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ACM_二分三分
专题二总结 搜索算法
搜索大致分为dfs,bfs,
二分三分
搜索,bfs用队列一层一层的压入压出,dfs用用递归。其中
二分三分
搜索比较简单,但要读懂题目,知道什么时候用。查找连续函数时结束条件要足够小。
baidu_34275752
·
2016-04-24 22:00
专题二 搜索总结
在本专题中,有令人熟悉的
二分三分
算法,还有让人头大的深
yjz_sdau
·
2016-04-24 17:00
专题二 搜索总结
在本专题中,有令人熟悉的
二分三分
算法,还有让人头大的深搜以及广搜。总体来说,可以掌握基本的内容。在今后的时光里,必须继续加强思维的锻炼并且逐步地去提高自己。
tansanity
·
2016-04-24 15:00
ACM
ACM_
程序设计竞赛:贪心算法:saruman army
/*saruman'sarmy给从左边开始,在圆内的最右边的点加上标记即可。*/ #include #include usingnamespacestd; #defineMAX_N1000 intr,n; intx[MAX_N]; voidsolve() { sort(x,x+n); inti=0,ans=0; while(i>n>>r) { for(inti=0;i>x[i]; if(
mijian1207mijian
·
2016-04-24 00:00
ACM_
程序设计竞赛:贪心算法:字典最小序
#include #include #include #include usingnamespacestd; constintN=8; charchs[N+1]="ADHCACBD"; char*solve(charchs[]) { intstart=0,end=N-1; boolisLeft=false; chardest[N]; while(startchs[end-i]) { isLe
mijian1207mijian
·
2016-04-21 15:00
搜索—
二分三分
搜索—1004
这个题主要是一个公式的推导还有二分法和三分法的综合运用。。思路:先三分出一个角度,让抛物线在x=x0出的y取得最大值。然后再到这个y一下去二分一个合适的值。代码:#include#include#include#include#definePIacos(-1.0)#defineG9.8#defineeps1e-12usingnamespacestd;doubleX,Y,v,sita;doublec
SLX_better
·
2016-04-21 08:00
the summary of search
掌握的知识:1
二分三分
法:通过计算机快速的运算,对答案的迭代,通过一次一次的精确,达到近似最优的结果
toy_block
·
2016-04-20 00:00
acm_
挑战编程:入门
3n+1问题/***************************************3n+1问题**************************************从整数n开始,如果n是偶数,把它除以2;如果n是奇数,把它乘以3加1.用新得到的值重复上述步骤。直到n=1为止。**************************************需要注意://1.中间计算过程会
mijian1207mijian
·
2016-04-18 00:00
ACM_
程序设计竞赛:贪心算法:区间问题
有n项工作,每项工作分别在si时间开始,ti时间结束。对于每项工作你选择参与与否,如果选择了参与,那么自始至终就必须全程参与。参与工作的时间段不可以重复。(开始的瞬间和结束重复也不可以)尽可能多的参与工作,可以参与多少。/*------------------------------------[输入]n=5,s={1,2,,4,6,8},t={3,5,7,9,10}---------------
mijian1207mijian
·
2016-04-12 01:00
ACM_
程序设计竞赛:贪心算法:硬币问题
贪心算法:就是贪心的选取当前最优策略的算法设计方法。有1元,5元,10元,50元,100元,500元的硬币,各C1,C5,C10,C50,C100,C500枚。现在要用来支付A元,最少需要各种硬币多少。/*[输入]c1=3,c5=2,c10=1,c50=3,c100=0,c500=2,A=620[输出]6(500的1个,50的2个,10的1个,5元的2个,合计6个)[算法]1.优先使用大面值*/
mijian1207mijian
·
2016-04-05 01:00
ACM_
程序设计竞赛:穷举法:BFS(广度优先)
类似于层次遍历首先访问起始顶点v,v出发,依次访问领接的顶点w1,w2,...,wi不后退,一步可以访问一批结点//结果:abcdefgh//算法: boolvisited[MAX_VERTEX_NUM];//访问标记数组 voidBFSTraverse(GraphG){ //图G进行广度优先遍历,设访问函数visit() for(i=0;i=0;w=NextNeighbor(G,v,w))//检
mijian1207mijian
·
2016-04-01 01:00
ACM_
程序设计竞赛:穷举法:DFS(深度优先)
DFS的伪码从顶点v出发;访问v相邻且未被访问的顶点w1依次w2,....,,直到不能继续退回到出发点v,若v的领域还有为访问结点,重复上述//结果:abdceghfboolvisited[MAX_VERTEX_NUM];//访问数组标记 voidDFSTraverse(GraphG) { //对图G深度遍历,访问函数是visit() for(v=0;v=0;w=NextNeighor(G,v,w
mijian1207mijian
·
2016-03-30 01:00
南阳理工
ACM_
题目23
描述一天,TT在寝室闲着无聊,和同寝的人玩起了取石子游戏,而由于条件有限,他/她们是用旺仔小馒头当作石子。游戏的规则是这样的。设有一堆石子,数量为N(102.usingnamespacestd;03.intmain()04.{05.intk;06.longm,n;07.cin>>k;08.while(k--)09.{10.cin>>n>>m;11.if(n%(m+1)==0)12.cout<<"L
不闹的疯子
·
2016-03-17 17:40
ACM练习题
ACM_
模板_求质因子
求一个数的质因子用到的方法就是有策略的暴力,没求出一个质因子,就让原数消除这一个因数。//求一个数的质因子 #include intmain() { __int64a[100],num,i,n; while(scanf("%I64d",&n)!=EOF) { num=0; for(i=2;i*i1) a[num++]=n; for(i=0;i
