HDU_1241 Oil Deposits

Problem 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

题目大意：

地质urvcomp地质勘测公司负责探测地下油层。GeoSurvComp每次处理一个大的矩形区域，并创建一个网格，将土地划分为许多方形地块。然后用传感设备分别分析每个地块，以确定该地块是否含有石油。一块含有石油的土地叫做口袋。如果两个储层相邻，则它们属于同一油层的一部分。石油储量可以相当大，可能包含许多口袋。你的工作是确定有多少不同的石油蕴藏在一个网格。

输入

输入文件包含一个或多个网格。每个网格都以包含m和n的一行开始，m和n是网格中行和列的数量，由一个空格分隔。如果m = 0，则表示输入结束;否则1 <= m <= 100, 1 <= n <= 100。下面是m行，每行有n个字符(不包括行尾字符)。每个角色对应一个情节，要么是“*”，代表没有油，要么是“@”，代表一个油袋。

输出

对于每个网格，输出不同的石油沉积物的数量。如果两个不同的储层在水平、垂直或对角方向相邻，那么它们就是同一个油层的一部分。一个石油矿床不会包含超过100个口袋。

样例输入

1 1

*

3 5

*@*@*

**@**

*@*@*

1 8

@@****@*

5 5

****@

*@@*@

*@**@

@@@*@

@@**@

0 0

样例输出

0 1 2 2

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3
 4 char martrix[110][110];
 5 int move[8][2]={1,0,-1,0,0,1,0,-1,1,1,-1,-1,1,-1,-1,1};//两个坐标一组 分为8组
 6 int h,w;
 7 void dfs(int x,int y)//定义dfs函数，主函数找到了@，dfs启动，寻找主函数找到的@八面存在的@
 8 {
 9     int next_x,next_y,i;
10     martrix[x][y]=‘*‘;//把找到的@变为*，以免重复搜索
11     for(i=0;i<8;i++)
12     {
13         next_x=x+move[i][0];//[0]表示两个坐标一组的第一个 [i]表示两个坐标一组的第几组
14         next_y=y+move[i][1];//[1]表示两个坐标一组的第二个 [i]表示两个坐标一组的第几组
15         if(next_x>=0&&next_x<h&&next_y>=0&&next_y<w)
16         {
17             if(martrix[next_x][next_y]==‘@‘)
18             {
19                 dfs(next_x,next_y);
20             }
21         }
22     }
23 }
24
25 int main()//主函数开始，寻找第一个@
26 {
27     freopen("input.txt","r",stdin);
28     int i,j,sum;
29     while(scanf("%d%d",&h,&w)&&(h!=0||w!=0))
30     {
31         for(i=0;i<h;i++)
32         scanf("%s",martrix[i]);
33         sum=0;
34         for(i=0;i<h;i++)
35         {
36             for(j=0;j<w;j++)
37             {
38                 if(martrix[i][j]==‘@‘)
39                 {
40                     dfs(i,j);//转移到dfs函数，由dfs函数开始搜索主函数找到@的八面存在的@
41                     sum++;
42                 }
43             }
44         }
45         printf("%d\n",sum);
46     }
47     fclose(stdin);
48     return 0;
49 }

原文地址：https://www.cnblogs.com/WindSun/p/10529416.html