1 # include<cstdio>
2 # include<iostream>
3
4 using namespace std;
5
6 # define MAX 123
7
8 char grid[MAX][MAX];
9 int nxt[8][2] = { {1,0},{0,-1},{-1,0},{0,1},{1,1},{1,-1},{-1,1},{-1,-1} };
10 int n,m;
11
12 int can_move( int x,int y )
13 {
14 if ( x>=0&&x<n&&y>=0&&y<m )
15 {
16 if ( grid[x][y]==‘W‘ )
17 return 1;
18 }
19 return 0;
20 }
21
22
23 void dfs ( int x,int y )
24 {
25 grid[x][y] = ‘.‘;
26 for ( int i = 0;i < 8;i++ )
27 {
28 int n_x = x+nxt[i][0], n_y = y+nxt[i][1];
29 if ( can_move(n_x,n_y) )
30 {
31 dfs(n_x,n_y);
32 }
33 }
34 return;
35 }
36
37
38 int main(void)
39 {
40 //int n,m;
41 int res;
42 while ( scanf("%d%d",&n,&m)!=EOF )
43 {
44 res = 0;
45 for ( int i = 0;i < n;i++ )
46 scanf("%s",grid[i]);
47 for ( int i = 0;i < n;i++ )
48 {
49 for ( int j = 0;j < m;j++ )
50 {
51 if ( grid[i][j]==‘W‘ )
52 {
53 dfs(i,j);
54 res++;
55 }
56 }
57 }
58 printf("%d\n",res);
59 }
60
61
62 return 0;
63 }
A - Lake Counting
Crawling in process... Crawling failed Time Limit:1000MS Memory Limit:65536KB 64bit IO Format:%I64d & %I64u
Submit Status Practice POJ 2386
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.
题目大意：
就是说一个n*m的网格中，求出联通块的个数。
解题思路：
就是一个dfs的基础题，从任意一个满足要求的点开始dfs，求出直到不能走动为止，res++。
复杂度：
O（8*m*n）
代码：