Description

The GeoSurvComp geologic survey company is responsible for detecting underground oil deposits. GeoSurvComp works with one large rectangular region of land at a time, and creates a grid that divides the land into numerous square plots. It then analyzes each plot separately, using sensing equipment to determine whether or not the plot contains oil. A plot containing oil is called a pocket. If two pockets are adjacent, then they are part of the same oil deposit. Oil deposits can be quite large and may contain numerous pockets. Your job is to determine how many different oil deposits are contained in a grid. 
 

Input

The input file contains one or more grids. Each grid begins with a line containing m and n, the number of rows and columns in the grid, separated by a single space. If m = 0 it signals the end of the input; otherwise 1 <= m <= 100 and 1 <= n <= 100. Following this are m lines of n characters each (not counting the end-of-line characters). Each character corresponds to one plot, and is either `*', representing the absence of oil, or `@', representing an oil pocket. 
 

Output

For each grid, output the number of distinct oil deposits. Two different pockets are part of the same oil deposit if they are adjacent horizontally, vertically, or diagonally. An oil deposit will not contain more than 100 pockets. 
 

Sample Input

1
1
*
3 5
*@*@*
**@**
*@*@*
1 8
@@****@*
5 5
****@
*@@*@
*@**@
@@@*@
@@**@
0 0
 

Sample Output

0 1 2 2
 
 
套模板,上代码:
 #include<stdio.h>
#include<string.h>
#include<algorithm>
#include<iostream>
using namespace std;
char map[][];
int vis[][];///标记数组
int dir[][]= {{,},{-,},{,},{,-},{,},{-,-},{,-},{-,}};///八面搜素
int n,m;
void DFS(int x,int y)
{
int a,b,i;
vis[x][y]=;
for(i=; i<; i++)
{
a=x+dir[i][];
b=y+dir[i][];
if(a>=&&a<n&&b>=&&b<m&&vis[a][b]==&&map[a][b]=='@')
{
DFS(a,b);
}
}
return ;
}
int main()
{
int count,i,j;
while(scanf("%d%d",&n,&m)!=EOF)
{
getchar();
if(n==&&n==)
break;
memset(map,,sizeof(map));
memset(vis,,sizeof(vis));
count=;
for(i=; i<n; i++)
{
scanf("%s",map[i]);
}
for(i=; i<n; i++)
{
for(j=; j<m; j++)
{
if(vis[i][j]==&&map[i][j]=='@')
{
count++;
DFS(i,j);
}
}
}
printf("%d\n",count);
}
return ;
}

Oil Deposits(DFS连通图)的更多相关文章

  1. HDOJ(HDU).1241 Oil Deposits(DFS)

    HDOJ(HDU).1241 Oil Deposits(DFS) [从零开始DFS(5)] 点我挑战题目 从零开始DFS HDOJ.1342 Lotto [从零开始DFS(0)] - DFS思想与框架 ...

  2. Oil Deposits(dfs)

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission( ...

  3. HDU 1241 Oil Deposits DFS(深度优先搜索) 和 BFS(广度优先搜索)

    Oil Deposits Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total ...

  4. UVa572 Oil Deposits DFS求连通块

      技巧:遍历8个方向 ; dr <= ; dr++) ; dc <= ; dc++) || dc != ) dfs(r+dr, c+dc, id); 我的解法: #include< ...

  5. HDU 1241 Oil Deposits (DFS/BFS)

    Oil Deposits Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Tota ...

  6. HDU-1241 Oil Deposits (DFS)

    Oil Deposits Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 65536/32768K (Java/Other) Total ...

  7. HDU_1241 Oil Deposits(DFS深搜)

    Problem Description The GeoSurvComp geologic survey company is responsible for detecting underground ...

  8. UVa 572 Oil Deposits(DFS)

     Oil Deposits  The GeoSurvComp geologic survey company is responsible for detecting underground oil ...

  9. [POJ] 1562 Oil Deposits (DFS)

    Oil Deposits Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 16655   Accepted: 8917 Des ...

  10. Oil Deposits(dfs水)

    题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1241 Oil Deposits Time Limit: 2000/1000 MS (Java/Othe ...

随机推荐

  1. 编程界失传秘术,SSO单点登录,什么是单点,如何实现登录?

    单点登录 多系统,单一位置登录,实现多系统同时登录的一种技术. 常出现在互联网应用和企业级平台中. 如:京东. 单点登录一般是用于互相授信的系统,实现单一位置登录,全系统有效的. 三方登录:某系统,使 ...

  2. 使用WIn10自带的Linux子系统

    最近一直有安装虚拟机的想法,今天刚刚知道win10有自带的Linux子系统,就准备试一下: 首先要保证自己的电脑处于开发者选项: 然后就要在控制面板的程序和功能页面点击“启用或者关闭WIndows功能 ...

  3. JSP/Servlet开发——第七章 Servel基础

    1.Servlet简介: ●Servlet是一个符合特定规范的 JAVA 程序 , 是一个基于JAVA技术的Web组件. ●Servlet允许在服务器端,由Servlet容器所管理,用于处理客户端请求 ...

  4. 浏览器端用JS实现创建和下载图片

    问题场景 在前端很多的项目中,文件下载的需求很常见.尤其是通过JS生成文件内容,然后通过浏览器端执行下载的操作.如图片,Execl 等的导出功能.日前,项目中就遇到了这类需求,在浏览器端实现保存当前网 ...

  5. JQuery制作网页—— 第七章 jQuery中的事件与动画

    1. jQuery中的事件: ●和WinForm一样,在网页中的交互也是需要事件来实现的,例如tab切换效果,可以通过鼠标单击事件来实现 ●jQuery事件是对JavaScript事件的封装,常用事件 ...

  6. chrome调试微信

    打开微信,设法打开网址 http://debugx5.qq.com (推荐直接把这个网址发给文件传输助手,然后就可以直接打开链接了) 在打开的网页中选择 [信息]->[TBS settings] ...

  7. 小白CSS学习日记-----杂乱无序记录(3)

    1.后代选择器 .antzone li { } class='antzone' 所有子孙后代中的li   2.子选择器 .antzone > li { } class='antzone' 的子一 ...

  8. php源码建博客2--实现单入口MVC结构

    主要: MVC目录结构 数据库工具类制作 创建公共模型类和公共控制器类 --------------文件结构:-------------------------------------- blog├─ ...

  9. 微信小程序通过api接口将json数据展现到小程序上

    实现知乎客户端的一个重要知识前提就是,要知道怎么通过知乎新闻的接口,来把数据展示到微信小程序端上. 那么我们这一就先学习一下,如何将接口获取到的数据展示到微信小程序上. 1.用到的知识点 <1& ...

  10. docker-compose入门示例:一键部署 Nginx+Tomcat+Mysql

    整体环境配置 整体环境的配置,如果一个一个 Dockerfile 去写,那么是相当麻烦的,好在 Docker 有一个名为 Docker-Compose 的工具提供,我们可以使用它一次性完成整体环境的配 ...