https://vjudge.ppsucxtt.cn/problem/HDU-1241 思路: 深搜找连通图数即可 代码: #include <iostream>#include<queue>#include<cstring>#include<algorithm>#include<cmath>#definefastio ios::sync_with_stdio(false), cin.tie(NULL), cout.tie(NULL)usingnamespacestd; typedeflonglongll; typedef pair<int,...
GeoSurvComp地质调查公司负责探测地下石油储藏。 GeoSurvComp现在在一块矩形区域探测石油,并把这个大区域分成了很多小块。他们通过专业设备,来分析每个小块中是否蕴藏石油。如果这些蕴藏石油 的小方格相邻,那么他们被认为是同一油藏的一部分。在这块矩形区域,可能有很多油藏。你的任务是确定有多少不同的油藏。 Input 输入...
原题链接:HDU-1241 题目大意:矩形平面内,有油田和普通土地,连在一起的油田算同一片油田(上下左右、左上左下右上右下相邻都算连在一起,8个方向),问一共有多少片油田。 找到一块油田,用bfs或dfs遍历与它相连的油田,全标记为访问过,然后记为一片油田。 bfs AC代码: AI检测代码解析 #include <iostream> #i...
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
@@@ @@**@ 0 0 Sample Output 0 1 2 2 #include<iostream>usingnamespacestd;chardp[105][105];intc;intk;intbfs(intx,inty){intsum=0;if(dp[x][y]=='@'){sum++;if(y<k){dp[x][y]='*';sum=sum+bfs(x,y+1);}if(y>0){dp[x][y]='*';sum=sum+bfs(x,y-1);}if(x<c)...
深度优先遍历图 VS 广度优先遍历图.gif importjava.io.BufferedReader;importjava.io.BufferedWriter;importjava.io.InputStreamReader;importjava.io.OutputStreamWriter;importjava.io.PrintWriter;importjava.util.Arrays;importjava.util.Scanner;publicclassMain{privatestaticfinalintMAXSIZE=105;privatestaticcharrooms[][...
2015-02-21 05:33 −HDU 1241 是深搜算法的入门题目,递归实现。 原题目传送门: http://acm.hdu.edu.cn/showproblem.php?pid=1241 代码仅供参考,c++实现: #include <iostream> using namespace std; c... 雨傾晨叶 0 748 Hdu_1241 Oil Deposits ...
简介: HDU-1241,Oil Deposits(DFS) 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 plots. It then...
1 /*思路:利用筛选法求得一个数的各因子之和,然后存入到数组中直接访问输出 2 //1.经典 3 #include <stdio.h> 4 #include <stdlib.h> 5 #define N 500000 6 7 int factors[500001] = {0}; 8 int main () 9 { 10 factors[0] = 0; factors[1] = 0; ...