POJ2386——Lake Counting

Lake Counting Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 33004   Accepted: 16459 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 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

题意：

找出水洼的数量，所有八连通的看做一个

解：

从任意‘w’开始，把能连接的部分用 '  .  ' 表示，一次dfs之后找出一个水洼，直到所有变成 ‘  .  ’ 为止。进行dfs次数为最终答案。

#include 
#include 
#include 

using namespace std;

char map[105][105];
int n,m;
int dis[8][2] = {{0,1},{0,-1},{-1,0},{1,0},{1,1},{1,-1},{-1,-1},{-1,1}};

void dfs(int x,int y)
{
	if(x<0 || y<0 || x>=n ||y>=m || map[x][y]=='.')
		return;
	map[x][y]='.';
	for(int i=0;i<8;i++){
		int newx = x+dis[i][0];
		int newy= y+dis[i][1];
		if(newx>=0 && newx=0 && newy>map[i][j];
			}
			getchar();
		}
		
		for(int i=0;i