POJ 2386 Lake Counting（搜索联通块）

Lake Counting

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 48370		Accepted: 23775

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

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】

对于一个图，八个方向代表相邻，求出相邻的（联通）块的个数

【分析】

以一个点W为入口将相邻的W 深搜一遍,同时将他改掉，避免重搜

【代码】

#include<cstdio>
using namespace std;
const int N=105;
int n,m,ans,dir[8][2]={{0,1},{0,-1},{1,0},{-1,0},{-1,-1},{1,-1},{1,1},{-1,1}};
char mp[N][N];
void dfs(int x,int y){
	mp[x][y]='.';
	for(int i=0;i<8;i++){
		int nx=x+dir[i][0];
		int ny=y+dir[i][1];
		if(nx<1||ny<1||nx>n||ny>m||mp[nx][ny]=='.') continue;
		dfs(nx,ny);
	}
}
inline void Init(){
	scanf("%d%d",&n,&m);
	for(int i=1;i<=n;i++) scanf("%s",mp[i]+1);
}
inline void Solve(){
	for(int i=1;i<=n;i++){
		for(int j=1;j<=m;j++){
			if(mp[i][j]=='W'){
				dfs(i,j);
				ans++;
			}
		}
	}
	printf("%d",ans);
}
int main(){
	Init();
	Solve();
	return 0;
} 

__EOF__

本文作者：shenben
本文链接：https://www.cnblogs.com/shenben/p/10353174.html
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