HDU - 1241 Oil Deposits

HDU - 1241 Oil Deposits

题目

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

题解

染色法求连通块数量但是这个稍微难一点,因为可以斜着走

/*************************************************************************
    > Problem: HDU - 1241 Oil Deposits
    > Author: ALizen_
    > Mail: hhu_zyh@qq.com
    > Blog: https://blog.youkuaiyun.com/Jame_19?spm=1001.2101.3001.5343
*************************************************************************/
#include<set>
#include<map>
#include<cmath>
#include<queue>
#include<stack>
#include<vector>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define re register int
#define sf(x) scanf("%d",&x)
#define pf(x) printf("%d\n",x)
#define slf(x) scanf("%lld",&x)
#define plf(x) printf("%lld\n",x)
#define MEM(a,b) memset((a),(b),sizeof(a))
#define ref(i,a,b) for(re i = a ; i >= b ; -- i)
#define fer(i,a,b) for(re i = a ; i <= b ; ++ i)
using namespace std;
typedef long long LL;
int dx[]={0,0,1,-1,1,-1,1,-1};
int dy[]={1,-1,0,0,1,1,-1,-1};
int book[N][N];
int num,n,m;
char mp[N][N];
void dfs(int x,int y){
	if(mp[x][y]=='*'||x<1||x>n||y<1||y>m||book[x][y]!=0)
	return ;
	book[x][y]=num;
	for(int j=0;j<=7;j++){
		int tx=x+dx[j];
		int ty=y+dy[j];
		if(mp[tx][ty]=='@'&&tx>=1&&tx<=n&&ty>=1&&ty<=m)
		dfs(tx,ty);
	}
} 
int main(){
    while(cin>>n>>m,n+m){
    	num=0;
    	memset(book,0,sizeof(book));
    	for(int i=1;i<=n;i++)
    	for(int j=1;j<=m;j++)
    	cin>>mp[i][j];
    	for( int i=1;i<=n;i++){
    		for(int j=1;j<=m;j++){
    			if(mp[i][j]=='@'&&book[i][j]==0)
    			        num++;
				dfs(i,j);
    			
			}
		}
		cout<<num<<endl;
	}


    return 0;
}

汉字字库存储芯片扩展实验 # 汉字字库存储芯片扩展实验 ## 实验目的 1. 了解汉字字库的存储原理和结构 2. 掌握存储芯片扩展技术 3. 学习如何通过硬件扩展实现大容量汉字字库存储 ## 实验原理 ### 汉字字库存储基础 - 汉字通常采用点阵方式存储(如16×16、24×24、32×32点阵) - 每个汉字需要占用32字节(16×16)到128字节(32×32)不等的存储空间 - 国标GB2312-80包含6763个汉字,需要较大存储容量 ### 存储芯片扩展方法 1. **位扩展**:增加数据总线宽度 2. **字扩展**:增加存储单元数量 3. **混合扩展**:同时进行位扩展和字扩展 ## 实验设备 - 单片机开发板(如STC89C52) - 存储芯片(如27C256、29C040等) - 逻辑门电路芯片(如74HC138、74HC373等) - 示波器、万用表等测试设备 - 连接线若干 ## 实验步骤 ### 1. 单芯片汉字存储实验 1. 连接27C256 EPROM芯片到单片机系统 2. 将16×16点阵汉字字库写入芯片 3. 编写程序读取并显示汉字 ### 2. 存储芯片字扩展实验 1. 使用地址译码器(如74HC138)扩展多片27C256 2. 将完整GB2312字库分布到各芯片中 3. 编写程序实现跨芯片汉字读取 ### 3. 存储芯片位扩展实验 1. 连接两片27C256实现16位数据总线扩展 2. 优化字库存储结构,提高读取速度 3. 测试并比较扩展前后的性能差异 ## 实验代码示例(单片机部分) ```c #include <reg52.h> #include <intrins.h> // 定义存储芯片控制引脚 sbit CE = P2^7; // 片选 sbit OE = P2^6; // 输出使能 sbit
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