백준 2573 빙산(dfs)

https://www.acmicpc.net/problem/2573

2573번: 빙산

#include <iostream>
#include <queue>

using namespace std;
const int MAX = 300;
int graph[MAX][MAX];
bool visited[MAX][MAX];
int N, M;
int result;
int tmpGraph[MAX][MAX];
int block;
typedef struct
{
    int x, y;
} Dir;
Dir dirMove[4] = {{-1, 0}, {1, 0}, {0, 1}, {0, -1}};

bool allZero(){
    for(int i=0;i<N;i++){
        for(int j=0;j<M;j++){
            if(graph[i][j]!=0) return false;
        }
    }
    return true;
}

void meltingIceberg(int x, int y)
{
    int cnt = 0;
    for (int i = 0; i < 4; i++)
    {
        int nextX = x + dirMove[i].x;
        int nextY = y + dirMove[i].y;
        if (0 <= nextX && nextX < N && 0 <= nextY && nextY < M && graph[nextX][nextY] == 0)
            cnt++;
    }
    if (graph[x][y] - cnt >= 0)
        tmpGraph[x][y] = graph[x][y] - cnt;
    else
        tmpGraph[x][y] = 0;
}

void dfs(int x,int y){
    visited[x][y]=true;
    for(int i=0;i<4;i++){
        int nextX=x+dirMove[i].x;
        int nextY=y+dirMove[i].y;
        if(0<=nextX&&nextX<N&&0<=nextY&&nextY<M&&!visited[nextX][nextY]&&graph[nextX][nextY]!=0)
        dfs(nextX,nextY);
    }
}

int main()
{
    ios::sync_with_stdio(false);
    cin.tie(NULL);
    cin >> N >> M;
    for (int i = 0; i < N; i++)
    {
        for (int j = 0; j < M; j++)
        {
            cin >> graph[i][j];
        }
    }
    while (1)
    {
        if(allZero()){
            cout<<0;
            return 0;
        }
        result++;

        for (int i = 0; i < N; i++)
        {
            for (int j = 0; j < M; j++)
            {
                meltingIceberg(i, j);
            }
        }

        for (int i = 0; i < N; i++)
        {
            for (int j = 0; j < M; j++)
            {
                graph[i][j] = tmpGraph[i][j];
            }
        }

        for(int i=0;i<N;i++){
            for(int j=0;j<M;j++){
                if(!visited[i][j]&&graph[i][j]!=0){
                    block++;
                    dfs(i,j);
                }
            }
        }

        if(block>=2){
            cout<<result;
            break;
        }
        for(int i=0;i<N;i++){
            for(int j=0;j<M;j++){
                visited[i][j]=false;
            }
        }
        block=0;    
    }
}

사실상 어려운 문제는 아니었지만, 체크해야하는 지점들이 여러개 있어서 꽤 시간이 걸렸다. 크게 코드에 대해 설명할 부분은 없는 것 같다.