RaAlGhul
·
2016-02-26 21:00
ACM_
模板_进制转换
有关于进制转换的问题,关键就是用C语言去模拟短除法的过程。#include//进制转换 intmain() { intN,R,k,a[20],i; while(scanf("%d%d",&N,&R)!=EOF) { k=0; if(N=10) printf("%c",'A'+a[i-1]-10); else printf("%d",a[i-1]); i--; } printf("\n"); } r
RaAlGhul
·
2016-02-26 20:00
ACM_
模板_最小公倍数
gcd函数的书写方式有很多种,很多人都是用循环的,而小编的模板用则是用函数的递归,从代码上很简介,且非常易于理解。#include//最小公倍数 intgcd(inta,intb) { if(a
RaAlGhul
·
2016-02-26 20:00
ACM_
模板_网络流
第一次接触网络流,感觉像是一堆的算法扑面而来,三天就要过去了,才刚刚对几个算法有了点初步的理解,感觉上离要熟练的做出题还很遥远,这里先给出小编对几个算法的模板总结。1.Edmond-Karp算法这种算法是最好理解的,网络上也有很多的对此算法的讲解,这里小编就不给出详细的讲解。算法的关键就是不停的在残留网络中找到增广路径,并不停的修改残留网络中的值,最后知道找不到增广路径为止,得到最大流。#incl
RaAlGhul
·
2016-02-25 22:00
动态规划初探 -- 背包问题
在为期一个星期的ACM集训之后,我就这样做了一个逃兵hhhh在这一个星期里面,学长讲了快速排序,
二分三分
搜索,矩阵快速幂,线段树,BFS(广度优先搜索)和DFS(深度优先搜索),邻接表和哈希表,结构体和优先队列
syuritsu
·
2016-01-24 12:00
nyoj--1087--摆方格(规律)
样例输入123样例输出1619上传者
ACM_
刘
qq_29963431
·
2015-11-13 20:00
ACM_
高次同余方程
/*poj3243 *解决高次同余方程的应用,已知X^Y=KmodZ,及X,Z,K的值,求Y的值 */ #include #include #include usingnamespacestd; #definelint__int64 #defineMAXN131071 structHashNode{lintdata,id,next;}; HashNodehash[MAXN=1) { if(b&1
xiaotan1314
·
2015-11-07 10:00
高次同余方程
ACM_
扩展欧几里德算法
/* 扩展欧几里德算法 基本算法:对于不完全为0的非负整数a,b,gcd(a,b)表示a,b的最大公约数,必然存在整数对x,y,使得gcd(a,b)=ax+by。 证明:设a>b。 1,显然当b=0,gcd(a,b)=a。此时x=1,y=0; 2,ab!=0时 设ax1+by1=gcd(a,b); bx2+(amodb)y2=gcd(b,amodb); 根据朴素的欧几里德原理有gcd
xiaotan1314
·
2015-11-06 14:00
算法
ACM
gcd
[
ACM_
模拟] UVA 12504 Updating a Dictionary [字符串处理 字典增加、减少、改变问题]
Updating a Dictionary In this problem, a dictionary is collection of key-value pairs, where keys are lower-case letters, and values are non-negative integers. Given
·
2015-11-01 10:30
字符串处理
[
ACM_
水题] UVA 12502 Three Families [2人干3人的活后分钱,水]
Three Families Three families share a garden. They usually clean the garden together at the end of each week, but last week, family C was on holiday, so family A sp
·
2015-11-01 10:29
ACM
[
ACM_
模拟] UVA 12503 Robot Instructions [指令控制坐标轴上机器人移动 水]
Robot Instructions You have a robot standing on the origin of x axis. The robot will be given some instructions. Your task is to predict its position after executin
·
2015-11-01 10:29
struct
[
ACM_
数据结构] POJ2352 [树状数组稍微变形]
Description Astronomers often examine star maps where stars are represented by points on a plane and each star has Cartesian coordinates. Let the level of a star be an amount of the stars tha
·
2015-10-31 11:37
数据结构
[
ACM_
图论] ZOJ 3708 [Density of Power Network 线路密度,a->b=b->a去重]
The vast power system is the most complicated man-made system and the greatest engineering innovation in the 20th century. The following diagram shows a typical 14 bus power system. In
·
2015-10-31 11:36
NetWork
[
ACM_
水题] ZOJ 3712 [Hard to Play 300 100 50 最大最小]
MightyHorse is playing a music game called osu!. After playing for several months, MightyHorse discovered the way of calculating score in osu!: 1. While p
·
2015-10-31 11:36
play
[
ACM_
暴力][
ACM_
几何] ZOJ 1426 Counting Rectangles (水平竖直线段组成的矩形个数,暴力)
