1751: [Usaco2005 qua]Lake Counting
Time Limit: 5 Sec Memory Limit: 64 MBSubmit: 168 Solved: 130
[Submit][Status]
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
OUTPUT DETAILS:
There are three ponds: one in the upper left, one in the lower left,
and one along the right side.
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#include<cstdio>
#include<iostream>
using namespace std;
const int mx[8]={1,1,1,0,0,-1,-1,-1};
const int my[8]={1,0,-1,1,-1,1,0,-1};
int n,m,ans;
int map[110][110];
void dfs(int x,int y)
{
map[x][y]=0;
for (int k=0;k<8;k++)
{
int nx=mx[k]+x;
int ny=my[k]+y;
if (map[nx][ny]) dfs(nx,ny);
}
}
int main()
{
scanf("%d%d",&n,&m);
for (int i=1;i<=n;i++)
for (int j=1;j<=m;j++)
{
char ch;
cin>>ch;
if (ch=='W')map[i][j]=1;
}
for (int i=1;i<=n;i++)
for (int j=1;j<=m;j++)
if(map[i][j]){dfs(i,j);ans++;}
printf("%d",ans);
}