백준 dfs와 bfs
풀이
bfs, dfs 구현만 해주면 되는 문제
- dfs (깊이 우선 탐색)을 통해 방문 하는 노드의 수 세주기
- bfs (너비 우선 탐색)을 통해 방문 하는 노드의 수 세주기
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import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.LinkedList;
import java.util.Queue;
import java.util.Stack;
import java.util.StringTokenizer;
class FastScanner {
BufferedReader br;
StringTokenizer st;
public FastScanner() {
br = new BufferedReader(new InputStreamReader(System.in));
}
String next() {
while (st == null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt() {
return Integer.parseInt(next());
}
long nextLong() {
return Long.parseLong(next());
}
}
public class Main{
static int map[][];
static int visited[];
static int n;
static int m;
static int max = 1000;
public static void main(String args[]) throws IOException{
FastScanner sc = new FastScanner();
n = sc.nextInt(); // 정점
m = sc.nextInt(); // 간선
int start = sc.nextInt(); // 시작점
map = new int[n+1][n+1];
for(int i=0; i<m; i++){
int a = sc.nextInt();
int b = sc.nextInt();
map[a][b] = 1;
map[b][a] = 1;
}
visited = new int[max+1];
dfs(start);
System.out.println();
for(int i=1; i<=max; i++){
visited[i] = 0;
}
bfs(start);
}
static void bfs(int point){
Queue<Integer> queue = new LinkedList<>();
queue.add(point);
visited[point] = 1;
System.out.print(point);
while(!queue.isEmpty()) {
int x = queue.poll();
for(int i=1; i <= n; i++)
if(visited[i] == 0 && map[x][i] == 1){
queue.add(i);
System.out.print(" " + i);
visited[i] = 1;
}
}
}
static void dfs(int point){
Stack<Integer> stack = new Stack<>();
stack.push(point);
visited[point] = 1;
System.out.print(point + " ");
while(!stack.isEmpty()){
for(int i=1; i<=n; i++){
if(map[point][i] == 1 && visited[i] == 0){
stack.push(i);
visited[i] = 1;
dfs(i);
}
}
stack.pop();
}
}
}