hdu 1241 dfs求连通块

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

题解：

紫书上的题。从每个’@’出发，递归遍历它周围的”@”格子。每次访问一个格子时就给它写上一个”连通分量编号”，这样就可以在访问之前检查它是否访问过了。避免了访问多次。
经典的种子填充算法。

代码：

#include <iostream>#include <cstdio>#include <cstring>const int maxn = 100+5;using namespace std;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) return;   if(idx[r][c]>0||pic[r][c]!='@') return;   idx[r][c]=id;   for(int dr=-1;dr<=1;dr++)    for(int dc=-1;dc<=1;dc++)      if(dr!=0||dc!=0) dfs(dr+r,dc+c,id);}int main(){    while(scanf("%d%d",&m,&n)==2&&m&&n)    {       for(int i=0;i<m;i++) scanf("%s",pic[i]);       memset(idx,0,sizeof(idx));       int cnt =0 ;       for(int i=0;i<m;i++)        for(int j=0;j<n;j++)         if(idx[i][j]==0&&pic[i][j]=='@') dfs(i,j,++cnt);       printf("%d\n",cnt);    }    return 0;}