Lake Counting
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 20958 | Accepted: 10561 |
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
#include<iostream>
#include<cstdlib>
#include<cstdio>
#include<cstring>
#include<string>
#include<cmath>
#include<algorithm>
using namespace std;
bool map[101][101];
char s[101][101];
int ans=0,n,m;
int dic[8][2]={{1,0},{-1,0},{0,1},{0,-1},{1,1},{1,-1},{-1,1},{-1,-1}};
void dfs(int l,int t)
{
int a,b;
for(int i=0;i<8;i++)
{
a=l+dic[i][0],b=t+dic[i][1];
if(a>=0&&a<m&&b>=0&&b<n&&map[a][b]==true)
map[a][b]=false,dfs(a,b);
}
}
int main()
{
scanf("%d%d",&m,&n);
for(int i=0;i<m;i++)
scanf("%s",s[i]);
for(int i=0;i<m;i++)
for(int j=0;j<n;j++)
{
if(s[i][j]==‘W‘)
map[i][j]=true;
else
map[i][j]=false;
}
for(int i=0;i<m;i++)
for(int j=0;j<n;j++)
if(map[i][j]==true)
map[i][j]=false,ans++,dfs(i,j);
printf("%d\n",ans);
return 0;
}