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 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
解题思路：
这个题目的大意是我们有许多的小分块，但是当这些小分块的水平连通，垂直连通或者是对角线连通时，这就是一个大的块。现在我们要求的就是这块油田有多少个分块。这是一个深度搜索的问题。当我们找到一个小块的时候，我们就将这个点的八个方向进行一次查找，当走过一个@后就把它标记为*，此时我们要用到递归的方法，再查找我们找到的@。
程序代码：
#include <iostream> using namespace std; #define N 111 char str[N][N]; int a,b; int next[8][2]={{0,1},{1,1},{1,0},{1,-1},{0,-1},{-1,-1},{-1,0},{-1,1}};//八个方向 void fun(int x,int y) { int m,n,k; for(k=0;k<8;k++) { n=x+next[k][0]; m=y+next[k][1]; if(n<0||n>a-1||m<0||m>b-1||str[n][m]=='*') continue; str[n][m]='*';//标记 fun(n,m);//递归 } } int main() { int flag; while(cin>>a>>b&&(a||b)) { flag=0; for(int i=0;i<a;i++) for(int j=0;j<b;j++) cin>>str[i][j]; for(int x=0;x<a;x++) { for(int y=0;y<b;y++) { if(str[x][y]=='@') { str[x][y]='*';//标记 fun(x,y); flag++;//累加连通的块数 } } } cout<<flag<<endl; } return 0; }