题目描述

blablabla

样例

blablabla

算法1

(暴力枚举) $O(n^2)$

blablabla

时间复杂度

参考文献

C++ 代码

#include<iostream>
#include<cstring>

using namespace std;

const int N = 22;

char g[N][N];
bool st[N][N];
int n, m;
int dx[] = {0, 1, 0, -1}, dy[] = {1, 0, -1, 0};

void dfs(int x, int y, int &u)
{
    u ++;
    st[x][y] = true;
    for (int i = 0; i < 4; i ++)
    {
        int a = x + dx[i], b = y + dy[i];
        if (a < 0 || a >= n || b < 0 || b >= m) continue;
        if (st[a][b]) continue;
        if (g[a][b] == '#') continue;
        dfs(a, b, u);
    }
}

int main()
{
    int T;
    while(cin >> m >> n , m || n)
    {
        for (int i = 0; i < n; i ++) cin >> g[i];
        memset(st, 0, sizeof st);
        int x, y;
        for (int i = 0; i < n; i ++)    
            for (int j = 0; j < m; j ++)
                if (g[i][j] == '@')
                {
                    x = i, y = j;
                    break;
                }

        int res = 0;
        dfs(x, y, res);
        cout << res << endl;
    }
    return 0;
}

算法2

(暴力枚举) $O(n^2)$

blablabla

时间复杂度

参考文献

C++ 代码

#include<iostream>
#include<cstring>

using namespace std;

const int N = 22;

char g[N][N];
bool st[N][N];
int n, m;
int dx[] = {0, 1, 0, -1}, dy[] = {1, 0, -1, 0};

int dfs(int x, int y)
{
    int cnt = 1;
    st[x][y] = true;
    for (int i = 0; i < 4; i ++)
    {
        int a = x + dx[i], b = y + dy[i];
        if (a < 0 || a >= n || b < 0 || b >= m) continue;
        if (st[a][b]) continue;
        if (g[a][b] == '#') continue;
        cnt += dfs(a, b);
    }
    return cnt;
}

int main()
{
    int T;
    while(cin >> m >> n , m || n)
    {
        for (int i = 0; i < n; i ++) cin >> g[i];
        memset(st, 0, sizeof st);
        int x, y;
        for (int i = 0; i < n; i ++)    
            for (int j = 0; j < m; j ++)
                if (g[i][j] == '@')
                {
                    x = i, y = j;
                    break;
                }


        cout << dfs(x, y) << endl;
    }
    return 0;
}