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
理解:
整道题就一个思路,那就是求有几个联通块 . 这也是我第一次就出DFS . 那时候感觉这个算法非常神奇 . 后来得知这其实和我们学习的离散数学的树的深度遍历是一个道理时...我就傻了.果然知识没学好在以后的日子里就会收到影响
做法
DFS (不知道的可以网上搜索或者看书,我参考的是《算法竞赛 入门经典》(第二版))P162.
代码部分
//第一个真正意义上的自己写的深搜算法...激动死啊..dfs
#include<iostream>
using namespace std;
int b=0,c=0;
char a[101][101];int aa[101][101];
void dfs(int i,int j)//这个函数虽然简单,但是绝大多数运算都被它占据了.
{
if(a[i][j]!='@'||aa[i][j]==1||i<0||j<0||i>b||j>c) return;
aa[i][j]=1;
dfs(i-1,j-1);//八个方向搜索
dfs(i-1,j);
dfs(i-1,j+1);
dfs(i,j-1);
dfs(i,j+1);
dfs(i+1,j-1);
dfs(i+1,j);
dfs(i+1,j+1);
}
int main()
{
while(cin>>b>>c)
{
if(b==0&&c==0)
return 0;
int sum=0;
for(int i=0;i<b;i++)
{
for(int j=0;j<c;j++)
{
cin>>a[i][j];
aa[i][j]=0;
}
}
for(int k=0;k<b;k++)
{
for(int l=0;l<c;l++)
{
if(!aa[k][l]&&a[k][l]=='@')
{
dfs(k,l);
sum++;
}
}
}
cout<<sum<<endl;
}
return 0;
}
