Oil Deposits

Time Limit : 2000/1000ms (Java/Other) Memory Limit : 65536/32768K (Java/Other)

Total Submission(s) : 61 Accepted Submission(s) : 25

Problem Description

The GeoSurvComp geologic survey company is responsible for detecting underground oil deposits. GeoSurvComp works with one large rectangular region of land at a time, and creates a grid that divides the land into numerous square plots. It then analyzes each plot separately, using sensing equipment to determine whether or not the plot contains oil. A plot containing oil is called a pocket. If two pockets are adjacent, then they are part of the same oil deposit. Oil deposits can be quite large and may contain numerous pockets. Your job is to determine how many different oil deposits are contained in a grid. <br>

Input

The input file contains one or more grids. Each grid
begins with a line containing m and n, the number of rows and columns in the
grid, separated by a single space. If m = 0 it signals the end of the input;
otherwise 1 <= m <= 100 and 1 <= n <= 100. Following this are m
lines of n characters each (not counting the end-of-line characters). Each
character corresponds to one plot, and is either `*‘, representing the absence
of oil, or `@‘, representing an oil pocket.<br>

Output

For each grid, output the number of distinct oil
deposits. Two different pockets are part of the same oil deposit if they are
adjacent horizontally, vertically, or diagonally. An oil deposit will not
contain more than 100 pockets.<br>

Sample Input

1 1

*

3 5

*@*@*

**@**

*@*@*

1 8

@@****@*

5 5

****@

*@@*@

*@**@

@@@*@

@@**@

0 0

Sample Output

0

1

2

2

简单题意： 求解@ 的连通块有几个，

思路分析：

　　dfs深度搜，每个点遍历搜索

# include <iostream>
# include <fstream>
# include <cstring>
using namespace std;
int n, m;
char map[1001][1001];
int is[1001][1001];
int dir[8][2] = { {-1,0},{1, 0},{0, 1},{0, -1},{1, 1},{1, -1},{-1, 1},{-1, -1}};

bool border(int i, int j)
{
    if(i >= 1 && i <= n && j <= m && j >= 1)
    return true;
    return false;
}
void bfs(int, int);

int main()
{
    //fstream cin("aaa.txt");
    while(cin >> n >> m)
    {
        if(n == 0 && m == 0)
        break;
        for(int i = 1; i <= n; i++)
            for(int j = 1; j <= m; j++)
                cin >> map[i][j];
        memset(is, 0, sizeof(is));
        int jishu = 0;
        for(int i = 1; i <= n; i++)
            for(int j = 1; j <= m; j++)
            {
                if(map[i][j] == ‘@‘ && is[i][j] == 0)
                {
                    is[i][j] = 1;
                    jishu++;
                    bfs(i, j);
                }
            }
        cout << jishu << endl;
    }
    return 0;
}

void bfs(int a, int b)
{
    int x, y;
    for(int i = 0; i < 8; i++)
    {
        x = a + dir[i][0];
        y = b + dir[i][1];
        if(border(x, y) && map[x][y] == ‘@‘ && is[x][y] == 0)
        {
            is[x][y] = 1;
            bfs(x, y);
        }
    }
}