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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

 0122 

 

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题意：

题目的意思就是在给出的图中@代表有石油,*代表没有石油,而在一个有石油的地方它的周围8个方向的地方如果也有石油,那么这2块石油是属于一块的,给出图,问图中有几块石油田.

思路：

类似我们课堂上的找药丸那个题，运用深度搜索，寻找符合条件的！详见代码：

#include<cstdio>using namespace std;char a[101][101];bool b[101][101]; //开一个bool数组，进行判断有没有访问过int n,m,i,j,s=0;int dx[8]={0,0,-1,1,-1,1,-1,1};int dy[8]={-1,1,0,0,1,-1,-1,1};//8个方向void f(int,int); //搜索int main(){     while(~scanf("%d%d\n",&n,&m)){s=0;if(n==0&&m==0)break;for (i=0;i<n;i++)    gets(a[i]);for (i=0;i<n;i++)    for (j=0;j<m;j++)     if(a[i][j]=='*')b[i][j]=0; //将b初始化else b[i][j]=1;for (i=0;i<n;i++)      for (j=0;j<m;j++)      if (b[i][j])      {         b[i][j]=0;         f(i,j);         s++;    //符合条件 加一   }   printf("%d\n",s);     }    return 0; }void f(int x,int y){  for (int i=0;i<8;i++)  if(x+dx[i]>-1&&x+dx[i]<n&&y+dy[i]>-1&&y+dy[i]<m&&b[x+dx[i]][y+dy[i]]) {                     b[x+dx[i]][y+dy[i]]=0;    f(x+dx[i],y+dy[i]);       //对找到的点的8个方向进行判断，并将它们置0                            }      }

心得：

搜索做起来还是有点吃力，继续刷题