Lake Countinge的s

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

Sample Output

3

解题思路：

题目大意就是问你图中有几个湖泊，W周围八个方向中的W都归为一个湖泊，是连通的。DFS水题，连回溯都不需要，直接往八个方向检索下去，只要是W，就把＇W＇转变成＇．＇。目的是为了下次检索的时候不去访问之前的W。

AC代码：

#include <iostream>#include <cstdio>using namespace std;const int maxn = 100 + 5;char map[maxn][maxn];int dir[8][2] = {{-1,-1},{-1,0},{-1,1},{0,1},{1,1},{1,0},{1,-1},{0,-1}};  // 用二维数组储存八个方向int dfs(int x,int y,int n,int m){     int a, b;     map[x][y] = '.';      // 讲访问过的"W"转化为"."     for(int i = 0;i < 8; i++)     {          a = x + dir[i][0];          b = y + dir[i][1];          if(a < n && a >= 0 && b < m && b >= 0 && map[a][b] == 'W')              dfs(a, b, n, m);     }     return 1;}int main(){    int n, m, ans = 0;    scanf("%d%d", &n, &m);    for(int i = 0; i < n; i++)        scanf("%s", map[i]);    for(int i = 0; i < n; i++)        for(int j = 0; j < m; j++)            if(map[i][j]=='W')       // 检索整个图                ans += dfs(i,j,n,m);    printf("%d\n", ans);    return 0;}