1349: Oil Deposits

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本文介绍了一种用于探测地下油藏分布的算法实现。通过构建网格模型并采用深度优先搜索方法来确定不同油藏间的连接关系，从而计算出特定区域内油藏的数量。

1349: Oil Deposits

Result	TIME Limit	MEMORY Limit	Run Times	AC Times	JUDGE
	3s	8192K	516	253	Standard

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

and

. 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

This problem is used for contest: 108 150

#include<stdio.h>
#include<string.h>
int xx[8]={-1,0,1,-1,1,-1,0,1};
int yy[8]={-1,-1,-1,0,0,1,1,1};
char a[1000][1000];
int row,colum;
int count;
void dfs(int i,int j)
{
    int k;
    a[i][j]='*';
    for(k=0;k<8;k++)
    {
        if(i+xx[k]<row&&i+xx[k]>=0&&j+yy[k]>=0&&j+yy[k]<colum&&a[i+xx[k]][j+yy[k]]=='@')
        {
            dfs(i+xx[k],j+yy[k]);
        }
    }
}
int main()
{
    int i,j;
    while(scanf("%d%d",&row,&colum)&&row)
    {
        for(i=0;i<row;i++)
        {
            scanf("%s",a[i]);
        }
        count=0;
        for(i=0;i<row;i++)
        {
            for(j=0;j<colum;j++)
            {
                if(a[i][j]=='@')
                {
                    dfs(i,j);
                    count++;
                }
            }
        }
        printf("%d/n",count);
    }
    return 0;
}