POJ - 2386：Lake Counting
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题目
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.
输入
- 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.
输出
- Line 1: The number of ponds in Farmer John’s field.
输入样例
10 12
W……..WW.
.WWW…..WWW
….WW…WW.
………WW.
………W..
..W……W..
.W.W…..WW.
W.W.W…..W.
.W.W……W.
..W…….W.
输出样例
3
提示
OUTPUT DETAILS:
There are three ponds: one in the upper left, one in the lower left,and one along the right side.
解题思路
参考代码
#include<stdio.h>
const int maxn=100+10;
char arr[maxn][maxn];
int vis[maxn][maxn];
int n,m;
void dfs(int x,int y)
{
vis[x][y]=1;//或者可以令arr[i][j]='.';
for(int i=-1;i<=1;i++)
{
for(int j=-1;j<=1;j++)
{
int nx=x+i,ny=y+j;
if(0<=nx && nx<n && 0<=ny && ny<m && arr[nx][ny]=='W' && !vis[nx][ny])
dfs(nx,ny);
}
}
}
int main()
{
scanf("%d%d",&n,&m);
int i,j;
for(i=0;i<n;i++)
scanf("%s",arr[i]);
int cnt=0;
for(i=0;i<n;i++)
{
for(j=0;j<m;j++)
{
if(arr[i][j]=='W' && !vis[i][j])
{
dfs(i,j);
cnt++;
}
}
}
printf("%d\n",cnt);
return 0;
}