ACM--DFS--poj 1562--Oil Deposits

 

poj地址：http://poj.org/problem?id=1562

                                                        Oil Deposits

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

0122

==============================天生骄傲的分割线==============================

描述：

       给出一个N*M的矩形区域和每个区域的状态 – @有/*没有石油，（定义）如果两个有石油的区域是相邻的（水平、垂直、斜）则认为这是属于同一个油田。求这块矩形区域一共有多少油田

输入：

       多组输入，每组输入包括两个整数n，m表示矩阵区域的列数和行数（1<=m,n<=100）和矩阵区域的状态，以

 n=m=-1结束输入

这个题是一个简单的DFS，没什么好说的

<pre name="code" class="cpp">#include<iostream>#include<stdio.h>#include<memory.h>using namespace std;char map[101][101];int result;int n,m;int move[8][2]={{-1,-1},{-1,0},{-1,1},{0,-1},{0,1},{1,-1},{1,0},{1,1}};//深度搜索void dfs(int x,int y){    int a,b;    map[x][y] = '*'; //访问过的都由@改为*，剪枝    for(int i=0;i<8;i++){       a=x+move[i][0];       b=y+move[i][1];       if(a>=0&&a<n&&b>=0&&b<m&&map[a][b]=='@'){           dfs(a,b);       }    }}int main(){   while(cin>>n>>m&&n||m){        int i,j;        //初始化        result=0;        for(i=0;i<n;i++){          cin>>map[i];        }        for(i=0;i<n;i++){            for(j=0;j<m;j++){                if(map[i][j]=='@'){                    dfs(i,j);                    result++;                }            }        }        //打印结果        printf("%d\n",result);   }   return 0;}

参考博客：http://blog.csdn.net/v5zsq/article/details/46591883

                  http://blog.csdn.net/keshacookie/article/details/22817433