Oil Deposits

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

0

1

2

2

题意：给你一个字符矩阵，求连在一起的@符有多少堆；

解题思路：深度优先搜多，找到一个节点开始往下搜，搜到的全都标记，然后搜不到了就停止，记录搜索到的次数；

感悟：第一次自己写的搜索代码；成功跑的那一刻有点兴奋！！！

代码：

#include
#include
#include
#define maxn 110

using namespace std;
char mapn[maxn][maxn];
int direction[8][2]={{-1,0},{1,0},{0,1},{0,-1},{-1,-1},{-1,1},{1,-1},{1,1}};
bool visit[maxn][maxn];
int m,n,sum=0;
bool isbound(int a,int b)//判断边界
{
    if(a<1||a>m||b<1||b>n)return true;
    return false;
}

void dfs(int x,int y)
{
    for(int i=0;i<8;i++)
    {
        if(mapn[x+direction[i][0]][y+direction[i][1]]=='*')
            continue;
        if(isbound(x+direction[i][0],y+direction[i][1]))
            continue;
        if(visit[x+direction[i][0]][y+direction[i][1]])
            continue;
        visit[x+direction[i][0]][y+direction[i][1]]=true;
        dfs(x+direction[i][0],y+direction[i][1]);
    }
}

int main()
{
    //freopen("in.txt","r",stdin);
    while(~scanf("%d%d\n",&m,&n)&&(m||n))//这地方应该有一个回车
    {
        //printf("m=%d n=%d\n",m,n);
        memset(visit,false,sizeof(visit));
        sum=0;
        for(int i=1;i<=m;i++)
        {
            for(int j=1;j<=n;j++)
            {
                scanf("%c",&mapn[i][j]);
            }
            scanf("\n");//每行输入结束都应该有一个回车
        }
        for(int i=1;i<=m;i++)
            for(int j=1;j<=n;j++)
            {
                if(mapn[i][j]=='@'&&!visit[i][j])
                {
                    dfs(i,j);
                    visit[i][j]=true;
                    sum++;
                }
            }
        printf("%d\n",sum);
    }
}