Oil Deposits Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 65536/32768K (Java/Other) Total Submission(s) : 4   Accepted Submission(s) : 3 Font: Times New Roman | Verdana | Georgia Font Size: ← → Problem Description The GeoSurvComp geologic s…

题目链接:Oil Deposits 解析:问有多少个"@"块.当中每一个块内的各个"@"至少通过八个方向之中的一个相邻. 直接从"@"的地方開始向相邻八个方向搜索,每搜到一个格子.就将它替换成".",一次搜索就会搜索完一个块,记录搜索的次数为答案. AC代码: #include <cstdio> #include <cstring> #include <queue> using namesp…

HDU 1241  Oil Deposits L -DFS Time Limit:1000MS     Memory Limit:10000KB     64bit IO Format:%I64d & %I64u   Description The GeoSurvComp geologic survey company is responsible for detecting underground oil deposits. GeoSurvComp works with one large r…

HDU 1241 Oil Deposits(石油储藏) 00 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)   Problem Description - 题目描述 The GeoSurvComp geologic survey company is responsible for detecting underground oil deposits. GeoSurvComp works with one large…

HDU 1241 题目大意:给定一块油田,求其连通块的数目.上下左右斜对角相邻的@属于同一个连通块. 解题思路:对每一个@进行dfs遍历并标记访问状态,一次dfs可以访问一个连通块,最后统计数量. /* HDU 1241 Oil Deposits --- 入门DFS */ #include <cstdio> int m, n; //n行m列 ][]; /* 将和i,j同处于一个连通块的字符标记出来 */ void dfs(int i, int j){ || j < || i >=…

题目:   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 a…

Problem 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 pl…

解题思路:第一道DFS的题目--- 参看了紫书和网上的题解-- 在找到一块油田@的时候,往它的八个方向找,直到在能找到的范围内没有油田结束这次搜索 可以模拟一次DFS,比如说样例 在i=0,j=1时,发现第一块油田,对它DFS,这样经过一次DFS,所有的油田都被找出来了,被记0 Oil Deposits Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submiss…

嗯... 题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1241 很经典的一道dfs,但是注意每次查到一个@之后,都要把它变成“ * ”,然后继续dfs,这样在dfs过程中一些@变成了“ * ”.这样也不需要flag数组,当继续遍历图的时候,再有@就是独立的,ans++.最后ans里边便是答案... AC代码: #include<cstdio> #include<iostream> #include<cstring> us…

题目链接 http://poj.org/problem?id=1562 题意 输入一个m行n列的棋盘,棋盘上每个位置为'*'或者'@',求'@'的连通块有几个(连通为8连通,即上下左右,两条对角线). 思路 floodfill问题,用dfs解决. 代码 #include <iostream> #include <cstdio> #include <cstring> using namespace std; ; char grid[N][N]; int visit[N][…

题意:求连通块个数. 分析:dfs. #include<cstdio> #include<cstring> #include<cstdlib> #include<cctype> #include<cmath> #include<iostream> #include<sstream> #include<iterator> #include<algorithm> #include<string&g…

最水的一道石油竟然改了一个小时,好菜好菜. x<=r  y<=c  x<=r  y<=c  x<=r  y<=c  x<=r y<=c #include<stdio.h> #include<algorithm> #include<string.h> using namespace std; ][]={,, ,-, ,, -,, ,, ,-, -,, -,-}; int r,c; ][]; int ans; void dfs…

题目链接 https://cn.vjudge.net/contest/65959#problem/L 题意 @表示油田 如果 @@是连在一起的 可以八个方向相连 那么它们就是 一块油田 要找出 一共有几块油田 思路 可以先遍历 一遍地图 遇到 @ 就更新答案 + 1 然后DFS 把与它相连 和与它相连的 相连的 油田 都变成 * 就可以了 AC代码 #include <cstdio> #include <cstring> #include <ctype.h> #incl…

HDOJ(HDU).1241 Oil Deposits(DFS) [从零开始DFS(5)] 点我挑战题目 从零开始DFS HDOJ.1342 Lotto [从零开始DFS(0)] - DFS思想与框架/双重DFS HDOJ.1010 Tempter of the Bone [从零开始DFS(1)] -DFS四向搜索/奇偶剪枝 HDOJ(HDU).1015 Safecracker [从零开始DFS(2)] -DFS四向搜索变种 HDOJ(HDU).1016 Prime Ring Problem (…

