1241

HDU 1241 Oil Deposits bfs 难度:0

pid=1241 对每个还未访问的点bfs,到达的点都标为一块,最后统计有多少块即可 #include <cstdio> #include <cstring> #include
·2015-10-27 14:46

hdu 1241 Oil Deposits (一次dfs搞定有某有)

1 #include<iostream> 2 #include<cstring> 3 #include<cstdio> 4 #include<algorithm> 5 using namespace std; 6 char map[105][105]; 7 8 int dir[8][2]={0, 1, 1, 0, -1
·2015-10-23 08:02

hdoj 1241 Oil Deposits

Oil Deposits Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 16542    Accepted Submission(s): 9500 Pro
·2015-10-21 13:38

UVa-Ecological Premium

option=com_onlinejudge&Itemid=8&page=show_problem&category=15&problem=1241&mosmsg
·2015-10-21 13:24

onNewIntent调用时机

onPause onStop onDestroy  onNewIntent一、其他应用发Intent，执行下列方法：I/@@@philn(12410): onCreateI/@@@philn(1241
·2015-10-21 11:35

hdu1241 Oil Deposits

                                                 Oil Deposits    
·2015-10-21 11:52

ecshop异常

错如信息如下： Notice: Undefined variable: data in E:\wwwroot\wuwangni\web\includes\lib_base.php on line 1241
·2015-10-21 10:02

九度OJ 1042：Coincidence（公共子序列） （DP）

时间限制：1秒内存限制：32兆特殊判题：否提交：2303解决：1241题目描述：Findalongestcommonsubsequenceoftwostrings.输入：Firstandsecondlineofeachinputcase
thudaliangrx·2015-10-18 18:00

关于MODBUS通讯心得之一：滞后性问题

关于滞后性，有网友提到：“我用CM1241挂15个Modbus设备（8台英威腾的CHE100变频器，7台欧姆龙的E5CZ温控器），每台从站设备有三
程金良·2015-10-16 20:57

[leetcode] 200 Number of Islands(DFS)

类似的题目还有hdu1016，hdu1241。intdir[4][2]={{-1,0},{1,0},{0,1},{0,-1}}; classSolution{ public: voiddfs
NK_test·2015-10-12 23:00

hihocoder1241 : Best Route in a Grid

题目链接：传送门题意：给定一个n*n的矩阵，每次只能往下或者往左走，而且0的位置不能走，求最后走的格子的数的乘积末尾连续0的最小个数。分析：很明显最后末尾连续0的个数和素因子2和素因子5的数量有关。那么我们就做两次dp,设dp[i][j]表示走的(i,j)的时候素因子2/5的最小的个数。然后那就是一个简单的dp了。Code:#include usingnamespacestd; constin
u013790563·2015-10-12 10:00

hiho #1241 : Best Route in a Grid 【dp 求解质因子2和5的最少匹配个数】

题目链接：点我题意：有一个n*n的图，每个位置都有一个非负数值。现在你想从左上角到达右下角，每次只能向下或向右走，已知数值为0的点不能走。我们累乘所走过位置的数值，得到一个数T。问T末尾连续0的个数最少是多少？思路：因为2*5=10得到一个0。我们先预处理质因子5的个数，从左上角到右下角跑一次dp，求出T中最少的5的个数ans1。然后预处理质因子2的个数，再跑一次dp，求出T中最少的2的个数ans
chenzhenyu123456·2015-10-11 21:00

hdu 1241 Oil Deposits(dfs)

OilDepositsTimeLimit:2000/1000MS(Java/Others)MemoryLimit:65536/32768K(Java/Others)TotalSubmission(s):18935AcceptedSubmission(s):10902ProblemDescriptionTheGeoSurvCompgeologicsurveycompanyisresponsiblef
Grit_ICPC·2015-10-03 16:00

1241：Oil Deposits

pid=1241方法：dfs思路：一看题觉得搜索倒是不难，难就难在怎样判断连通图。后查阅资料，发现了一种与dfs巧妙结合的神奇方法，真是拜服啊！
freedom098·2015-09-30 19:00

