POJ 2386 Lake Counting DFS灌水

Lake Counting

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 17638 Accepted: 8951

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 12W........WW..WWW.....WWW....WW...WW..........WW..........W....W......W...W.W.....WW.W.W.W.....W..W.W......W...W.......W.

Sample Output

3

Hint

OUTPUT DETAILS: 

There are three ponds: one in the upper left, one in the lower left,and one along the right side.

Source

USACO 2004 November

 这个DFS是深度优先遍历的意思.

挺好的例题 感谢@吴烨琪 童鞋的推荐

#include <iostream>using namespace std;const int dx[8]={-1,-1,-1, 0, 0, 1, 1, 1};const int dy[8]={-1, 0, 1,-1, 1,-1, 0, 1};int i,j,n,m,map[110][110]={0},count=0;string s;void dfs(int x,int y){map[x][y]=0;for (int i=0;i<8;++i)if (map[x+dx[i]][y+dy[i]]==1 && x+dx[i]>=0 && x+dx[i]<n && y+dy[i]>=0 && y+dy[i]<m)dfs(x+dx[i],y+dy[i]);}int main(){cin>>n>>m;for (i=0;i<n;++i){cin>>s;for (j=0;j<m;++j)if (s[j]=='W') map[i][j]=1; }for (i=0;i<n;++i)for (j=0;j<m;++j)if (map[i][j]){count++;dfs(i,j);}cout<<count<<endl;return 0;}

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