【B007】Lake Counting【难度B】——————————————————————————————————————————
【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】
【分析】
不要被那么一大串英文吓到，这就是一个统计八连快。。。。。。果断DFS，秒A。。。。。。
【代码】
#include<iostream>
using namespace std;
const int maxn=101;
const int maxm=101;
int n,m;
char field[maxn][maxm];
void dfs(int x,int y)
{
field[x][y]='.';
for(int dx=-1;dx<=1;dx++)
{
for(int dy=-1;dy<=1;dy++)
{
int nx=x+dx,ny=y+dy;
if(0<=nx && nx<n && 0<=ny && ny<m && field[nx][ny]=='W') dfs(nx,ny);
}
}
return ;
}
int main()
{
int res=0;
cin>>n>>m;
for(int i=0;i<n;i++)
for(int j=0;j<m;j++)
cin>>field[i][j];
for(int i=0;i<n;i++)
{
for(int j=0;j<m;j++)
{
if(field[i][j]=='W')
{
dfs(i,j);
res++;
}
}
}
cout<<res;
return 0;
}