POJ-1562 Oil Deposits

油藏探测算法解析

最新推荐文章于 2021-08-15 08:36:35 发布

转载最新推荐文章于 2021-08-15 08:36:35 发布 · 48 阅读

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原文链接：http://www.cnblogs.com/wiklvrain/p/8179448.html

本文介绍了一种通过深度优先搜索（DFS）算法来确定矩形区域内不同油藏数量的方法。该算法将区域划分为多个正方形地块，并针对含有油藏的地块进行分析，通过递归搜索相邻地块来识别独立的油藏。输入包括区域大小和地块状态，输出为独立油藏的数量。

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

解题思路：

dfs

代码：

#include <cstdio>
#include <cstring>
using namespace std;
const int maxn = 100 + 5;

int n, m;
char s[maxn][maxn];

void dfs(int row, int col){
    s[row][col] = '*';
    for(int i = -1; i <= 1; ++i){
        for(int j = -1; j <= 1; ++j){
            if(i == 0 && j == 0) continue;
            int nx = row + i;
            int ny = col + j;
            if(nx >= 0 && nx < n && ny >= 0 && ny < m && s[nx][ny] == '@'){
                dfs(nx, ny);
            }
        }
    }
}
int main()
{
    while(~scanf("%d%d", &n, &m)){
        if(m == 0) break;
        for(int i = 0; i < n; ++i){
            for(int j = 0; j < m; ++j){
                scanf(" %c", &s[i][j]);
            }
        }
        
        int ans = 0;
        for(int i = 0; i < n; ++i){
            for(int j = 0; j < m; ++j){
                if(s[i][j] == '@'){
                    dfs(i, j);
                    ++ans;
                }
            }
        }
        printf("%d\n", ans);
    }
    return 0;
}

转载于:https://www.cnblogs.com/wiklvrain/p/8179448.html