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
Given a diagram of Farmer John's field, determine how many ponds he has.
* 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
* 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.
* Line 1: The number of ponds in Farmer John's field.
Sample Input
10 12W........WW..WWW.....WWW....WW...WW..........WW..........W....W......W...W.W.....WW.W.W.W.....W..W.W......W...W.......W.Sample Output
3Hint
OUTPUT DETAILS:
There are three ponds: one in the upper left, one in the lower left,and one along the right side.
There are three ponds: one in the upper left, one in the lower left,and one along the right side.
这题就是让你找有几个连通域，这题是向8个方向发散
#include
#include
#include
using namespace std;char a[110][110];int n,m;int s=0;void bfs(int i,int j){
a[i][j]='.';
if(j+1
bfs(i,j+1);
if(i+1
bfs(i+1,j);
if(i+1
bfs(i+1,j+1);
if(i-1>=0&&j+1
bfs(i-1,j+1);
if(i-1>=0&&a[i-1][j]=='W')
bfs(i-1,j);
if(j-1>=0&&a[i][j-1]=='W')
bfs(i,j-1);
if(i-1>=0&&j-1>=0&&a[i-1][j-1]=='W')
bfs(i-1,j-1);
if(i+1
=0&&a[i+1][j-1]=='W')
bfs(i+1,j-1);}int main(){
scanf("%d%d",&n,&m);
char c;
scanf("%c",&c);
for(int i=0;i
{
for(int j=0;j
scanf("%c",&a[i][j]);
scanf("%c",&c);
}
for(int i=0;i
{
for(int j=0;j
{
if(a[i][j]=='W')
{
//
printf("%d %d\n",i,j);
s++;
bfs(i,j);
}
}
}
printf("%d",s);
return 0;}
