isEmpty()) {
int cur = q.poll();
for (int nxt : adj[cur]) {
if (vis[nxt]) continue;
q.offer(nxt);
vis[nxt] = true;
}
}
}
}
|
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
| #include <bits/stdc++.h>
using namespace std;
const int MAX_N = 1 + 1000;
vector<int> adj[MAX_N];
bool vis[MAX_N];
void bfs(int start) {
queue<int> q;
q.push(start);
vis[start] = true;
while (!q.empty()) {
int cur = q.front();
q.pop();
for (int nxt : adj[cur]) {
if (vis[nxt]) continue;
q.push(nxt);
vis[nxt] = true;
}
}
}
int main() {
ios::sync_with_stdio(0);
cin.tie(0);
int n, m;
cin >> n >> m;
while (m--) {
int u, v;
cin >> u >> v;
adj[u].push_back(v);
adj[v].push_back(u);
}
int cnt = 0;
for (int node = 1; node <= n; node++) {
if (vis[node]) continue;
bfs(node);
cnt++;
}
cout << cnt;
}
|
풀이 2: DFS [Java][C++]
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
| import java.io.*;
import java.util.*;
public class Main {
static List<Integer>[] adj;
static boolean[] vis;
public static void main(String[] args) throws IOException {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
StringTokenizer st = new StringTokenizer(br.readLine());
int n = Integer.parseInt(st.nextToken());
int m = Integer.parseInt(st.nextToken());
adj = new ArrayList[1 + n];
for (int i = 1; i <= n; i++) {
adj[i] = new ArrayList<>();
}
while (m-- > 0) {
st = new StringTokenizer(br.readLine());
int u = Integer.parseInt(st.nextToken());
int v = Integer.parseInt(st.nextToken());
adj[u].add(v);
adj[v].add(u);
}
vis = new boolean[1 + n];
int cnt = 0;
for (int node = 1; node <= n; node++) {
if (vis[node]) continue;
dfs(node);
cnt++;
}
System.out.println(cnt);
}
static void dfs(int cur) {
vis[cur] = true;
for (int nxt : adj[cur]) {
if (vis[nxt]) continue;
dfs(nxt);
}
}
}
|
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
| #include <bits/stdc++.h>
using namespace std;
const int MAX_N = 1 + 1000;
vector<int> adj[MAX_N];
bool vis[MAX_N];
void dfs(int cur) {
vis[cur] = true;
for (int nxt : adj[cur]) {
if (vis[nxt]) continue;
dfs(nxt);
}
}
int main() {
ios::sync_with_stdio(0);
cin.tie(0);
int n, m;
cin >> n >> m;
while (m--) {
int u, v;
cin >> u >> v;
adj[u].push_back(v);
adj[v].push_back(u);
}
int cnt = 0;
for (int node = 1; node <= n; node++) {
if (vis[node]) continue;
dfs(node);
cnt++;
}
cout << cnt;
}
|
풀이 3: 유니온 파인드 [Java][C++]
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
| import java.io.*;
import java.util.*;
public class Main {
static int[] p;
static void make(int n) {
p = new int[1 + n];
for (int i = 1; i <= n; i++) {
p[i] = i;
}
}
static int find(int x) {
if (p[x] == x) return x;
return p[x] = find(p[x]);
}
static void union(int x, int y) {
p[find(y)] = find(x);
}
public static void main(String[] args) throws IOException {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
StringTokenizer st = new StringTokenizer(br.readLine());
int n = Integer.parseInt(st.nextToken());
int m = Integer.parseInt(st.nextToken());
make(n);
while (m-- > 0) {
st = new StringTokenizer(br.readLine());
int u = Integer.parseInt(st.nextToken());
int v = Integer.parseInt(st.nextToken());
union(u, v);
}
int cnt = 0;
for (int node = 1; node <= n; node++) {
if (node == find(node)) cnt++;
}
System.out.println(cnt);
}
}
|
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
| #include <bits/stdc++.h>
using namespace std;
vector<int> p(1 + 1000, -1);
int find(int x) {
if (p[x] < 0) return x;
return p[x] = find(p[x]);
}
void unite(int x, int y) {
x = find(x);
y = find(y);
if (x != y) p[y] = x;
}
int main() {
ios::sync_with_stdio(0);
cin.tie(0);
int n, m;
cin >> n >> m;
while (m--) {
int u, v;
cin >> u >> v;
unite(u, v);
}
int cnt = 0;
for (int node = 1; node <= n; node++) {
if (node == find(node)) cnt++;
}
cout << cnt;
}
|