Description We are given a figure consisting of only horizontal and vertical line segments. Our goal is to count the number of all different rectangles formed by these segments. As an example, the n
·
2015-10-31 11:36
count
[
ACM_
动态规划] UVA 12511 Virus [最长公共递增子序列 LCIS 动态规划]
Virus We have a log file, which is a sequence of recorded events. Naturally, the timestamps are strictly increasing. However, it is infected by a virus, so random
·
2015-10-31 11:36
动态规划
[
ACM_
动态规划] hdu 1176 免费馅饼 [变形数塔问题]
Problem Description 都说天上不会掉馅饼,但有一天gameboy正走在回家的小径上,忽然天上掉下大把大把的馅饼。说来gameboy的人品实在是太好了,这馅饼别处都不掉,就掉落在他身旁的10米范围内。馅饼如果掉在了地上当然就不能吃了,所以gameboy马上卸下身上的背包去接。但由于小径两侧都不能站人,所以他只能在小径上接。由于gameboy平时老
·
2015-10-31 11:36
动态规划
[
ACM_
模拟] ACM - Draw Something Cheat [n个长12的大写字母串,找出交集,按字母序输出]
Description Have you played Draw Something? It's currently one of the hottest social drawing games on Apple iOS and Android Devices! In this game, you and your friend play in turn. You
·
2015-10-31 11:36
ACM
[
ACM_
数学] LA 3708 Graveyard [墓地雕塑 圈上新加点 找规律]
Description Programming contests became so popular in the year 2397 that the governor of New Earck -- the largest human-inhabited planet of the galaxy -- opened a special Alley of
·
2015-10-31 11:36
ACM
[
ACM_
水题] UVA 11292 Dragon of Loowater [勇士斗恶龙 双数组排序 贪心]
Once upon a time, in the Kingdom of Loowater, a minor nuisance turned into a major problem. The shores of Rellau Creek in central Loowater had always been a prime breeding ground for g
·
2015-10-31 11:36
water
[
ACM_
模拟][
ACM_
数学] LA 2995 Image Is Everything [由6个视图计算立方体最大体积]
Description Your new company is building a robot that can hold small lightweight objects. The robot will have the intelligence to determine if an object is light enough to hold. It
·
2015-10-31 11:36
image
[
ACM_
图论] The Perfect Stall 完美的牛栏(匈牙利算法、最大二分匹配)
描述 农夫约翰上个星期刚刚建好了他的新牛棚,他使用了最新的挤奶技术。不幸的是,由于工程问题,每个牛栏都不一样。第一个星期,农夫约翰随便地让奶牛们进入牛栏,但是问题很快地显露出来:每头奶牛都只愿意在她们喜欢的那些牛栏中产奶。上个星期,农夫约翰刚刚收集到了奶牛们的爱好的信息(每头奶牛喜欢在哪些牛栏产奶)。一个牛栏只能容纳一头奶牛,当然,一头奶牛只能在一个牛栏中产奶。 给出奶牛们的爱好的信息,计算最
·
2015-10-31 11:35
ACM
[
ACM_
搜索] ZOJ 1103 || POJ 2415 Hike on a Graph (带条件移动3盘子到同一位置的最少步数 广搜)
Description "Hike on a Graph" is a game that is played on a board on which an undirected graph is drawn. The graph is complete and has all loops, i.e. for any two locations there is exactly
·
2015-10-31 11:35
Graph
[
ACM_
图论] Fire Net (ZOJ 1002 带障碍棋盘布炮,互不攻击最大数量)
Suppose that we have a square city with straight streets. A map of a city is a square board with n rows and n columns, each representing a street or a piece of wall. A blockhouse is a small cas
·
2015-10-31 11:35
ACM
[
ACM_
图论] Sorting Slides(挑选幻灯片,二分匹配,中等)
Description Professor Clumsey is going to give an important talk this afternoon. Unfortunately, he is not a very tidy person and has put all his transparencies on one big heap. Before giving the talk
·
2015-10-31 11:35
sort
[
ACM_
搜索] Triangles(POJ1471,简单搜索,注意细节)
Description It is always very nice to have little brothers or sisters. You can tease them, lock them in the bathroom or put red hot chili in their sandwiches. But there is also a time when all meanne
