[bzoj1751][Usaco2005 qua]Lake Counting

1751: [Usaco2005 qua]Lake Counting

Time Limit: 5 Sec  Memory Limit: 64 MB
Submit: 261  Solved: 212
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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

3

OUTPUT DETAILS:

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

HINT

noip普及组T1难度

#include<bits/stdc++.h>using namespace std;bool vis[105][105],mp[105][105]; char ch[105][105];int dx[8] = {0,0,1,1,1,-1,-1,-1},dy[8] = {1,-1,0,-1,1,0,-1,1};int n,m,ans;void dfs( int x, int y ){vis[x][y] = 1;for( int i = 0; i < 8; i++ ){int xx = x + dx[i], yy = y + dy[i];if( !vis[xx][yy] && mp[xx][yy] ) dfs( xx, yy ); }}int main(){scanf("%d%d", &n, &m);for( int i = 1; i <= n; i++ ) scanf("%s", ch[i]+1);for( int i = 1; i <= n; i++ )for( int j = 1; j <= m; j++ )mp[i][j] = ( ch[i][j] == 'W' );for( int i = 1; i <= n; i++ )for( int j = 1; j <= m; j++ )if( !vis[i][j] && mp[i][j] ){ans++; dfs( i, j );}printf("%d\n", ans);return 0;}