Lake Counting (dfs–深度搜索)
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
题意:大致为求水洼的个数 很容易理解
"> #include <iostream>
#include<cstdio>
#include"string.h"
#include"math.h"
using namespace std;
int m,n;//全局变量
char a[105][105];
void dfs(int x,int y)
{
a[x][y]='.';
for(int i=-1; i<=1; i++)//
for(int j=-1; j<=1; j++)//对某一位置的周围8块位置进行判断
{
int nx=x+i,ny=y+j;
if(nx>=0&&nx<m&&a[nx][ny]=='W'&&ny<n&&ny>=0)//递归条件
{
dfs(nx,ny);
}
}
return ;
}
int main()
{
int i,j,k;
while(~scanf("%d %d",&m,&n))
{
k=0;
for(i=0; i<m; i++)
scanf("%s",a[i]);
for(i=0; i<m; i++)
for(j=0; j<n; j++)
{
if(a[i][j]=='W')
{
dfs(i,j);
k++;
}
}
printf("%d\n",k);
}
return 0;
}
</textarea>
思路:先对输入的图形进行遍历,找到W,然后对其进行深搜将与其相连的所有W变成‘’.‘’,最后深搜的次数就是所要求的水洼的个数