Vjudge Oil Deposits 油田问题（dfs）

最新推荐文章于 2022-03-09 12:58:56 发布

柏油

最新推荐文章于 2022-03-09 12:58:56 发布

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分类专栏： # 图论 DFS

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本文介绍了一种用于探测地下油田分布的算法实现。通过将土地划分为网格，并使用递归深度优先搜索的方法来确定不同油田的数量。每个含有石油的地块被视为一个口袋，如果两个口袋相邻，则属于同一油田。

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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

AC代码：

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int maxn = 100 + 2;
char g[maxn][maxn];
int d[maxn][maxn];
int m,n;
	
void dfs(int a ,int b){
	d[a][b] = 1;
	for(int i = -1; i <= 1; i++){
		for(int j = -1; j <=1; j++){
			int x = a+i;
			int y = b+j; 
			if(x < 0 || x >= m || y < 0 || y >= n) continue;
			if(g[x][y] == '@' && !d[x][y]){
					dfs(x,y);
			}
		}
	}
}

int main(){
	while(scanf("%d%d",&m,&n) == 2 && m != 0){
		getchar();
		for(int i = 0; i < m; i++){
			scanf("%s",g[i]);
		}
		
		memset(d,0,sizeof(d));
		int cnt = 0;
		for(int i = 0; i < m; i++){
			for(int j = 0; j < n; j++){
				if(g[i][j] =='@' && !d[i][j]){
					dfs(i,j);
					cnt++;
				}
			}
		}
		
		printf("%d\n",cnt);
	}
	return 0;
}