天涯明月刀中秋活动所有诗人的位置分享

开封(2110,1241)乞丐位置开封(2022,1370)诗句一开封(2018,1369)诗句二开封(2214,1270)诗句三开封(2276,1281)诗句四开封(2018,1368)诗句五酒仙居2
佚名·2015-09-26 15:49

hdu1241 Oil Deposits

pid=1241ProblemDescription TheGeoSurvCompgeologicsurveycompanyisresponsiblefordetectingundergroundoildeposits.GeoSurvCompworkswithonelargerec
aonaigayiximasi·2015-09-15 20:00

HDU1241 Oil Deposits

pid=1241/*HDU1241OilDepositshttp://acm.hdu.edu.cn/showproblem.php?
qq_26437925·2015-09-04 09:00

《搜索》hdoj 1241 dfs模版

#include chars[150][150];intmove[8][2]={{1,0},{-1,0},{0,1},{0,-1},{1,1},{1,-1},{-1,1},{-1,-1}};intn,m,k,ans;voiddfs(intx,inty){inti;for(i=0;i=0&&tx=0&&ty
a709743744·2015-09-02 22:00

POJ 1562 && HDU 1241 Oil Deposits（dfs）

Description给出一个N*M的矩形区域和每个区域的状态–@有/*没有石油，（定义）如果两个有石油的区域是相邻的（水平、垂直、斜）则认为这是属于同一个油田。求这块矩形区域一共有多少油田Input多组输入，每组输入包括两个整数n，m表示矩阵区域的列数和行数（1 #include #include usingnamespacestd; #definemaxn105 charmap[maxn][m
V5ZSQ·2015-08-25 10:00

hdu1241Oil Deposits

pid=1241八个方向如果有@则为同一油田，问共多少油田。
aonaigayiximasi·2015-08-18 12:00

DFS-HDU-1241-Oil Deposits

OilDepositsTimeLimit:2000/1000MS(Java/Others)MemoryLimit:65536/32768K(Java/Others)TotalSubmission(s):18633AcceptedSubmission(s):10736ProblemDescriptionTheGeoSurvCompgeologicsurveycompanyisresponsiblef
Roy_Yuan·2015-08-17 19:00

ZOJ 1241 Geometry Made Simple

GeometryMadeSimpleTimeLimit:2000/1000ms(Java/Other)   MemoryLimit:65536/32768K(Java/Other)TotalSubmission(s):105   AcceptedSubmission(s):16ProblemDescriptionMathematicscanbesoeasywhenyouhaveacomputer.
qq_24653023·2015-08-15 22:00

HDU 1241 Oil Deposits DFS

HDU1241OilDepositsProblemDescriptionTheGeoSurvCompgeologicsurveycompanyisresponsiblefordetectingundergroundoildeposits.GeoSurvCompworkswithonelargerectangularregionoflandatatime
qq_21120027·2015-08-10 19:00

hdu 1241 AC广搜

点击打开链接，原地址packagecn.hncu.search; importjava.util.Scanner; publicclasssearchbfs{ publicstaticvoidmain(String[]args){ Scannersc=newScanner(System.in); intcount=0; while(sc.hasNext()){ intm=sc.nextInt
yangxin_blog·2015-08-10 17:00

hdu 1241 Oil Deposits 深搜 Ac

1241hdu原链接处packagecn.hncu.start; importjava.util.Scanner; publicclassp1241{ staticintn,m; finalstaticintb
yangxin_blog·2015-08-10 17:00

hdu1241 Oil Deposits裸的搜哦哦哦哦

rt题目#include #include usingnamespacestd; charch[105][105]; intn,m; intdir[8][2]={{1,0},{-1,0},{0,1},{0,-1},{1,1},{1,-1},{-1,1},{-1,-1}}; voiddfs(intx,inty) { ch[x][y]='*'; for(inti=0;in)continue; if(n
zhou_yujia·2015-08-10 09:00

hdu 1241 Oil Deposits(经典dfs)

pid=1241大意：寻找一块区域中有多少的‘@’域，左右上下还有四个斜45度的相邻点都算作连起来的。
theArcticOcean·2015-08-06 20:00

杭电 1241 Oil Deposits 递归 DFS

OilDepositsTimeLimit:2000/1000MS(Java/Others)    MemoryLimit:65536/32768K(Java/Others)TotalSubmission(s):18322    AcceptedSubmission(s):10568ProblemDescriptionTheGeoSurvCompgeologicsurveycompanyisresp
qq_24678203·2015-08-06 11:00

