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基础教程三 —常见函数
查看>>
mysql基础教程二
查看>>
mysql基础教程四 --连接查询
查看>>
MySQL基础知识:创建MySQL数据库和表
查看>>
MySQL基础系列—SQL分类之一
查看>>
MySQL处理千万级数据分页查询的优化方案
查看>>
mysql备份
查看>>
mysql备份与恢复
查看>>
mysql备份工具xtrabackup
查看>>
mysql备份恢复出错_尝试备份/恢复mysql数据库时出错
查看>>
mysql复制内容到一张新表
查看>>
mysql复制表结构和数据
查看>>
mysql复杂查询,优质题目
查看>>
MySQL外键约束
查看>>
MySQL多表关联on和where速度对比实测谁更快
查看>>
MySQL多表左右连接查询
查看>>
mysql大批量删除(修改)The total number of locks exceeds the lock table size 错误的解决办法
查看>>
mysql如何做到存在就更新不存就插入_MySQL 索引及优化实战(二)
查看>>
mysql如何删除数据表,被关联的数据表如何删除呢
查看>>
MySQL如何实现ACID ?
查看>>