·
2015-10-31 11:35
ACM
[
ACM_
动态规划] ZOJ 1425 Crossed Matchings(交叉最大匹配 动态规划)
Description There are two rows of positive integer numbers. We can draw one line segment between any two equal numbers, with values r, if one of them is located in the first row and the other one is
·
2015-10-31 11:35
match
[
ACM_
搜索] POJ 1096 Space Station Shielding (搜索 + 洪泛算法Flood_Fill)
Description Roger Wilco is in charge of the design of a low orbiting space station for the planet Mars. To simplify construction, the station is made up of a series of Airtight Cubical Modules (ACM's
·
2015-10-31 11:35
ACM
[
ACM_
几何] Transmitters (zoj 1041 ,可旋转半圆内的最多点)
Description In a wireless network with multiple transmitters sending on the same frequencies, it is often a requirement that signals don't overlap, or at least that they don't conflict. One way of a
·
2015-10-31 11:35
ACM
[
ACM_
模拟] The Willy Memorial Program (poj 1073 ,联通水管注水模拟)
Description Willy the spider used to live in the chemistry laboratory of Dr. Petro. He used to wander about the lab pipes and sometimes inside empty ones. One night while he was in a pipe, he fell as
·
2015-10-31 11:35
ACM
[
ACM_
其他] Square Ice (poj1099 规律)
Description Square Ice is a two-dimensional arrangement of water molecules H2O, with oxygen at the vertices of a square lattice and one hydrogen atom between each pair of adjacent oxygen atoms. The
·
2015-10-31 11:35
ACM
[
ACM_
图论] Highways (变形说法的最小生成树)
http://acm.hust.edu.cn/vjudge/contest/view.action?cid=28972#problem/C 题目给出T种情况,每种情况有n个城镇,接下来每一行是第i个城镇到所有城镇的距离(其实就是个可达矩阵)。 求建设一条公路联通所有城镇并且要求最长的一段最小(其实就是最小生成树)!代码如下: #include<
·
2015-10-31 11:34
最小生成树
[
ACM_
数学] Counting Solutions to an Integral Equation (x+2y+2z=n 组合种类)
http://acm.hust.edu.cn/vjudge/contest/view.action?cid=27938#problem/E 题目大意:Given, n, count the number of solutions to the equation x+2y+2z=n, where x,y,z,n are non negative inte
·
2015-10-31 11:34
count
[
ACM_
几何] The Deadly Olympic Returns!!! (空间相对运动之最短距离)
http://acm.hust.edu.cn/vjudge/contest/view.action?cid=28235#problem/B 题目大意: 有两个同时再空间中匀速运动的导弹,告诉一个时间以及各自的初始坐标和该时间时的坐标,求运动过程中的最短距离 解题思路: 求出相对初位置、相对速度,则答案就是原点到射线型轨迹的距离,注意是射线!!!
·
2015-10-31 11:34
return
[
ACM_
几何] F. 3D Triangles (三维三角行相交)
http://acm.hust.edu.cn/vjudge/contest/view.action?cid=28235#problem/A 题目大意:给出三维空间两个三角形三个顶点,判断二者是否有公共点,三角形顶点、边、内部算三角形的一部分。 解题思路:见模板 //**********************************************
·
2015-10-31 11:34
ACM
[
ACM_
动态规划] 轮廓线动态规划——铺放骨牌(状态压缩1)
Description Squares and rectangles fascinated the famous Dutch painter Piet Mondriaan. One night, after producing the drawings in his 'toilet series' (where he had to use his toilet paper to draw on
·
2015-10-31 11:34
动态规划
[
ACM_
图论] 棋盘问题 (棋盘上放棋子的方案数)
不能同行同列,给定形状和大小的棋盘,求摆放k个棋子的可行方案 Input 2表示是2X2的棋盘,1表示k,#表示可放,点不可放(-1 -1 结束) Output 输出摆放的方案数目C Sample Input 2 1 #. .# 4 4 ...# ..#. .#.. #... -1 -1 Sample Output 2 1
·
2015-10-31 11:34
ACM
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