uva-572 - Oil Deposits

最新推荐文章于 2025-12-25 18:59:07 发布

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#图论

GeoSurvComp地质勘测公司负责地下油藏探测，通过在单一矩形区域内创建网格划分土地，利用感应设备单独分析每个地块，判断是否含有土壤。发现含油地块称为油囊，相邻的油囊构成同一油藏。本文旨在确定网格中不同油藏的数量。

Oil Deposits

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

    简单的图论题，统计八连通块数目，直接遍历搜索统计即可

#include <iostream>

using namespace std;

char g[100][100];

int dir[8][2] = { {-1,1},{-1,0},{-1,-1},{0,1},{0,-1},{1,0},{1,1},{1,-1}};
int m, n;

void dfs( int x, int y)
{
    g[x][y] = '*';

    for( int i = 0; i < 8; i++){
        int nx = x+dir[i][0], ny = y+dir[i][1];
        if( g[nx][ny] == '@' && nx >= 0 && ny >=0 && nx < m && ny < n)
            dfs( nx, ny);
    }
}

int main()
{

    while( cin >> m >> n){
        if( !m && !n) break;
        for( int i = 0; i < m; i++)
            for( int j = 0; j < n; j++)
            cin >> g[i][j];

        int cnt = 0;
        for( int i = 0; i < m; i++)
            for( int j = 0; j < n; j++)
        if( g[i][j] == '@'){
            dfs( i, j);
            cnt++;
        }

        cout << cnt << endl;
    }
    return 0;
}