poj1562 Oil Deposits(深搜dfs)
来源:互联网 发布:23种设计模式 java 编辑:程序博客网 时间:2024/05/22 19:26
转载请注明出处:http://blog.csdn.net/u012860063?viewmode=contents
题目链接:http://poj.org/problem?id=1562
----------------------------------------------------------------------------------------------------------------------------------------------------------
欢迎光临天资小屋:http://user.qzone.qq.com/593830943/main
----------------------------------------------------------------------------------------------------------------------------------------------------------
Description
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 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
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
0122
Source
Mid-Central USA 1997
代码如下:
#include <cstring>#include <iostream>#include <algorithm>using namespace std;#define TM 117int n, m, ans, k;char map[TM][TM];int vis[TM][TM];int xx[8] = {1,1,1,0,-1,-1,-1,0};int yy[8] = {1,0,-1,-1,-1,0,1,1};void dfs(int x, int y){vis[x][y] = 1;for(int i = 0; i < 8; i++){int dx = x+xx[i];int dy = y+yy[i];if(dx>=0&&dx<n&&dy>=0&&dy<m&&map[dx][dy]=='@'&&!vis[dx][dy]){vis[dx][dy] = 1;dfs(dx,dy);}}}int main(){int i, j;while(cin>>n>>m){if(n==0 && m==0)break;memset(vis,0,sizeof(vis));k = 0;for(i = 0; i < n; i++){cin>>map[i];}for(i = 0; i < n; i++){for(j = 0; j < m; j++){if(map[i][j] == '@' && !vis[i][j]){dfs(i,j);k++;}}}cout<<k<<endl;}return 0;}
1 0
- poj1562 Oil Deposits(深搜dfs)
- POJ1562:Oil Deposits(DFS)
- poj1562 Oil Deposits(DFS)
- poj1562--Oil Deposits(DFS)
- poj1562 & zoj1709 - Oil Deposits (DFS)
- poj1562 Oil Deposits(dfs求联通分量)
- UVA572 HDU1241 POJ1562 Oil Deposits【DFS】
- poj1562 Oil Deposits
- poj1562 Oil Deposits
- Oil Deposits POJ1562
- poj1562 Oil Deposits BFS
- POJ1562.Oil Deposits
- 【POJ1562】Oil Deposits
- Oil Deposits poj1562
- UVA572 POJ1562 oil deposits(DFS求连通块问题)
- POJ1562 Oil Deposits (比较水的dfs)
- DFS:POJ1562-Oil Deposits(求连通块个数)
- Oil Deposits(dfs深搜)
- 树状数组
- 请教C语言语法, 老外写的代码
- 1. awk, gawk命令使用
- uva6468
- 我的淘宝历程
- poj1562 Oil Deposits(深搜dfs)
- hdu 1007 Quoit Design 最近点对(分治)
- 面试- 字符串反转 单词不反转
- C++ 从类型转换到文件读入数组
- 概率dp 投掷色子走一定步数 期望
- Eclipse插件的安装方法三则
- Android开发中在一个Activity中关闭另一个Activity
- Eclipse中创建并运行Servlet项目
- 初探linux子系统集之led子系统(三)