P2960 写了 $100$ 行代码来完成 $30$ 行就能完成的事，是我不适合 OOP 吗？

#include <bits/stdc++.h>
using namespace std;
class slove{
    static const int N = 105;
    int n,m;
    int mx,my;
    char maps[N][N];
    int vis[N][N];
    int cnt;
    const int dir[8][2] = 
    {
    {-1,-1},
    {-1,0},
    {-1,1},
    {0,1},
    {0,-1},
    {1,0},
    {1,1},
    {1,-1}
    };
    public:
    class nodes_of_bfs
    {
        public:
        int x,y,step;
        nodes_of_bfs(int xx = 0,int yy = 0,int stepp = 0):x(xx),y(yy),step(stepp){}
    };
    void input(){
        scanf("%d%d",&m,&n);
        scanf("%d%d",&mx,&my);
        int temp = mx;
        mx = n - my + 1;
        my = temp;
        cnt = 0;
        memset(maps,'*',sizeof(maps));
        memset(vis,0,sizeof(vis));
        for(int i = 1;i <= n;i ++)
        {
            scanf("%s",maps[i]);
            for(int j = m-1;j >= 0;j --){
                maps[i][j+1] = maps[i][j];
            }
            maps[i][0] = '*';
        }
        for(int i = 1;i <= n;i ++)
        {
            for(int j = 1;j <= m;j ++){
                if(maps[i][j] == '.')
                {
                    cnt ++;
                }
            }
        }
    }
    int bfs(){
        nodes_of_bfs start(mx,my);
        queue<nodes_of_bfs> que;
        que.push(start);
        vis[mx][my] = 1;
        cnt --;
        while(!que.empty())
        {
            nodes_of_bfs now = que.front();
            que.pop();
            for(int i = 0;i < 8;i ++)
            {
                nodes_of_bfs next = now;
                next.x += dir[i][0];
                next.y += dir[i][1];
                next.step ++;
                if(maps[next.x][next.y] == '.')
                {
                    if(vis[next.x][next.y] == 0)
                    {
                        vis[next.x][next.y] = 1;
                        que.push(next);
                        cnt --;
                    }
                    if(cnt == 0)
                    {
                        return next.step;
                    }
                }
            }
        }
        return 0;
    }
    slove()
    {
        input();
        printf("%d\n",bfs());
    }
    ~slove()
    {
        exit(0);
    }
};
int main() {
    slove slover;
    return 0;
}