博客
关于我
HDU 1241 Oil Deposits
阅读量:787 次
发布时间:2019-03-23

本文共 5108 字,大约阅读时间需要 17 分钟。

    

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 ≤ 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    

C++ Implementation

        #include 
#include
#include
#include
using namespace std; const int MAXN = 105; char maze[MAXN][MAXN]; bool vis[MAXN][MAXN]; int n, m; bool judge(int x, int y) { return x >= 0 && x < n && y >= 0 && y < m; } void dfs(int x, int y) { vis[x][y] = true; for (int i = -1; i <= 1; ++i) { for (int j = -1; j <= 1; ++j) { int tx = x + i; int ty = y + j; if (judge(tx, ty) && !vis[tx][ty] && maze[tx][ty] == '@') { dfs(tx, ty); } } } } int main() { while (scanf("%d%d", &n, &m) != EOF) { if (n == 0 && m == 0) break; for (int i = 0; i < n; ++i) { scanf("%s", maze[i]); } memset(vis, false, sizeof(vis)); int ans = 0; for (int i = 0; i < n; ++i) { for (int j = 0; j < m; ++j) { if (maze[i][j] == '@' && !vis[i][j]) { ans++; dfs(i, j); } } } cout << ans << endl; } return 0; }

Java Implementation

        import java.util.Scanner;        public class Main {            static int MAXN = 105;            static boolean vis[][] = new boolean[MAXN][MAXN];            static char maze[][] = new char[MAXN][MAXN];            static int n, m;            static boolean judge(int x, int y) {                if (x < 0 || x >= n || y < 0 || y >= m) return false;                return true;            }            public static void main(String args[]) {                Scanner cin = new Scanner(System.in);                while (cin.hasNext()) {                    n = cin.nextInt();                    m = cin.nextInt();                    cin.nextLine();                    if (n == 0 && m == 0) break;                    for (int i = 0; i < n; ++i) {                        for (int j = 0; j < m; ++j) {                            vis[i][j] = false;                        }                    }                    for (int i = 0; i < n; ++i) {                        String s = cin.nextLine();                        maze[i] = s.toCharArray();                    }                    int ans = 0;                    for (int i = 0; i < n; ++i) {                        for (int j = 0; j < m; ++j) {                            if (maze[i][j] == '@' && !vis[i][j]) {                                ans++;                                dfs(i, j);                            }                        }                    }                    System.out.println(ans);                }                cin.close();            }            static void dfs(int x, int y) {                vis[x][y] = true;                for (int i = -1; i <= 1; ++i) {                    for (int j = -1; j <= 1; ++j) {                        int tx = x + i;                        int ty = y + j;                        if (judge(tx, ty) && !vis[tx][ty] && maze[tx][ty] == '@') {                            dfs(tx, ty);                        }                    }                }            }        }    

转载地址:http://olhzk.baihongyu.com/

你可能感兴趣的文章
MySQL底层概述—9.ACID与事务
查看>>
Mysql建立中英文全文索引(mysql5.7以上)
查看>>
mysql建立索引的几大原则
查看>>
Mysql建表中的 “FEDERATED 引擎连接失败 - Server Name Doesn‘t Exist“ 解决方法
查看>>
mysql开启bin-log日志,用于canal同步
查看>>
MySQL开源工具推荐,有了它我卸了珍藏多年Nactive!
查看>>
MySQL异步操作在C++中的应用
查看>>
MySQL引擎讲解
查看>>
Mysql当前列的值等于上一行的值累加前一列的值
查看>>
MySQL当查询的时候有多个结果,但需要返回一条的情况用GROUP_CONCAT拼接
查看>>
MySQL必知必会(组合Where子句,Not和In操作符)
查看>>
MySQL必知必会总结笔记
查看>>
MySQL快速入门
查看>>
MySQL快速入门——库的操作
查看>>
mysql快速复制一张表的内容,并添加新内容到另一张表中
查看>>
mysql快速查询表的结构和注释,字段等信息
查看>>
mysql怎么删除临时表里的数据_MySQL中关于临时表的一些基本使用方法
查看>>
mysql性能优化
查看>>
mysql性能优化学习笔记-存储引擎
查看>>
MySQL性能优化必备25条
查看>>