随笔分类 -  实用数据结构

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hdu 1087 最大上升子序列的和(dp或线段树)
摘要:Super Jumping! Jumping! Jumping!Time Limit: 2000/1000 MS (Java/Others)Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 23328Accepted Subm... 阅读全文

posted @ 2014-11-13 02:51 雄.. 阅读(593) 评论(0) 推荐(0)

hdu 1558 线段相交+并查集路径压缩
摘要:Segment setTime Limit: 3000/1000 MS (Java/Others)Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 3457Accepted Submission(s): 1290Problem... 阅读全文

posted @ 2014-08-22 23:41 雄.. 阅读(202) 评论(0) 推荐(0)

hdu 3992 AC自动机上的高斯消元求期望
摘要:Crazy TypewriterTime Limit: 2000/1000 MS (Java/Others)Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 391Accepted Submission(s): 109Prob... 阅读全文

posted @ 2014-08-15 19:25 雄.. 阅读(303) 评论(0) 推荐(0)

codeforces 446C DZY Loves Fibonacci Numbers 数论+线段树成段更新
摘要:DZY Loves Fibonacci NumbersTime Limit:4000MSMemory Limit:262144KB64bit IO Format:%I64d & %I64uSubmitStatusAppoint description:DescriptionIn mathematic... 阅读全文

posted @ 2014-07-18 12:16 雄.. 阅读(665) 评论(0) 推荐(0)

hust 1605 - Gene recombination(bfs+字典树)
摘要:1605 - Gene recombinationTime Limit:2sMemory Limit:64MBSubmissions: 264 Solved: 46DESCRIPTIONAs a gene engineer of a gene engineering project, Enigma ... 阅读全文

posted @ 2014-05-28 14:03 雄.. 阅读(205) 评论(0) 推荐(0)

ZOJ 3772 Calculate the Function 线段树+矩阵
摘要:Calculate the FunctionTime Limit:2000MSMemory Limit:65536KB64bit IO Format:%lld & %lluSubmitStatusAppoint description:DescriptionYou are given a list ... 阅读全文

posted @ 2014-05-26 13:53 雄.. 阅读(223) 评论(0) 推荐(0)

hdu 1277 AC自动机
摘要:全文检索Time Limit: 2000/1000 MS (Java/Others)Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 1134Accepted Submission(s): 357Problem Description我们大家经常用google检索信息,但是检索信息的程序是很困难编写的;现在请你编写一个简单的全文检索程序。问题的描述是这样的:给定一个信息流文件,信息完全有数字组成,数字个数不超过60000个,但也不少于60个;再给定一个关键字集合,其中关键字个数不超过10000个,每个关键字的信息数字不超 阅读全文

posted @ 2014-04-08 16:43 雄.. 阅读(254) 评论(0) 推荐(0)

poj 3468 线段树成段更新
摘要:A Simple Problem with IntegersTime Limit:5000MSMemory Limit:131072KTotal Submissions:54012Accepted:16223Case Time Limit:2000MSDescriptionYou haveNintegers,A1,A2, ... ,AN. You need to deal with two kinds of operations. One type of operation is to add some given number to each number in a given interv 阅读全文

posted @ 2014-03-09 17:25 雄.. 阅读(259) 评论(0) 推荐(0)

hdu 1698 线段树成段更新
摘要:#include#includeusing namespace std;#define lson left,mid,i0) { f[i>1; bulid(lson); bulid(rson); pushup(i); return ;}void update(int left,int right,int set,int i){ if(f[i].left==left && f[i].right==right) { f[i].set=set; f[i].sum=(right-left+1)*set; r... 阅读全文

posted @ 2014-03-09 16:47 雄.. 阅读(116) 评论(0) 推荐(0)

hdu 1754 线段树(点修改)
摘要:I Hate ItTime Limit: 9000/3000 MS (Java/Others)Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 31979Accepted Submission(s): 12649Problem Description很多学校流行一种比较的习惯。老师们很喜欢询问,从某某到某某当中,分数最高的是多少。这让很多学生很反感。不管你喜不喜欢,现在需要你做的是,就是按照老师的要求,写一个程序,模拟老师的询问。当然,老师有时候需要更新某位同学的成绩。Input本题目包含多组测试,请处理到文件结束。在每 阅读全文

posted @ 2014-03-09 13:49 雄.. 阅读(341) 评论(0) 推荐(0)

hdu 2888 二维RMQ
摘要:Check CornersTime Limit: 2000/10000 MS (Java/Others)Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 1646Accepted Submission(s): 597Problem DescriptionPaul draw a big m*n matrix A last month, whose entries Ai,j are all integer numbers ( 1 #include#includeusing namespa... 阅读全文

posted @ 2014-03-07 15:38 雄.. 阅读(301) 评论(0) 推荐(0)

