<algorithm> 2. 10026번 RGB

RGB

RRR
GGG -> 답:3 2(색명은 R-G 하나로 보면 된다) 
BBB

주어지면 그 해당 영역을 구하는 문제. 
DFS를 통해 모든 경로를 탐색하고 영역수를 증가시키면 된다.

Public class RGB {
 
    public static void main(String[] args) throws IOException {
        // TODO Auto-generated method stub
 
        BufferedReader in=new BufferedReader
                (new InputStreamReader(System.in));
 
        StringTokenizer token=new StringTokenizer(in.readLine());
 
 
        int N=Integer.parseInt(token.nextToken());
        //System.out.println(N);
        char[][] graph=new char[N][N];
 
        for(int i=0;i<N;i++){
            token=new StringTokenizer(in.readLine());
            String a=token.nextToken();
            for(int j=0;j<N;j++){
                graph[i][j]=a.charAt(j);
            }
        }
 
 
        System.out.println(new Area(graph).getCount()+" "+new Area2(graph).getCount());
    }
 
}
class Area{
 
    char[][] graph;
    boolean[][] visited;
    char start;
    int count=0;
 
    public Area(char[][] graph) {
        // TODO Auto-generated constructor stub
        this.graph=graph;
        visited=new boolean[graph.length][graph.length];
 
 
        for(int i=0;i<graph.length;i++){
            for(int j=0;j<graph.length;j++){
                if(!visited[i][j]){
                    if(graph[i][j]=='R'){
                        r(i,j);
                        count++;
                    }else if(graph[i][j]=='G'){
                        g(i,j);
                        count++;
                    }else{
                        b(i,j);
                        count++;
                    }
                }
            }
        }
    }
 
 
    public void r(int i,int j){
        visited[i][j]=true;
        if((i+1)<graph.length&&!visited[i+1][j]&&graph[i+1][j]=='R'){
            r(i+1,j);
        }if((i-1)>=0&&!visited[i-1][j]&&graph[i-1][j]=='R'){
            r(i-1,j);
        }if((j+1)<graph.length&&!visited[i][j+1]&&graph[i][j+1]=='R'){
            r(i,j+1);
        }if((j-1)>=0&&!visited[i][j-1]&&graph[i][j-1]=='R'){
            r(i,j-1);
        }
 
    }
    public void g(int i,int j){
        visited[i][j]=true;
        if((i+1)<graph.length&&!visited[i+1][j]&&graph[i+1][j]=='G'){
            g(i+1,j);
        }if((i-1)>=0&&!visited[i-1][j]&&graph[i-1][j]=='G'){
            g(i-1,j);
        }if((j+1)<graph.length&&!visited[i][j+1]&&graph[i][j+1]=='G'){
            g(i,j+1);
        }if((j-1)>=0&&!visited[i][j-1]&&graph[i][j-1]=='G'){
            g(i,j-1);
        }
    }
    public void b(int i,int j){
        visited[i][j]=true;
        if((i+1)<graph.length&&!visited[i+1][j]&&graph[i+1][j]=='B'){
            b(i+1,j);
        }if((i-1)>=0&&!visited[i-1][j]&&graph[i-1][j]=='B'){
            b(i-1,j);
        }if((j+1)<graph.length&&!visited[i][j+1]&&graph[i][j+1]=='B'){
            b(i,j+1);
        }if((j-1)>=0&&!visited[i][j-1]&&graph[i][j-1]=='B'){
            b(i,j-1);
        }
    }
    public int getCount(){
        return count;
    }
 
}
 
class Area2{
 
    char[][] graph;
    boolean[][] visited;
    char start;
    int count=0;
 
    public Area2(char[][] graph) {
        // TODO Auto-generated constructor stub
        this.graph=graph;
        visited=new boolean[graph.length][graph.length];
 
 
        for(int i=0;i<graph.length;i++){
            for(int j=0;j<graph.length;j++){
                if(!visited[i][j]){
                    if(graph[i][j]=='R'||graph[i][j]=='G'){
                        rg(i,j);
                        count++;
                    }else{
                        b(i,j);
                        count++;
                    }
 
                }
            }
        }
 
    }
 
    public void rg(int i,int j){
        visited[i][j]=true;
        if((i+1)<graph.length&&!visited[i+1][j]&&(graph[i+1][j]=='R'||graph[i+1][j]=='G')){
            rg(i+1,j);
        }if((i-1)>=0&&!visited[i-1][j]&&(graph[i-1][j]=='R'||graph[i-1][j]=='G')){
            rg(i-1,j);
        }if((j+1)<graph.length&&!visited[i][j+1]&&(graph[i][j+1]=='R'||graph[i][j+1]=='G')){
            rg(i,j+1);
        }if((j-1)>=0&&!visited[i][j-1]&&(graph[i][j-1]=='R'||graph[i][j-1]=='G')){
            rg(i,j-1);
        }
 
    }
    public void b(int i,int j){
        visited[i][j]=true;
        if((i+1)<graph.length&&!visited[i+1][j]&&graph[i+1][j]=='B'){
            b(i+1,j);
        }if((i-1)>=0&&!visited[i-1][j]&&graph[i-1][j]=='B'){
            b(i-1,j);
        }if((j+1)<graph.length&&!visited[i][j+1]&&graph[i][j+1]=='B'){
            b(i,j+1);
        }if((j-1)>=0&&!visited[i][j-1]&&graph[i][j-1]=='B'){
            b(i,j-1);
        }
    }
    public int getCount(){
        return count;
    }
 
}

링크  https://www.acmicpc.net/problem/10026

DFS 기본문제로 DFS 원리만 이해한다면 쉽게 풀 수 있다.