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优化之BTree索引使用规则
查看>>
MySQL优化之推荐使用规范
查看>>
Webpack Critical CSS 提取与内联教程
查看>>
mysql优化概述(范式.索引.定位慢查询)
查看>>
MySQL优化的一些需要注意的地方
查看>>
mysql优化相关
查看>>
MySql优化系列-优化版造数据(存储过程+函数+修改存储引擎)-2
查看>>
MySql优化系列-进阶版造数据(load data statment)-3
查看>>
MySql优化系列-造数据(存储过程+函数)-1
查看>>
MySQL优化配置详解
查看>>
Mysql优化高级篇(全)
查看>>
mysql会员求积分_MySql-统计所有会员的最高前10次的积分和
查看>>
mysql会对联合索性排序优化_MySQL索引优化实战
查看>>
MySQL作为服务端的配置过程与实际案例
查看>>