hdu 1166 树状数组(线段树)
摘要:敌兵布阵Time Limit: 2000/1000 MS (Java/Others)Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 35307Accepted Submission(s): 14974Problem DescriptionC国的死对头A国这段时间正在进行军事演习,所以C国间谍头子Derek和他手下Tidy又开始忙乎了。A国在海岸线沿直线布置了N个工兵营地,Derek和Tidy的任务就是要监视这些工兵营地的活动情况。由于采取了某种先进的监测手段,所以每个工兵营地的人数C国都掌握的一清二楚,每个工兵营地的人 阅读全文

posted @ 2014-03-06 20:24 雄.. 阅读(170) 评论(0) 推荐(0)

hdu 3874 树状数组
摘要:NecklaceTime Limit: 15000/5000 MS (Java/Others)Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 2447Accepted Submission(s): 865Problem DescriptionMery has a beautiful necklace. The necklace is made up of N magic balls. Each ball has a beautiful value. The balls with the same beautiful v 阅读全文

posted @ 2014-03-06 19:55 雄.. 阅读(146) 评论(0) 推荐(0)

hdu 1189 并查集
摘要:Farm IrrigationTime Limit: 2000/1000 MS (Java/Others)Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 4802Accepted Submission(s): 2073Problem DescriptionBenny has a spacious farm land to irrigate. The farm land is a rectangle, and is divided into a lot of samll squares. W... 阅读全文

posted @ 2014-03-05 22:00 雄.. 阅读(220) 评论(0) 推荐(0)

hdu 1272 并查集
摘要:题目大意:给一个图,不能出现环,而且只能有一个联通子图。好惆怅,又是爆栈又是wrong answer的。#include#includeusing namespace std;const int maxn=100005;int f[maxn],ans;bool v[maxn];void init(){for(int i=1;ifb) f[fa]=fb; else f[fb]=fa;}int main(){ int a,b,i,sum; while(scanf("%d %d",&a,&b) && !(a==-1 && b==-1 阅读全文

posted @ 2014-03-05 21:02 雄.. 阅读(164) 评论(0) 推荐(0)

hdu 1856 并查集
摘要:More is betterTime Limit: 5000/1000 MS (Java/Others)Memory Limit: 327680/102400 K (Java/Others)Total Submission(s): 11865Accepted Submission(s): 4395Problem DescriptionMr Wang wants some boys to help him with a project. Because the project is rather complex,the more boys come, the bett... 阅读全文

posted @ 2014-03-05 19:09 雄.. 阅读(299) 评论(0) 推荐(0)

hdu 1232 并查集
摘要:#include#includeusing namespace std;const int maxn=1005;int f[maxn];bool v[maxn];int findset(int x){ return f[x] != x ? f[x]=findset(f[x]) : x;}void merge(int a,int b){ int fa=findset(a); int fb=findset(b); if(fa!=fb) f[fa]=fb;}int main(){ int n,m,i,a,b,ans; while(scanf("%d",&n),n) ... 阅读全文

posted @ 2014-03-04 22:05 雄.. 阅读(121) 评论(0) 推荐(0)

uva 11235 RMQ范围最大值
摘要:题目大意:给一个整数上升序列,对于一系列询问区间(i,j),回答这段区间出现次数最多值所出现的次数。分析:一个上升序列,相同的值聚集在一起,把相同的值的区间看作一个整体,假设这样的整体有n个,把他们编号。num[p]表示原下标下的数所在段的编号,left[p]、right[p]表示所在段原下标的左右端点位置。编号后的数组用RMQ求范围最大值。对于询问的区间(L,R)在编号后的数组中以以下三个部分的最大值:从L到L所在段的结束处的元素个数(即right[L]-L+1)、从R所在段的开始处到R处的元素的个数(即R-left[R]+1)、中间第num[L]+1段到num[R]-1段的最大值(用RMQ 阅读全文

posted @ 2014-03-04 20:52 雄.. 阅读(243) 评论(0) 推荐(0)

LA 4329 树状数组
摘要:N(3N20000)ping pong players live along a west-east street(consider the street as a line segment). Each player has a unique skill rank. To improve their skill rank, they often compete with each other. If two players want to compete, they must choose a referee among other ping pong players and hold th 阅读全文

posted @ 2014-02-25 16:32 雄.. 阅读(215) 评论(0) 推荐(0)

LA 并查集路径压缩
摘要:题目大意:有n个节点,初始时每个节点的父亲节点都不存在。有两种操作I u v:把点节点u的父亲节点设为v,距离为|u-v|除以1000的余数。输入保证执行指令前u没有父亲节点。E u:询问u到根节点的距离。分析:并查集加路径压缩。#include#include#include#includeusing namespace std;const int maxn=20010;int f[maxn],d[maxn];int findset(int x){ if(f[x] !=x) { int root=findset(f[x]); d[x]+=d[f[x]]... 阅读全文

posted @ 2014-02-24 20:55 雄.. 阅读(148) 评论(0) 推荐(0)

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