POJ 2386 Lake Counting（水淹菜地，DFS，八连通，连通分量）

Lake Counting

同类型博文：Oil Deposits

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 29064 Accepted: 14543

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

题意：

让我们求出被水淹没的区域有几块。（字母 ’ W ‘ 表示水淹没的区域）

思路：

简单的DFS，求出八连通的连通分量就行了。

代码：

#include<stdio.h>#include<string.h>#define MYDD 1103int N,M;char map[128][128];//记录积水的坐标点int dx[]= {0,0,-1,1,1,1,-1,-1};//八连通的图int dy[]= {-1,1,0,0,-1,1,-1,1};//当前位置周围的八个方向void DFS(int x,int y) {map[x][y]='.';//置当前位置没有积水for(int j=0; j<8; j++) {int gx=x+dx[j];int gy=y+dy[j];if(gx>=0&&gx<=N&&gy>=0&&gy<=M&&map[gx][gy]=='W') {DFS(gx,gy);//满足条件继续查询}}return ;}int main() {scanf("%d%d",&N,&M);for(int j=0; j<N; j++)scanf("%s",map[j]);//键入地图int ans=0;//记录答案for(int j=0; j<N; j++) {for(int k=0; k<M; k++) {if(map[j][k]=='W') {//从有积水的地方遍历DFS(j,k);ans++;}}}printf("%d\n",ans);return 0;}