Due to recent rains, water has pooled in various places in Farmer John's field, which is represented by a rectangle of N x M (1 <= N <= 100; 1 <= M <= 100) squares. Each square contains either water ('W') or dry land ('.'). Farmer John would like to figure out how many ponds have formed in his field. A pond is a connected set of squares with water in them, where a square is considered adjacent to all eight of its neighbors.?
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Given a diagram of Farmer John's field, determine how many ponds he has.
Input
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* Line 1: Two space-separated integers: N and M?
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* Lines 2..N+1: M characters per line representing one row of Farmer John's field. Each character is either 'W' or '.'. The characters do not have spaces between them.
Output
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* Line 1: The number of ponds in Farmer John's field.
Sample Input
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10 12
W........WW.
.WWW.....WWW
....WW...WW.
.........WW.
.........W..
..W......W..
.W.W.....WW.
W.W.W.....W.
.W.W......W.
..W.......W.
Sample Output
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3
Hint
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OUTPUT DETAILS:?
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There are three ponds: one in the upper left, one in the lower left,and one along the right side.
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题解：输入m行n列的字符矩阵，统计字符“W”组成多少个八连块。如果两个字符“W”所在的格子相邻（横，竖或者对角线方向），就说它们属于同一个八连块，采用二重循环来找。
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? ? ? ? 这就是连通块原理；每访问一次“W”，就给它写上标记的编号，方便检查。
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AC代码：
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#include
#include
const int maxn=1000+5;
char tu[maxn][maxn]; //输入图的数组
int m,n,idx[maxn][maxn]; //标记数组
void dfs(int r,int c,int id)
{
if(r<0||r>=m||c<0||c>=n)
return;
if(idx[r][c]>0||tu[r][c]!='W')
return;
idx[r][c]=id;
for(int dr=-1; dr<=1; dr++)
for(int dc=-1; dc<=1; dc++) // 寻找周围八块
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
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