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数据库导入导出_windows系统以及linux系统下的操作---linux工作笔记042
查看>>
mysql数据库导出导入
查看>>
MySQL数据库工具类之——DataTable批量加入MySQL数据库(Net版)
查看>>
mysql数据库常用命令
查看>>
MySQL数据库必会的增删查改操作(CRUD)
查看>>
MySQL数据库性能分析与调优实践
查看>>
mysql数据库扫盲,你真的知道什么是数据库嘛
查看>>
mysql数据库批量插入数据shell脚本实现
查看>>
MySQL数据库操作
查看>>
MySQL数据库故障排错
查看>>
MySQL数据库无法远程连接的解决办法
查看>>
mysql数据库时间类型datetime、bigint、timestamp的查询效率比较
查看>>
MySQL数据库服务器端核心参数详解和推荐配置(一)
查看>>
mysql数据库死锁的产生原因及解决办法
查看>>
MySQL数据库的事务管理
查看>>
mysql数据库的备份与恢复
查看>>
Mysql数据库的条件查询语句
查看>>
MySQL数据库的高可用
查看>>
Mysql数据库相关各种类型的文件
查看>>
MYSQL数据库简单的状态检查(show processlist)
查看>>