HDU 1241

最新推荐文章于 2025-09-05 16:34:31 发布

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最新推荐文章于 2025-09-05 16:34:31 发布

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文章标签： java

原文链接：http://www.cnblogs.com/XuYiting/p/9354504.html

本文介绍了一种用于探测地下油藏分布的算法实现。通过深度优先搜索方法来确定不同油藏间的连接情况，并最终统计出特定区域内独立油藏的数量。算法采用网格划分的方式，将土地划分为多个正方形区域进行独立分析。

Oil Deposits

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 41481 Accepted Submission(s): 24011

Problem 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 <= 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

0 1 2 2

// ConsoleApplication10.cpp: 定义控制台应用程序的入口点。
//

//#include "stdafx.h"
#include<iostream>
using namespace std;
int n, m;
const int MAX_N = 105, MAX_M = 105;
char field[MAX_N][MAX_M];
void dfs(int x, int y)
{
    field[x][y] = '*';
    for (int dx = -1; dx <= 1; dx++)
    {
        for (int dy = -1; dy <= 1; dy++)
        {
            int nx = x + dx, ny = y + dy;
            if (0 <= nx && nx < n && 0 <= ny && ny < m&&field[nx][ny] == '@') dfs(nx, ny);
        }
    }
    return;
}
void solve()
{
    int res = 0;
    for (int i = 0; i < n; i++)
    {
        for (int j = 0; j < m; j++)
        {
            if (field[i][j] == '@')
            {
                dfs(i, j);
                res++;
            }
        }
    }
    cout << res << endl;
}
int main()
{
    while (cin >> n >> m)
    {
        if (n == 0 && m == 0)
            break;
        for (int i = 0; i < n; i++)
        {
            for (int j = 0; j < m; j++)
                cin >> field[i][j];
        }
        solve();
    }
    return 0;
}