洛谷 P1596 [USACO10OCT]湖计数Lake Counting 题解

题目描述

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.

由于近期的降雨，雨水汇集在农民约翰的田地不同的地方。我们用一个NxM(1<=N<=100;1<=M<=100)网格图表示。每个网格中有水(‘W‘) 或是旱地(‘.‘)。一个网格与其周围的八个网格相连，而一组相连的网格视为一个水坑。约翰想弄清楚他的田地已经形成了多少水坑。给出约翰田地的示意图，确定当中有多少水坑。

输入输出格式

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.

第1行：两个空格隔开的整数：N 和 M 第2行到第N+1行：每行M个字符，每个字符是‘W‘或‘.‘，它们表示网格图中的一排。字符之间没有空格。

Line 1: The number of ponds in Farmer John‘s field.

一行：水坑的数量

输入输出样例

10 12
W........WW.
.WWW.....WWW
....WW...WW.
.........WW.
.........W..
..W......W..
.W.W.....WW.
W.W.W.....W.
.W.W......W.
..W.......W.

3

分析：

深度优先搜索。

AC代码：

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cmath>
 4 #include<algorithm>
 5 
 6 using namespace std;
 7 char mp[110][110];
 8 int m,n,ans = 0;
 9 void dfs(int x,int y)
10 {
11     mp[x][y] = ‘.‘;
12         //通过将水坑标记为旱地，避免再次遍历
13     for(int i = -1;i <= 1;i ++)
14         for(int j = -1;j <= 1;j ++)
15         {
16             int nx = x + i;
17             int ny = y + j;
18             if(nx > 0 && nx <= n && ny > 0 && ny <= m && mp[nx][ny] == ‘W‘)
19                 dfs(nx,ny);
20         }
21     return;
22 }
23 
24 int main()
25 {
26     scanf("%d%d",&n,&m);
27     for(int i = 1;i <= n;i ++)
28             scanf("%s",mp[i] + 1);
29     for(int i = 1;i <= n;i ++)
30         for(int j = 1;j <= m;j ++)
31         {
32             if(mp[i][j] == ‘W‘)
33                 dfs(i,j),ans ++;    
34         } 
35     printf("%d",ans);
36     return 0;
37 }