hdu 1241 Oil Deposits(dfs)

Oil Deposits

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 42771 Accepted Submission(s): 24819

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.

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

题意：探测地下的石油储量，含有油的地区被称为口袋。如果两个口袋是相邻的，那么它们是相同的石油矿床的一部分。石油储量可能非常大，可能含有大量的口袋。你的任务是要确定不同的石油储量有多少包含在一个网格。

输入文件包含一个或多个网格。每个网格开始用含有m和n，行和列的网格中的号码的线路，由一个空格分开。如果m = 0它标志着输入的结束; 否则1 <= M <= 100，1 <= N <= 100。在此之后是每个n字符（不包括端部的行的字符）m条。每个字符对应一个情节，'*'，代表没有油，'@'，是有油。

对于每一个网格，输出不同油沉积物的数量。一堆@连在一起的算一个大油田（对角线，横竖都算）。问有几个大油田。

//HD1241
#include <stdio.h>
#include<iostream>
char a[111][111];
int dir[8][2] = { { 1, 0 },{ -1, 0 },{ 0, 1 },{ 0, -1 },{ 1, 1 },{ -1, -1 },{ 1, -1 },{ -1, 1 } };
int m, n;
int flag;
int dfs(int x, int y)
{
    if (a[x][y] == '*')
        return flag;
    flag = 1;
    a[x][y] = '*';//记得标记，这样不会重复计数
    int tx, ty;
    for (int i = 0; i<8; i++)
    {
        tx = x + dir[i][0];
        ty = y + dir[i][1];
        if (tx >= 0 && tx<m&&ty >= 0 && ty<n)
            dfs(tx, ty);
    }
    return flag;
}
int main()
{
    while (scanf("%d %d", &m, &n) && m + n)
    {
        int num = 0;
        for (int i = 0; i<m; i++)
            scanf("%s", a[i]);
        for (int i = 0; i<m; i++)
        {
            for (int j = 0; j<n; j++)//把每个点都进去搜索一遍
            {
                flag = 0;
                if (dfs(i, j))
                {
                    num++;
                }
            }
        }
        printf("%d
", num);
    }
}