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 mand 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  and . 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 思路:
直接用dfs爆搜
实现代码:
#include<iostream>
#include<cstring>
using namespace std;
const int M = 1e2+;
int vis[M][M];
char mp[M][M];
int n,m;
void dfs(int u,int v,int d){
if(u < ||u >= m||v < ||v >= n) return;
if(vis[u][v]||mp[u][v]!='@') return;
vis[u][v] = ;
for(int i = -;i <= ;i ++){
for(int j = -;j <= ;j ++){
if(i==&&j==) continue;
dfs(u+i,v+j,d);
}
}
return;
} int main()
{
ios::sync_with_stdio(false);
cin.tie();
cout.tie();
while(cin>>m>>n){
if(n==||m==) break;
for(int i = ;i < m;i ++)
for(int j = ;j < n;j ++)
cin>>mp[i][j];
memset(vis,,sizeof(vis));
int ans = ;
for(int i = ;i < m;i ++){
for(int j = ;j < n;j ++){
if(!vis[i][j]&&mp[i][j]=='@'){
dfs(i,j,++ans);
}
}
}
cout<<ans<<endl;
}
return ;
}

UVa 572 油田 (dfs)的更多相关文章

  1. UVa 572 油田(DFS求连通块)

    https://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem& ...

  2. UVA 572 油田连通块-并查集解决

    题意:8个方向如果能够连成一块就算是一个连通块,求一共有几个连通块. 分析:网上的题解一般都是dfs,但是今天发现并查集也可以解决,为了方便我自己理解大神的模板,便尝试解这道题目,没想到过了... # ...

  3. UVA 572 Oil Deposits油田(DFS求连通块)

    UVA 572     DFS(floodfill)  用DFS求连通块 Time Limit:1000MS     Memory Limit:65536KB     64bit IO Format: ...

  4. UVA 572 -- Oil Deposits(DFS求连通块+种子填充算法)

    UVA 572 -- Oil Deposits(DFS求连通块) 图也有DFS和BFS遍历,由于DFS更好写,所以一般用DFS寻找连通块. 下述代码用一个二重循环来找到当前格子的相邻8个格子,也可用常 ...

  5. 2018 Spring Single Training B (uva 572,HihoCoder 1632,POJ 2387,POJ 2236,UVA 10054,HDU 2141)

    这场比赛可以说是灰常的水了,涨信心场?? 今下午义务劳动,去拿着锄头发了将近一小时呆,发现自己实在是干不了什么,就跑到实验室打比赛了~ 之前的比赛补题补了这么久连一场完整的都没补完,结果这场比完后一小 ...

  6. UVA 572 (dfs)

    题意:找出一块地有多少油田.'@'表示油田.找到一块就全部标记. #include<cstdio> #define maxn 110 char s[maxn][maxn]; int n,m ...

  7. UVA 572 dfs求连通块

    The GeoSurvComp geologic survey company is responsible for detecting underground oil deposits. GeoSu ...

  8. UVA - 572 Oil Deposits(dfs)

    题意:求连通块个数. 分析:dfs. #include<cstdio> #include<cstring> #include<cstdlib> #include&l ...

  9. 暴力求解——UVA 572(简单的dfs)

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

随机推荐

  1. 微服务RPC框架选美

    原文:http://p.primeton.com/articles/59030eeda6f2a40690f03629 1.RPC 框架谁最美? Hello,everybody!说到RPC框架,可能大家 ...

  2. WFP loading 窗口显示 SplashScreen

    public partial class App : Application { protected override void OnStartup(StartupEventArgs e) { Spl ...

  3. 20155318 《网络攻防》Exp4 恶意代码分析

    20155318 <网络攻防>Exp4 恶意代码分析 基础问题 如果在工作中怀疑一台主机上有恶意代码,但只是猜想,所有想监控下系统一天天的到底在干些什么.请设计下你想监控的操作有哪些,用什 ...

  4. adb连接不上手机的解决方案

    一.确认手机的USB调试接口是打开的:----------打开开发者模式,暴击手机版本号多次,直到提示已打开开发者模式. 二.使用USB线连接电脑和手机,可以首先执行adb remount(重新挂载系 ...

  5. QT要点

    1. QT设计器最终会被解释为ui_**.h. 2. QString与init之间的转换: QString转int: bool bIsOk; int a = str.toInt( &bIsOk ...

  6. [CERC2017]Intrinsic Interval[scc+线段树优化建图]

    题意 给定一个长度为 \(n\) 的排列,有 \(q\) 次询问,每次询问一个区间 \([l,r]\) ,找到最小的包含 \([l,r]\) 的区间,满足这个区间包含了一段连续的数字. \(n\leq ...

  7. kafka0.8--0.11各个版本特性预览介绍

    kafka-0.8.2 新特性 producer不再区分同步(sync)和异步方式(async),所有的请求以异步方式发送,这样提升了客户端效率.producer请求会返回一个应答对象,包括偏移量或者 ...

  8. git 报错 error: insufficient permission for adding an object to repository database ./objects

    参照:http://stackoverflow.com/questions/1918524/error-pushing-to-github-insufficient-permission-for-ad ...

  9. Vxlan抓包

    实验目的:验证Openstack  vxlan组网模式验证虚拟机数据是否通过物理网卡流出 一. 同网段不同主机间虚拟机通讯 (同网段通讯直接通过物理机隧道口链接对端物理机隧道口,不需要通过网络节点): ...

  10. OPENSTACK重装系统失败导致虚拟机状态为error

    重装系统失败导致虚拟机状态为error DASHBOARD查看虚拟机状态: 查看日志: 磁盘不足导致下载新镜像失败. Virsh list -all 无法发现虚拟机,底层盘消失(因为重装系统时nova ...