HDU 1241:Oil Deposits【递归】

OilDepositsTimeLimit:2000/1000MS(Java/Others)    MemoryLimit:65536/32768K(Java/Others)TotalSubmission(s):18221    AcceptedSubmission(s):10503ProblemDescriptionTheGeoSurvCompgeologicsurveycompanyisresp
lin14543·2015-08-04 10:00

HDOJ 1241 Oil Deposits(dfs)

OilDepositsTimeLimit:2000/1000MS(Java/Others)    MemoryLimit:65536/32768K(Java/Others)TotalSubmission(s):18204    AcceptedSubmission(s):10495ProblemDescriptionTheGeoSurvCompgeologicsurveycompanyisresp
zwj1452267376·2015-08-03 22:00

hdu1241详解 Java广搜搞定

importjava.util.Scanner; publicclassMain{ publicstaticvoidmain(String[]args){ Scannersc=newScanner(System.in); while(sc.hasNext()){ intm=sc.nextInt();//输入地图的行数 intn=sc.nextInt();//输入地图的列数 if(m==0){/
u011459262·2015-08-03 21:00

素数判定 SDUT 1241

题目描述对于表达式n2+n+41，当n在（x,y）范围内取整数值时（包括x,y）(-39#include#includeintmain(){inta[10010],b[100100];intn,m;intk=-39;memset(b,0,sizeof(b));for(inti=0;i<=89;i++){a[i]=k*k+k+41;k++;}for(inti=2;i<3000;i++){for(in
Sunmh_AC·2015-08-03 20:45

hdoj 1241 Oil Deposits【DFS】

OilDepositsTimeLimit:2000/1000MS(Java/Others)    MemoryLimit:65536/32768K(Java/Others)TotalSubmission(s):18094    AcceptedSubmission(s):10431ProblemDescriptionTheGeoSurvCompgeologicsurveycompanyisresp
liu6886·2015-08-03 16:00

hdu1241（Oil Deposits）

点击打开hdu1241ProblemDescriptionTheGeoSurvCompgeologicsurveycompanyisresponsiblefordetectingundergroundoildeposits.GeoSurvCompworkswithonelargerectangularregionoflandatatime
u011479875·2015-08-02 21:00

暑假集训第三周周六赛 搜索 F - Oil Deposits 油块

OilDepositsTimeLimit:1000MS     MemoryLimit:32768KB     64bitIOFormat:%I64d&%I64uSubmit Status Practice HDU1241DescriptionTheGeoSurvCompgeologicsurveycompanyisresponsiblefordetectingundergroundoild
linyuxilu·2015-08-02 14:00

HDOJ 1241 Oil Deposits（8方位dfs）

OilDepositsTimeLimit:2000/1000MS(Java/Others)    MemoryLimit:65536/32768K(Java/Others)TotalSubmission(s):17828    AcceptedSubmission(s):10250ProblemDescriptionTheGeoSurvCompgeologicsurveycompanyisresp
helloiamclh·2015-07-29 10:00

HDU 1241 Oil Deposits

题目地址：点击打开链接思路：DFSAC代码：#include usingnamespacestd; intm,n; charlol[110][110]; intx[8]={-1,-1,-1,0,0,1,1,1}; inty[8]={-1,0,1,-1,1,-1,0,1}; voiddfs(inta,intb) { lol[a][b]='*'; inti,newx,newy; for(i=0;i=0
qq_25605637·2015-07-26 21:00

ZZUOJ - 1241 - 斐波那契数

郑州大学第八届ACM大学生程序设计竞赛正式赛C.斐波那契数Descriptionxq最近迷恋上了斐波那契数列，因为当n趋向于无穷大时，后一项与前一项的比值越来越逼近黄金分割0.618。不过令xq不开心的是，斐波那契数的增长速度太快了，以至于当n很大时，用计算机中的32位整数甚至64位整数都已经不能表示这个数了。基于这个原因，xq遇到了这个难题，xq想要知道第K（0=2），其中F（0）=0，F（1）
violet_echo_0908·2015-07-23 16:00

