#HDU 1241 Oil Deposits (dfs求连通块)

最新推荐文章于 2021-02-16 20:21:13 发布

olamins_

最新推荐文章于 2021-02-16 20:21:13 发布

阅读量140

点赞数 1

CC 4.0 BY-SA版权

分类专栏： DFS

本文链接：https://blog.youkuaiyun.com/weixin_43851525/article/details/92235504

DFS 专栏收录该内容

7 篇文章

订阅专栏

本文介绍了一种用于探测地下油藏分布的算法。通过将土地划分为网格，并使用深度优先搜索（DFS）来识别和计数不同的油藏沉积物。算法在遇到油藏标记时，会遍历并标记所有相邻油藏，最终确定独立油藏数量。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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

题目大意：给出一张图，求出所有以 “ @” 为联通块的数量

思路：每碰到 “@”，就从该坐标开始遍历，把所有与之相邻的@全部改为‘*’,同时，连通块的数量 + 1；

AC代码：

#include<iostream>
#include<cstring>
#include<set>
using namespace std;
const int maxn = 1e2 + 5;

struct node
{
    int x, y;
};
char p[maxn][maxn];
int n, m, sum;
int dir[8][2] = {0, -1, 1, -1, 1, 0, 1, 1, 0, 1, -1, 1, -1, 0, -1, -1};
void dfs(int x, int y) {
    for (int i = 0; i < 8; i++) {
        int xx = x + dir[i][0];
        int yy = y + dir[i][1];
        if (xx >= 0 && yy >= 0 && xx < n && yy < m && p[xx][yy] == '@') {
            p[xx][yy] = '*';
            dfs(xx, yy);
        }
    }
}

int main()
{
    while (cin >> n >> m && n + m) {
        sum = 0;
        for (int i = 0; i < n; i++) cin >> p[i];
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {
                if (p[i][j] == '@') {
                    sum++;
                    dfs(i, j);
                }
            }
        }
        cout << sum << endl;
    }
    return 0;
}