Lake Counting
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 18201 | Accepted: 9192 |
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.
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.
* 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.
There are three ponds: one in the upper left, one in the lower left,and one along the right side.
Source
简单深搜。
遍历迷宫中所有的点，如果是‘W’，开始dfs搜索，将它临近的八个方向是‘W’的点全部走遍，并将走过的点变为‘.’，这样这一块‘W’区域就全部走完。一共走过了多少个这样的区域，就是结果。
代码：
1 #include <iostream>
2
3 using namespace std;
4 int n,m;
5 char a[105][105];
6 int dx[8] = {0,1,1,1,0,-1,-1,-1}; //八个方向
7 int dy[8] = {1,1,0,-1,-1,-1,0,1};
8 bool judge(int x,int y)
9 {
10 if(x<1 || x>n || y<1 || y>m)
11 return 1;
12 if(a[x][y]!='W')
13 return 1;
14 return 0;
15 }
16 void dfs(int x,int y)
17 {
18 a[x][y] = '.'; //将‘W’转化为‘.’
19 for(int i=0;i<8;i++){
20 int nx = x + dx[i];
21 int ny = y + dy[i];
22 //如果这一步是‘W’，且没有越界，可以走。
23 if(judge(nx,ny))
24 continue;
25 dfs(nx,ny);
26 }
27 }
28 int main()
29 {
30 while(cin>>n>>m){
31 int sum = 0;
32 for(int i=1;i<=n;i++)
33 for(int j=1;j<=m;j++)
34 cin>>a[i][j];
35 for(int i=1;i<=n;i++)
36 for(int j=1;j<=m;j++)
37 if(a[i][j]=='W'){
38 sum++;
39 dfs(i,j);
40 }
41 cout<<sum<<endl;
42 }
43 return 0;
44 }
45
Freecode : www.cnblogs.com/yym2013