HDU - 1241 - Oil Deposits

题意：给你一个二维数组，输出有几块油田，由样例可知，八个方向只要相连，就算是一个油田，所有八个方向都要搜索到。#include #defineMAXN150 chars[MAXN][MAXN]; intmove[8][2]={{1,0},{-1,0},{0,1},{0,-1},{1,1},{-1,1},{1,-1},{-1,-1}}; intm,n,k,ans; voiddfs(intx,in
violet_echo_0908·2015-07-22 11:00

HDOJ 1241 Oil Deposits（8方位简单dfs）

OilDepositsTimeLimit:2000/1000MS(Java/Others)    MemoryLimit:65536/32768K(Java/Others)TotalSubmission(s):17290    AcceptedSubmission(s):9961ProblemDescriptionTheGeoSurvCompgeologicsurveycompanyisrespo
helloiamclh·2015-07-22 10:00

ZOJ1241--Geometry Made Simple

Geometry Made Simple Time Limit: 2 Seconds      Memory Limit: 65536 KB Mathematics can be so easy when you have a computer. Consider the following example. You probably know
·2015-07-21 10:00

【算法渣渣的逆袭之路】最近在做DFS的专题，贴道题吧。

HDU-1241-OilDepositsDescriptionTheGeoSurvCompgeologicsurveycompanyisresponsiblefordetectingundergroundoildeposits.GeoSurvCompworkswithonelargerectangularregionoflandatatime
ydjcs567·2015-07-21 10:00

【暑假集训专题#搜索 HDU1241】

【题目链接】clickhere~~【题目大意】'@'代表油田位置，'*'代表地面，八个方向相邻的油田视为一个，求给定地图里油田数目【解题思路】八个方向搜索即可代码：#include #include #include #include usingnamespacestd; constintN=1010; intdir4[4][2]={{1,0},{0,1},{-1,0},{0,-1}}; intd
u013050857·2015-07-07 14:00

struts 模型驱动的发射原理

http://www.jikexueyuan.com/course/1241.html packagecom.jikexueyuan.servlet; importjava.io.IOException
c_c123·2015-06-28 16:00

bzoj1798线段树。。调的要死

Time Limit: 30 Sec  Memory Limit: 64 MB Submit: 3349  Solved: 1241
·2015-06-11 20:00

L_Oil Deposits(HDU_1241)

DescriptionTheGeoSurvCompgeologicsurveycompanyisresponsiblefordetectingundergroundoildeposits.GeoSurvCompworkswithonelargerectangularregionoflandatatime,andcreatesagridthatdividesthelandintonumeroussq
highmath_Final·2015-05-11 11:00

HDU1241 Oil Deposits DFS

OilDepositsTimeLimit:1000MS    MemoryLimit:32768KB    64bitIOFormat:%I64d&%I64uSubmitStatusPracticeHDU1241DescriptionTheGeoSurvCompgeologicsurveycompanyisresponsiblefordetectingundergroundoildeposits
wr132·2015-05-01 15:00

hdu 1241Oil Deposits

竞赛的时候一直吧题意错了。。。adjacent相邻的。。。妈的英语我最头痛啊啊啊~~~ 哎如果理解了题意就多对一个了 #include #include intn,m,num,visit[105][105],a[10000],q=0; charmap[105][105]; voiddfs(intx,inty) { if(x>=m||x=n||y<0||visit[x][y]||map[x][y]=
su20145104009·2015-04-10 19:00

HDU-1241 Oil Deposits

pid=1241         OilDepositsTimeLimit:2000/1000MS(Java/Others)    MemoryLimit:65536/32768K(Java/Others
·2015-04-02 21:00

51Nod 1241 特殊的排序

题目传送门 分析：之前想的是求LIS，后面发现这组数据会出问题：312456。其实这里的LIS还应当满足前后两个元素的值相差为1。比如上面的最长子序列为：3456，而不是12456。即只需移动1、2即可。即如果n个数最长连续上升子序列（这里的连续指数值上连续，即前后相差1）长度为k，则最少移动次数为n-k 简略证明：首先需要证明：最少移动次数和移动数的个数是相等的，即每个数最多只移动一次。因为最终
u014679804·2015-04-01 22:00
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