UVA - 572 Oil Deposits

最新推荐文章于 2021-05-26 22:19:09 发布

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GeoSurvComp地质调查公司通过检测地下油藏进行工作。它使用感应设备在单一大矩形区域的土地上创建网格，将土地划分为多个正方形地块，并单独分析每个地块来确定是否含有土壤。油气藏可能非常大，包含多个油藏口袋。任务是确定网格中不同数量的油气藏。

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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 $1 \le m \le 100$ and $1 \le n \le 100$ . 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

分析：和二叉树一样，图的处理也是DFS和BFS，这些都是套路，由于DFS更容易写，一般都是DFS找连通块：从每个’@‘格子出发，递归遍历它周围的’@‘格子。每次访问一个格子时就给他写上一个连通分量编号，这样就可以在访问之前检查它是否已经有了编号，从而避免同一个格子访问多次。

#include <cstdio>
#include <cstring>

using namespace std;
const int maxn = 110;
int idx[maxn][maxn];
char pic[maxn][maxn];
int m, n;

void dfs(int row, int col, int id) {
	if(row<0 || row>m-1 || col<0 || col>n-1) return;
	if(idx[row][col]>0 || pic[row][col]!='@') return;
	idx[row][col] = id;
	for(int dr=-1; dr<=1; dr++)
		for(int dc=-1; dc<=1; dc++)
			if(dr!=0 || dc!=0)dfs(row+dr, col+dc, id);
}

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