Lake Counting
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 34735 | Accepted: 17246 |
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
题解
这道题显然是一道入门搜索题
【涨涨自信2333】
只要八个方向dfs就好
代码如下：
#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std; int n,m,ans;
int xx[]={,,,,,,-,-,-},
yy[]={,,-,,,-,,,-};
char a[][];
bool map[][],vis[][]; void dfs(int a,int b)
{
for(int i=;i<=;++i)
{
int x=xx[i]+a,y=yy[i]+b;
if(map[x][y]&&!vis[x][y])
{
vis[x][y]=;
dfs(x,y);
}
}
} int main()
{
scanf("%d%d",&n,&m);
for(int i=;i<=n;++i)
{
scanf("%s",a[i]);
for(int j=;j<m;++j)
if(a[i][j]=='W')map[i][j+]=;
}
for(int i=;i<=n;++i)
for(int j=;j<=m;++j)
if(map[i][j]&&!vis[i][j])
{
vis[i][j]=;
dfs(i,j);
ans++;
}
printf("%d",ans);
}