poj 1562 Oil Deposits

最新推荐文章于 2025-09-13 23:06:41 发布

转载最新推荐文章于 2025-09-13 23:06:41 发布 · 76 阅读

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GeoSurvComp公司通过地质调查技术，在矩形区域内划分土地为多个正方形地块，利用感应设备检测每块地是否含有土壤。若两块含油地块相邻，则属于同一油藏。本文提供了一个算法，用于确定网格中不同油藏的数量。

2019独角兽企业重金招聘Python工程师标准>>>

/*Oil Deposits
Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 10708		Accepted: 5802

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

0
1
2
2

Source
Mid-Central USA 1997*/

#include <cstdio>
#include <cstdlib>
#include <iostream>
#include <string>
#include <cstring>

using namespace std;
int m = 0, n = 0, sum_ = 0;
bool mp[105][105] = {0};
int color[105][105] = {0};
int dx[8] = {0, 1,1,1,0,-1,-1,-1}, dy[8] = {-1,-1,0,1,1,1,0,-1};

void dfs(int x, int y, int c)
{
    for(int i = 0; i < 8; ++i)
    {
        if(y + dy[i] >= 0 && y + dy[i] < n && mp[x + dx[i]][y + dy[i]] == 1 && color[x + dx[i]][y + dy[i]] == 0)
        {
            color[x + dx[i]][y + dy[i]] = c;
            dfs(x + dx[i], y + dy[i], c);
        }
    }
    return;
}


int main()
{
    //while(scanf("%d%d", &m, &n))
    while(true)
    {
        cin >> m >> n;
        if(m == 0 || n == 0) break;
        memset(mp, 0, sizeof(mp));
        memset(color,0,sizeof(color));
        sum_ = 0;
        //scanf("\n");
        char str[105];
        for(int i = 1; i <= m; ++i)
        {
            cin >> str;
            //scanf("%s", str);
            for(int j = 0; j < n; ++j)
            {
                if(str[j] == '@') mp[i][j] = 1;
                else mp[i][j] = 0;
            }
        }
        for(int i = 1; i <= m; ++i)
        for(int j = 0; j < n; ++j)
        {
            if(mp[i][j] == 1 && color[i][j] == 0)
            {
                sum_++;
                dfs(i, j, sum_);
            }
        }
        cout << sum_ << endl;
        //printf("%d\n", sum_);
    }
    return 0;
}

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