水洼数——蓝桥

#include <iostream>
using namespace std;

char arr[200][200];
int ans = 0, n, m;
int nx[9] = {0, -1, -1, 0, 1, 1, 1, 0, -1};
int ny[9] = {0,  0, 1,  1, 1, 0,-1, -1,-1};

void dfs(int i, int j)
{
    arr[i][j] = '#';
//    printf("arr[%d][%d] = %c\n", i, j, arr[i][j]);
    for(int k = 1; k <= 8; k++)    //注意此局部变量 不能和参数同一个标识符
    {
        int row = i + nx[k];
        int col = j + ny[k];
        if((row >= 1) && (row <= n) && (col >= 1) && (col <= m)&&(arr[row][col] == '@'))
        {
//            printf("row = %d, col = %d\n", row, col);
            dfs(row, col);
        }
    }
}

int main()
{
    scanf("%d%d", &n, &m);
    for(int i = 1; i <= n; i++)
    {
        scanf("%s", arr[i]+1);
    }
    for(int i = 1; i <= n; i++)
    {
        for(int j = 1; j <= m; j++)
        {
            if(arr[i][j] == '@')
            {
                dfs(i, j);
                ans++;
//                printf("i = %d, j = %d\n", i, j);
            }
        }
    }
    printf("%d\n", ans);
    return 0;
}

#include <cstdio>
#include <cstring>

const int maxn = 100 + 5;

char pic[maxn][maxn];

int m, n, idx[maxn][maxn];

void dfs(int r, int c, int id)
{
    if(r < 0 || r >= m || c < 0 || c >= n) return;      //“出界”的格子
    if(idx[r][c] > 0 || pic[r][c] != '@') return ;  //不是“@”或者已经访问过的格子
    idx[r][c] = id; //连通分量编号
    for(int dr = -1; dr <= 1; ++dr)
    {
        for(int dc = -1; dc <= 1; ++dc)
        {
            if(dr != 0 || dc != 0) dfs(r+dr, c+dc, id);
        }
    }
}

int main()
{
    while(scanf("%d%d", &m, &n) == 2 && m && n)
    {
        for(int i = 0; i < m; ++i) scanf("%s", pic[i]);
        memset(idx, 0, sizeof(idx));
        
        int cnt = 0;
        for(int i = 0; i < m; ++i)
        {
            for(int j = 0; j < n; ++j)
            {
                if(idx[i][j] == 0 && pic[i][j] == '@')
                    dfs(i, j, ++cnt);
            }
        }
        printf("%d\n", cnt);
    }
    return 0;
}

DFS基础模版题。。。。