UVa 572 油田 (dfs)
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
- UVa 572 油田(DFS求连通块)
https://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem& ...
- 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... # ...
- 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: ...
- 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 ...
- 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 ...
- 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 ...
- UVA 572 dfs求连通块
The GeoSurvComp geologic survey company is responsible for detecting underground oil deposits. GeoSu ...
- UVA - 572 Oil Deposits(dfs)
题意:求连通块个数. 分析:dfs. #include<cstdio> #include<cstring> #include<cstdlib> #include&l ...
- 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
- Base64Util 工具类
package com.org.utils; import java.io.ByteArrayOutputStream; public class Base64Util { private stati ...
- stop-hbase.sh一直处于等待状态
今天关闭HBase时,输入stop-hbase.sh一直处于等待状态 解决方法: 先输入:hbase-daemon.sh stop master 再输入:stop-hbase.sh就可以关闭HBase ...
- 2015520吴思其 基于《Arm试验箱的国密算法应用》课程设计个人报告
20155200吴思其 基于<Arm试验箱的国密算法应用>课程设计个人报告 课程设计中承担的任务 完成试验箱测试功能4,5,6以及SM3加密实验的实现 测试四 GPIO0按键中断实验 实验 ...
- 20155307《网络对抗》MSF基础应用
实验过程 实验系统 所需设备: 靶机1:Windows XP Professional SP2 ,IP地址:192.168.1.128 靶机2:Windows XP Professional SP3 ...
- 记一次SpringMVC碰到的坑
在SpringMVC中,我们Controller中接收比如表单的参数,只要保证方法的形参的名字和表单中input元素的的name一样就可以接收到参数. 但是,我开发的一 ...
- mfc CSpinButton
知识点: CSliderCtrl(滑块)控件 CSliderCtrl常用属性 CSliderCtrl类常用成员函数 CSliderCtrl运用示例 一.CSliderCtr常用属性 Orientati ...
- shell脚本事例 -- 获取当前日期的前一天日期
记录一个shell脚本事例,事例中包括shell的一些语法(函数定义.表达式运算.if.case...) #!/bin/sh #获取当前时间 RUN_TIME=`date +%H%M%S` #取当前日 ...
- 记一次Spring的aop代理Mybatis的DAO所遇到的问题
由来 项目中需要实现某个订单的状态改变后然后推送给第三方的功能,由于更改状态的项目和推送的项目不是同一个项目,所以为了不改变原项目的代码,我们考虑用spring的aop来实现. 项目用的是spring ...
- UWP简单示例(一):快速合成音乐MV
说明 本文发布时间较早,内容可能已过时.最新动态请关注 TypeScript 版本.(2019 年 3 月 注) 在线演示: 音频可视化(TypeScript) 准备 IDE:Visual Studi ...
- 整理一些常用的前端CND加速库,VUE,Jquery,axios
VUE https://cdn.staticfile.org/vue/2.2.2/vue.min.js Jquery https://cdn.bootcss.com/jquery/3.4.0/jque ...