| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 41340 | Accepted: 20504 |
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
//深度优先的典型题，大致意思是说找水坑
#include<iostream>
using namespace std;
char field[][];
int n, m;
void dfs(int i, int j)
{
field[i][j] = '.';
for(int dx=-;dx<=;dx++)
for (int dy = -; dy <= ; dy++)
{
int k = i + dx, kk = j + dy;
if (k >= && k < n&& kk < m&&kk >= && field[k][kk] == 'W')
dfs(k, kk);
}
return;
}
int main()
{
int mark = ;
cin >> n >> m;
for (int i = ; i < n; i++)
for (int j = ; j < m; j++)
cin >> field[i][j];
for(int i=;i<n;i++)
for (int j = ; j < m; j++)
if (field[i][j] == 'W')
{
dfs(i, j);
mark++;
}
cout << mark << endl;
return ;
}
真正意义上的第一个深度题，代码不是很难理解，可是感觉掌握起来有点小麻烦；
还需多练习