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。
输出格式
Line 1: The number of ponds in Farmer John's field.
一行：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.
AC代码及注释如下
#include<bits/stdc++.h>
using namespace std;
int cnt = 0,n,m;
int col[102][102];//用来染色
char a[102][102];
int to[8][2] = { {1,0},{0,1},{0,-1},{-1,0},{1,1},{-1,-1},{1,-1},{-1,1}};//八个方向
void dfs(int x, int y, int k) {//x行y列，第k个水池
if (x<1 || x>n || y<1 || y>m||a[x][y]=='.')return;//如果遇到边界或者'.'便返回
for (int i = 0; i < 8; i++) {//八个方向搜索
int tx = x + to[i][0];
int ty = y + to[i][1];
//八个方向中相邻的水都属于同一个水池
if (a[tx][ty] == 'W' && col[tx][ty] == 0) {
col[tx][ty] = k;//用当前水池数对每个水池中的水格进行染色
dfs(tx, ty,k);//继续搜索
}
}
}
int main(){
cin >> n >> m;
memset(col, 0, sizeof(col));
for (int i = 1; i <= n; i++)
for (int j = 1; j <= m; j++)
cin >> a[i][j];
for (int i = 1; i <=n; i++) {
for (int j = 1; j <= m; j++) {
if (a[i][j] == 'W' && col[i][j] == 0) {//新水池的一部分
cnt++;//水池数++
col[i][j] = cnt;//染色
dfs(i, j, cnt);//开始搜索
}
}
}
cout << cnt;//完美输出
return 0;
}
