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
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
输入样例#1
10 12
W........WW.
.WWW.....WWW
....WW...WW.
.........WW.
.........W..
..W......W..
.W.W.....WW.
W.W.W.....W.
.W.W......W.
..W.......W.
输出样例#1
3思路
水题，就是连通块的纯模版，下面用dfs做。
#include <stdio.h>
#include <iostream>
using namespace std;
int n,m,s;
char a[101][101];
int tox[9]={0,1,0,-1,0,1,1,-1,-1};
int toy[9]={0,0,1,0,-1,-1,1,1,-1};
inline void dfs(int x,int y)
{
register int i;
a[x][y]='.';
for(i=1;i<=8;i++)
{
int x1=x+tox[i];
int y1=y+toy[i];
if(x1>=1 && x1<=n && y1>=1 && y1<=m && a[x1][y1]=='W')
{
dfs(x1,y1);
}
}
return;
}
int main()
{
ios::sync_with_stdio(false);
int i,j;
cin>>n>>m;
for(i=1;i<=n;i++)
{
for(j=1;j<=m;j++)
{
cin>>a[i][j];
}
}
for(i=1;i<=n;i++)
{
for(j=1;j<=m;j++)
{
if(a[i][j]=='W')
{
s++;
dfs(i,j);
}
}
}
cout<<s<<endl;
return 0;
}