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***Lake Counting***Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 24201 Accepted: 12216 DescriptionDue 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. Given a diagram of Farmer John's field, determine how many ponds he has.Input* Line 1: Two space-separated integers: N and M * 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* Line 1: The number of ponds in Farmer John's field.Sample Input10 12W........WW..WWW.....WWW....WW...WW..........WW..........W....W......W...W.W.....WW.W.W.W.....W..W.W......W...W.......W.Sample Output3 解题思路：
dfs ，注意从8个方面进行搜索： 上代码：#include#include #include using namespace std;int n,m,ans;char map[110][110];bool judge(int i,int j){ if(i>=n || i<0) return false; if(j>=m || j<0) return false; if(map[i][j] == '.') return false; return true;}void dfs(int i,int j){ map[i][j] = '.'; if(judge(i+1, j)) dfs(i+1, j); if(judge(i, j+1)) dfs(i, j+1); if(judge(i-1, j)) dfs(i-1, j); if(judge(i, j-1)) dfs(i, j-1); if(judge(i+1, j+1)) dfs(i+1, j+1); if(judge(i-1, j-1)) dfs(i-1, j-1); if(judge(i+1, j-1)) dfs(i+1, j-1); if(judge(i-1, j+1)) dfs(i-1, j+1);}int main(){ while(cin>>n>>m) { getchar(); ans=0; for(int i=0; i
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