POJ 2386 Lake Counting

时间：2016-07-25 16:07:22 阅读：174 评论：0 收藏：0 [点我收藏+]

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Lake Counting

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 28966		Accepted: 14505

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

Hint

OUTPUT DETAILS:

There are three ponds: one in the upper left, one in the lower left,and one along the right side.

Source

USACO 2004 November

解析：DFS。从一个‘W‘开始，每次把‘W‘连通的部分消掉，经过多次这种操作之后，图中不再有‘W‘，操作的次数就是结果。

 1 #include <cstdio>
 2 
 3 int n, m;
 4 char s[105][105];
 5 
 6 bool inField(int r, int c)
 7 {
 8     return r >= 0 && r < n && c >= 0 && c < m;
 9 }
10 
11 void dfs(int x, int y)
12 {
13     s[x][y] = ‘.‘;
14     for(int i = -1; i <= 1; ++i){
15         for(int j = -1; j <= 1; ++j){
16             int tx = x+i, ty = y+j;
17             if(inField(tx, ty) && s[tx][ty] == ‘W‘)
18                 dfs(tx, ty);
19         }
20     }
21 }
22 
23 int main()
24 {
25     scanf("%d%d", &n, &m);
26     for(int i = 0; i < n; ++i)
27         scanf("%s", s[i]);
28     int res = 0;
29     for(int i = 0; i < n; ++i){
30         for(int j = 0; j < m; ++j){
31             if(s[i][j] == ‘W‘){
32                 ++res;
33                 dfs(i, j);
34             }
35         }
36     }
37     printf("%d\n", res);
38     return 0;
39 }

POJ 2386 Lake Counting

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原文地址：http://www.cnblogs.com/inmoonlight/p/5703767.html

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