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.
输入格式
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.
输出格式
Line 1: The number of ponds in Farmer John's field.
题意翻译
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 �×�(1≤�≤100,1≤�≤100)N×M(1≤N≤100,1≤M≤100) 的网格图表示。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。
输入第 11 行：两个空格隔开的整数：�N 和 �M。
第 22 行到第 �+1N+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。
输出一行，表示水坑的数量。
输入输出样例
输入 #1复制
10 12 W........WW. .WWW.....WWW ....WW...WW. .........WW. .........W.. ..W......W.. .W.W.....WW. W.W.W.....W. .W.W......W. ..W.......W.
输出 #1复制
3
说明/提示
OUTPUT DETAILS: There are three ponds: one in the upper left, one in the lower left, and one along the right side.
解题思路
就是找W连通块的数量，具体怎么找，看下面这个图
找到上面三个符合要求的点就用无尽的水去灌溉（DFS），知道没有水域为止。
#include<iostream>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<string>
#include<vector>
#include<math.h>
#include<iomanip>
#include<set>
#include<queue>
#include<stack>
#include<map>
#include<list>
#include <stdlib.h>
#include<deque>
using namespace std;
int n, m, ans, v[105][105];//地图大小，连通块计数，记录水域状态
char g[105][105];//图
int a[8][2] = { -1,0,0,1,1,0,0,-1,-1,1,1,1,1,-1,-1,-1 };//八连通，八个反方向灌溉
void dfs(int x, int y)
{for (int i = 0; i < 8; i++){int cx = x + a[i][0], cy = y + a[i][1];if (g[cx][cy] == 'W' && cx < n && cx >= 0 && cy >= 0 && cy < m&&v[cx][cy]==0)判断是否满足灌溉条件{v[cx][cy] = 1;dfs(cx, cy);//不用恢复上一步，因为已经灌溉过了，不用再抽干再灌溉}}
}
int main()
{cin >> n >> m;for (int i = 0; i < n; i++){for (int j = 0; j < m; j++){cin >> g[i][j];}}for (int i = 0; i < n; i++){for(int j=0;j<m;j++){if (g[i][j] == 'W' && v[i][j] == 0)//找到一个适合的点，开始灌溉{dfs(i, j);ans++;}}}cout << ans;
}
