POJ 2386 Lake Counting

Lake Counting

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 41344		Accepted: 20507

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

题目链接

AC代码：

#include<cstdio>
#include<cstring>
using namespace std;

char field[200][200];
int N,M;
void dfs(int i, int j){
	//将 当前位置 W 替换为 . 
	field[i][j] = '.';
	//遍历8个方向
	for(int dirx =-1;dirx <= 1; dirx++)
		for(int diry =-1; diry <= 1; diry++){
			int row = i + dirx;
			int col= j + diry;
			//row, col 在园子范围内且 filed[row][col] 位置有积水 
			if((row >= 0 && row < N) && (col >= 0 && col < M) && (field[row][col]=='W')){
				dfs(row, col);
			}
		}
}

int main(){
	
	scanf("%d%d",&N,&M);
	getchar();
	for(int i = 0; i < N; i++){
		gets(field[i]);
	}

	int res = 0;
	//遍历园子每一个位置 
	for(int i = 0; i < N; i++)
		for(int j = 0; j < M; j++){
			if(field[i][j]=='W'){
				res++;
				dfs(i, j);
			//	printf("%d %d\n", i, j);
			}
		}
	printf("%d\n",res);
	
	return 0;
}