大家好，又见面了，我是全栈君，今天给大家准备了Idea注册码。
转载请注明出处：题目链接：http://poj.org/problem?id=1562
----------------------------------------------------------------------------------------------------------------------------------------------------------
欢迎光临天资小屋：http://user.qzone.qq.com/593830943/main
----------------------------------------------------------------------------------------------------------------------------------------------------------
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 12 W........WW. .WWW.....WWW ....WW...WW. .........WW. .........W.. ..W......W.. .W.W.....WW. W.W.W.....W. .W.W......W. ..W.......W.
Sample Output
3
代码例如以下：
#include <iostream>
#include <algorithm>
using namespace std;
#include <cstring>
#define TM 100+17
int N, M;
char map[TM][TM];
bool vis[TM][TM];
int xx[8]={0,1,1,1,0,-1,-1,-1};
int yy[8]={1,1,0,-1,-1,-1,0,1};
void DFS(int x, int y)
{
vis[x][y] = true;
for(int i = 0; i < 8; i++)
{
int dx = x+xx[i];
int dy = y+yy[i];
if(dx>=0&&dx<N&&dy>=0&&dy<M&&!vis[dx][dy]&&map[dx][dy] == 'W')
{
vis[dx][dy] = true;
DFS(dx,dy);
}
}
}
int main()
{
int i, j;
while(cin>>N>>M)
{
int count = 0;
memset(vis,false,sizeof(vis));
for(i = 0; i< N; i++)
{
cin>>map[i];
}
for(i = 0; i < N; i++)
{
for(j = 0; j < M; j++)
{
if(map[i][j] == 'W' && !vis[i][j])
{
count++;
DFS(i,j);
}
}
}
cout<<count<<endl;
}
return 0;
}
版权声明：本文博客原创文章，博客，未经同意，不得转载。
版权声明：本文内容由互联网用户自发贡献，该文观点仅代表作者本人。本站仅提供信息存储空间服务，不拥有所有权，不承担相关法律责任。如发现本站有涉嫌侵权/违法违规的内容， 请发送邮件至 举报，一经查实，本站将立刻删除。
发布者：全栈程序员-用户IM，转载请注明出处：https://javaforall.cn/117525.html原文链接：https://javaforall.cn
【正版授权，激活自己账号】： Jetbrains全家桶Ide使用，1年售后保障，每天仅需1毛
【官方授权 正版激活】： 官方授权 正版激活 支持Jetbrains家族下所有IDE 使用个人JB账号...