poj1562 Oil Deposits(简单的深搜)

Oil Deposits

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 14391 Accepted: 7823

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

Source

Mid-Central USA 1997

一道简单的深搜。只要认识这个单词就行adjacent，邻近的。也就是说油田的八个方向都是邻近的，而不是上下左右四个方向。小心这里就行了、

#include <stdio.h>#include <string.h>char map[105][105];int n,m,dir[8][2]={0,1,0,-1,1,0,-1,0,-1,-1,1,1,1,-1,-1,1};//dir保存八个方向bool limit(int x,int y)//判断是否出边界{if(x>=0&&y>=0&&x<m&&y<n&&map[x][y]=='*')return false;elsereturn true;}void dfs(int x,int y){for(int i=0;i<8;i++){int xx=x+dir[i][0];int yy=y+dir[i][1];if(map[xx][yy]=='@'&&limit(xx,yy)){map[xx][yy]='*';//遇到油田设置为*，避免下次遍历、dfs(xx,yy);}}}int main(){while(scanf("%d %d",&m,&n)!=EOF){if(m==0&&n==0)break;memset(map,0,sizeof(map));for(int i=0;i<m;i++)scanf("%s",map[i]);int count=0;for(int i=0;i<m;i++)for(int j=0;j<n;j++){if(map[i][j]=='@'){count++;//计数dfs(i,j);}}printf("%d\n",count);}return 0;}