POJ 1562 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 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

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

题意描述：
输入n和m为矩阵的大小（1 <= m <= 100 and 1 <= n <= 100），以及n*m的矩阵
计算并输出该矩阵中@的组成的油田有多少个（上下左右斜对角都能相连）
解题思路：
遍历字符数组，遇到'@'对其进行一遍深搜，搜索过程中标记已经走过的地方，直到遍历完所有元素。
代码实现：
 1 #include<stdio.h>
 2 #include<string.h>
 3 char map[110][110];
 4 int m,n,book[110][110];
 5 void dfs(int x,int y,int c);
 6 int main() 
 7 {
 8     int i,j,num;
 9     while(scanf("%d%d",&m,&n),m||n)
10     {
11         memset(map,'0',sizeof(map));//初始化的位置 
12         for(i=1;i<=m;i++)
13         {
14             for(j=1;j<=n;j++) 
15                 scanf(" %c",&map[i][j]);//处理行列数后多余的空格 
16             getchar();
17         }
18         memset(book,0,sizeof(book));
19         num=0;
20         for(i=1;i<=m;i++)
21         {
22             for(j=1;j<=n;j++)
23             {
24                 if(map[i][j]=='@'&&book[i][j]==0)
25                 {
26                     num--;
27                     book[i][j]=num;
28                     dfs(i,j,num);
29                 }
30             }
31         }
32         printf("%d\n",-num);
33     }
34     return 0;
35 }
36 void dfs(int x,int y,int c)
37 {
38     int next[8][2]={0,1,1,1,1,0,1,-1,0,-1,-1,-1,-1,0,-1,1};//注意搜索的方向个数 
39     int k,tx,ty;
40     book[x][y]=c;
41     for(k=0;k<=7;k++)
42     {
43         tx=x + next[k][0];//位置在原来的基础上变化 
44         ty=y + next[k][1];
45         if(tx<1||tx>m||ty>n||ty<1)
46             continue;
47         if(map[tx][ty]=='@' && book[tx][ty]==0)
48         {
49             book[tx][ty]=c;
50             
51             dfs(tx,ty,c);
52         }
53     }
54     return ;
55 }

易错分析：

1、处理地图的时候注意吃掉空格

2、注意搜索方向的个数

3、tx要在原来的基础上进行变化

posted @ 2017-08-01 22:00  Reqaw  阅读(666)  评论(0)  编辑  收藏  举报