Lake Counting
Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 36208 Accepted: 17982 Description
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.
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 Input
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
3Hint
OUTPUT DETAILS:
There are three ponds: one in the upper left, one in the lower left,and one along the right side.
题意：
八连通的'W'为一片水洼，问总共有多少片水洼。
对于每个'W'，搜索其八连通区域，将所有找到的'W'都变成'.'，直至所有点都变成'.'。
期间所进行dfs的次数就是水洼的数目。
AC代码：
1 #include<bits/stdc++.h>//POJ提交注意更换头文件 2 using namespace std; 3 4 char mp[110][110]; 5 6 int n,m; 7 8 void dfs(int i,int j){ 9 mp[i][j]='.'; 10 for(int dx=-1;dx<=1;dx++){ 11 for(int dy=-1;dy<=1;dy++){ 12 int nx=i+dx,ny=j+dy; 13 if(0<=nx&&nx<n&&0<=ny&&ny<m&&mp[nx][ny]=='W') 14 dfs(nx,ny); 15 } 16 } 17 return ; 18 } 19 20 int main(){ 21 ios::sync_with_stdio(false); 22 while(cin>>n>>m){ 23 int ans=0; 24 for(int i=0;i<n;i++){ 25 for(int j=0;j<m;j++){ 26 cin>>mp[i][j]; 27 } 28 } 29 for(int i=0;i<n;i++){ 30 for(int j=0;j<m;j++){ 31 if(mp[i][j]=='W'){ 32 dfs(i,j); 33 ans++; 34 } 35 } 36 } 37 cout<<ans<<endl; 38 } 39 return 0; 40 }