Oil Deposits UVA - 572

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.

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.

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.

Sample Input

1 1
*
3 5
*@*@*
**@**
*@*@*
1 8
@@****@*
5 5
****@
*@@*@
*@**@
@@@*@
@@**@
0 0

Sample Output

0
1
2
2

HINT

此题重点在于对dfs算法的掌握，一个是记录原始数据的数组，一个是是否已经访问过的标志数组，在主函数里面对每一个点进行判断，如果没有被访问并且是@那么进行dfs，并将访问下标加一，以和上一个进行区别。那么在dfs里面的内容就是以给定的点为中心，将所有的可能的点进行判断访问。这样从主函数进入dfs再回来意味着将一个连通分量判断完成了。那么主函数将每一个元素都判断，肯定会求出所有的连通分量。

Accepted

#include <cstdio>
#include <cstring>
const int maxn = 105;
const int dx[] = {-1, -1, -1, 0, 1, 1, 1, 0};
const int dy[] = {-1, 0, 1, 1, 1, 0, -1, -1};
int n, m, s[maxn][maxn], idx[maxn][maxn];

void dfs(int x, int y, int cnt) {
    if (idx[x][y] != -1 ||  x < 0 || x >= n || y < 0 || y >= m) return;
    if (s[x][y] == '@') {
        idx[x][y] = cnt;
        for (int i = 0; i < 8; i++) {
            int x1 = dx[i] + x, y1 = dy[i] + y;
            dfs(x1, y1, cnt);
        }
    }
}

int main() {
    while (scanf("%d%d\n", &n, &m) == 2 && n && m) {
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++)
                s[i][j] = getchar();
            getchar();
        }
        int cnt = 0;
        memset(idx, -1, sizeof(idx));
        for (int i = 0; i < n; i++)
            for (int j = 0; j < m; j++)
                if (s[i][j] == '@' && idx[i][j] == -1)
                    dfs(i, j, ++cnt);
        printf("%d\n", cnt);
    }
    return 0;
}