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. wireshark利用正则表达式过滤http协议中的jpg png zip等无用的数据包

    主要工具:小度随身wifi热点 + wireshark抓包工具.(强烈不建议使用360的产品,非常垃圾,而且干扰代理#墙IP,搞得你不能***) 利用wireshark这个强大的协议分析利器.去分析某 ...

  2. 模拟MBR Grub故障修复

    1.  MBR故障修复 备份 mkdir /pp mount /dev/sdb1 /pp dd if=/dev/sda of=/pp/mrb.bak bs=512 count=1   破坏mrb dd ...

  3. day 20 约束 异常处理 MD5

    1.类的约束(重点): 写一个父类.  父类中的某个方法要抛出一个异常  NotImplementError # 项目经理 class Base:     # 对子类进行了约束. 必须重写该方法    ...

  4. 大数据学习--day11(抽象类、接口、equals、compareTo)

    抽象类.接口.equals.compareTo 什么是抽象方法  ?     区分于正常的方法       1.使用了 abstract 修饰符          该修饰符修饰方法 则该方法就是抽象方 ...

  5. Linux下Bash shell学习笔记

    原文地址: http://www.cnblogs.com/NickQ/p/8870423.html 1.shell下没有变量类型和定义的概念. 变量直接使用不用定义 所有值都视为字符串. 在对变量取值 ...

  6. 863. All Nodes Distance K in Binary Tree

    /** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode ...

  7. 【EXCEL】簡単に合計をとる方法

    下記のような表があるとして.合計を取るときみんなSUM関数を使用しています. その方法もよいですが.もっと簡単の方法を説明します. ①合計する部分を選択します. ②ALT+=を押します. ※ノートパソ ...

  8. C++中引用的本质分析

    引用的意义 引用作为变量别名而存在,因此在一些场合可以代替指针 引用相对于指针来说具有更好的可读性和实用性 swap函数的实现对比: void swap(int* a, int* b) { int t ...

  9. 优步UBER司机全国各地奖励政策汇总 (3月21日-3月27日)

    滴快车单单2.5倍,注册地址:http://www.udache.com/ 如何注册Uber司机(全国版最新最详细注册流程)/月入2万/不用抢单:http://www.cnblogs.com/mfry ...

  10. 北京Uber优步司机奖励政策(12月29日)

    滴快车单单2.5倍,注册地址:http://www.udache.com/ 如何注册Uber司机(全国版最新最详细注册流程)/月入2万/不用抢单:http://www.cnblogs.com/mfry ...