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服务器安装(Linux)
查看>>
mysql服务器查询慢原因分析方法
查看>>
mysql服务无法启动的问题
查看>>
MySQL杂谈
查看>>
mysql权限
查看>>
mysql条件查询
查看>>
MySQL条件查询
查看>>
MySQL架构与SQL的执行流程_1
查看>>
MySQL架构与SQL的执行流程_2
查看>>
MySQL架构介绍
查看>>
MySQL架构优化
查看>>
mysql架构简介、及linux版的安装
查看>>
MySQL查看数据库相关信息
查看>>
MySQL查看表结构和表中数据
查看>>
MySQL查询优化:LIMIT 1避免全表扫描
查看>>
MySQL查询优化之索引
查看>>
mysql查询储存过程,函数,触发过程
查看>>
mysql查询总成绩的前3名学生信息
查看>>
mysql查询慢排查
查看>>
MySQL查询报错ERROR:No query specified
查看>>