题目传送门 /* DFS:油田问题,一道经典的DFS求连通块.当初的难题,现在看上去不过如此啊 */ /************************************************ Author :Running_Time Created Time :2015-8-4 10:11:11 File Name :HDOJ_1241.cpp ************************************************/ #include <cstdio> #i…

UVA 572 -- Oil Deposits(DFS求连通块) 图也有DFS和BFS遍历,由于DFS更好写,所以一般用DFS寻找连通块. 下述代码用一个二重循环来找到当前格子的相邻8个格子,也可用常量数组或者写8条DFS调用. 下述算法是:种子填充(floodfill) 两种连通区域 四连通区域:从区域内一点出发,可通过上.下.左.右四个方向的移动组合,在不越出区域的前提下,能到达区域内的任意像素 八连通区域:从区域内每一像素出发,可通过八个方向,即上.下.左.右.左上.右上.左下.右下移动的…

Oil Deposits Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 8274 Accepted Submission(s): 4860 Problem Description The GeoSurvComp geologic survey company is responsible for detecting underground…

Oil Deposits Descriptions: 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 sq…

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…

Oil Deposits Time Limit: 2 Seconds      Memory Limit: 65536 KB 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 th…

Oil Deposits Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 15533    Accepted Submission(s): 8911 Problem Description The GeoSurvComp geologic survey company is responsible for detecting underg…

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…

题目: The GeoSurvComp geologic survey company is responsible for detecting underground oil deposits. GeoSurvCompworkswithonelargerectangularregionoflandatatime,andcreatesagridthatdivides the land into numerous square plots. It then analyzes each plot s…

题目链接 Problem Description There is a tree with n nodes, each of which has a type of color represented by an integer, where the color of node i is ci. The path between each two different nodes is unique, of which we define the value as the number of di…

Oil Deposits Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 45017    Accepted Submission(s): 25972 Problem Description The GeoSurvComp geologic survey company is responsible for detecting under…

排列2 Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 5817    Accepted Submission(s): 2229 Problem Description Ray又对数字的列产生了兴趣:现有四张卡片,用这四张卡片能排列出很多不同的4位数,要求按从小到大的顺序输出这些4位数.   Input 每组数据占一行,代表四张卡片上的数…

解题思路: 1. 遍历扫描二维数组,遇到‘@’,结果ans++; 2. 将当前 i,j 位置置为‘*’,将当前‘@’的 i,j 传人到DFS函数中,开始遍历八个方向的字符 如果碰到 '@' 则先将当前置为‘*’,然后再次递归传递,直到超出界限或者扫描不到‘@’,结束递归 3. DFS()的作用是将i,j为开始周围连续的“@”全部改为‘*’ 4. 最后输出 ans 即可: Ac code : #include<bits/stdc++.h> using namespace std; ][]; st…

上一次基本了解了下BFS,这次又找了个基本的DFS题目来试试水,DFS举个例子来说就是 一种从树的最左端开始一直搜索到最底端,然后回到原端再搜索另一个位置到最底端,也就是称为深度搜索的DFS--depth first search,话不多说,直接上题了解: Description:某石油勘探公司正在按计划勘探地下油田资源,工作在一片长方形的地域中.他们首先将该地域划分为许多小正方形区域,然后使用探测设备分别探测每一块小正方形区域内是否有油.若在一块小正方形区域中探测到有油,则标记为’@’,否则标…

链接 : Here! 思路 : 搜索判断连通块个数, 所以 $DFS$ 或则 $BFS$ 都行喽...., 首先记录一下整个地图中所有$Oil$的个数, 然后遍历整个地图, 从油田开始搜索它所能连通多少块其他油田, 只需要把它所连通的油田个数减去, 就ok了 /************************************************************************* > File Name: E.cpp > Author: > Mail: >…

#include<iostream> #include<cstring> #include<cstdio> #include<algorithm> using namespace std; ][]; ][]={, , , , -, , , -, , , , -, -, , -, -}; int n, m; int ans; bool judge(int x, int y){ || x>n || y< || y>m) return false…