hdu 1241 Oil Deposits（DFS求连通块）

 

                       HDU 1241  Oil Deposits

L -DFS

Time Limit:1000MS     Memory Limit:10000KB     64bit IO Format:%I64d & %I64u

 

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 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

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

 

一道经典的dfs搜索题，搜索连通块个数

 

AC代码：

#include<cstdio>
#include<cstring>
const int maxn=100+5;
char pic[maxn][maxn];
int m,n,idx[maxn][maxn];

void dfs(int r,int c,int id)
{
    if(r<0||r>=m||c<0||c>=n)          //判定出界
        return;
    if(idx[r][c]>0||pic[r][c]!='@')   //不是'@'或者已经访问过的格子
        return;
    idx[r][c]=id;                     // 连通分量编号
    for(int dr=-1; dr<=1; dr++)
        for(int dc=-1; dc<=1; dc++)
            if(dr!=0||dc!=0)
                dfs(r+dr,c+dc,id);
}
int main()
{
    int i,j;
    while(scanf("%d%d",&m,&n)==2&&m&&n)
    {
        for(i =0; i<m; i++)
            scanf("%s",pic[i]);
        memset(idx,0,sizeof(idx));
        int q=0;
        for(i=0; i<m; i++)
            for(j=0; j<n; j++)
                if(idx[i][j]==0&&pic[i][j]=='@')
                    dfs(i,j,++q);
        printf("%d\n",q);
    }
    return 0;
}