hihocoder offer收割编程练习赛12 C 矩形分割

思路：

模拟，深搜。

实现：

#include <iostream>
#include <cstdio>
#include <string>
using namespace std;

const int dx[4] = { 0, 1, 0, -1 };
const int dy[4] = { -1, 0, 1, 0 };

int n, m, cnt = 0;
int a[305][305];
bool vis[305][305];

void dfs(int x, int y)
{
    vis[x][y] = true;
    for (int i = 0; i < 4; i++)
    {
        int nx = x + dx[i];
        int ny = y + dy[i];
        if (nx >= 0 && nx < 3 * n && ny >= 0 && ny < 3 * m && !a[nx][ny] && !vis[nx][ny])
        {
            dfs(nx, ny);
        }
    }
}

int main()
{
    scanf("%d %d", &n, &m);
    getchar();
    for (int i = 0; i < n; i++)
    {
        for (int j = 0; j < m; j++)
        {
            char c = getchar();
            if (c == '/')
            {
                a[3 * i][3 * j] = 0;
                a[3 * i][3 * j + 1] = 0;
                a[3 * i][3 * j + 2] = 1;
                a[3 * i + 1][3 * j] = 0;
                a[3 * i + 1][3 * j + 1] = 1;
                a[3 * i + 1][3 * j + 2] = 0;
                a[3 * i + 2][3 * j] = 1;
                a[3 * i + 2][3 * j + 1] = 0;
                a[3 * i + 2][3 * j + 2] = 0;
            }
            else if (c == '\')
            {
                a[3 * i][3 * j] = 1;
                a[3 * i][3 * j + 1] = 0;
                a[3 * i][3 * j + 2] = 0;
                a[3 * i + 1][3 * j] = 0;
                a[3 * i + 1][3 * j + 1] = 1;
                a[3 * i + 1][3 * j + 2] = 0;
                a[3 * i + 2][3 * j] = 0;
                a[3 * i + 2][3 * j + 1] = 0;
                a[3 * i + 2][3 * j + 2] = 1;
            }
        }
        getchar();
    }

    for (int i = 0; i < 3 * n; i++)
    {
        for (int j = 0; j < 3 * m; j++)
        {
            if (!a[i][j] && !vis[i][j])
            {
                dfs(i, j);
                cnt++;
            }
        }
    }
    cout << cnt << endl;
    return 0;
}