UVA 12594 Naming Babies

$dp$,斜率优化. 设$dp[j][i]$表示前$i$个位置分成$j$段的最小值,递推式很好写,预处理几个前缀和就可以了,然后斜率优化即可. #pragma comment(linker, "/STACK:1024000000,1024000000") #include<cstdio> #include<cstring> #include<cmath> #include<algorithm> #include<vector>…

[kuangbin带你飞]专题二十斜率DP

ID Origin Title 20 / 60 Problem A HDU 3507 Print Article 13 / 19 Problem B HDU 2829 Lawrence 1 / 5 Problem C HDU 4528 小明系列故事――捉迷藏 5 / 6 Problem D HDU 1300 Pearls 0 / 42 Problem E HDU 2993 MAX Average Problem 1 / 20 Problem F UVALi…

KUANGBIN带你飞

KUANGBIN带你飞全专题整理 https://www.cnblogs.com/slzk/articles/7402292.html 专题一简单搜索 POJ 1321 棋盘问题 //2019.3.18 POJ 2251 Dungeon Master POJ 3278 Catch That Cow //4.8 POJ 3279 Fliptile POJ 1426 Find The Multiple //4.8 POJ 3126 Prime Path POJ 3087 Shuffle…

[kuangbin带你飞]专题1-23题目清单总结

[kuangbin带你飞]专题1-23 专题一简单搜索 POJ 1321 棋盘问题POJ 2251 Dungeon MasterPOJ 3278 Catch That CowPOJ 3279 FliptilePOJ 1426 Find The MultiplePOJ 3126 Prime PathPOJ 3087 Shuffle'm UpPOJ 3414 PotsFZU 2150 Fire GameUVA 11624 Fire!POJ 3984 迷宫问题HDU 1241 Oil Deposit…

ACM--[kuangbin带你飞]--专题1-23

专题一简单搜索 POJ 1321 棋盘问题POJ 2251 Dungeon MasterPOJ 3278 Catch That CowPOJ 3279 FliptilePOJ 1426 Find The MultiplePOJ 3126 Prime PathPOJ 3087 Shuffle'm UpPOJ 3414 PotsFZU 2150 Fire GameUVA 11624 Fire!POJ 3984 迷宫问题HDU 1241 Oil DepositsHDU 1495 非常可乐HDU 26…

算法笔记--斜率优化dp

斜率优化是单调队列优化的推广用单调队列维护递增的斜率参考:https://www.cnblogs.com/ka200812/archive/2012/08/03/2621345.html 以例1举例说明: 转移方程为:dp[i] = min(dp[j] + (sum[i] - sum[j])^2 + C) 假设k < j < i, 如果从j转移过来比从k转移过来更优那么 dp[j] + (sum[i] - sum[j])^2 + C < dp[k] + (sum[i] - sum[…

【java】Naming.bind和Registry.bind区别

Naming类和Registry类均在java.rmi包 Naming类通过解析URI绑定远程对象,将URI拆分成主机.端口和远程对象名称,使用的仍是Registry类. public static Remote lookup(String name) throws NotBoundException, java.net.MalformedURLException, RemoteException { ParsedNamingURL parsed = parseURL(name); Regist…

uva 1354 Mobile Computing ——yhx

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UVA 10564 Paths through the Hourglass[DP 打印]

UVA - 10564 Paths through the Hourglass 题意: 要求从第一层走到最下面一层,只能往左下或右下走问有多少条路径之和刚好等于S? 如果有的话,输出字典序最小的路径. f[i][j][k]从下往上到第i层第j个和为k的方案数上下转移不一样,分开处理没必要判断走出沙漏打印方案倒着找下去行了,尽量往左走沙茶的忘注释掉文件WA好多次 #include <iostream> #include <cstdio> #include <a…

UVA 11404 Palindromic Subsequence[DP LCS 打印]

UVA - 11404 Palindromic Subsequence 题意:一个字符串,删去0个或多个字符,输出字典序最小且最长的回文字符串不要求路径区间DP都可以做然而要字典序最小倒过来求LCS,转移同时维护f[i][j].s为当前状态字典序最小最优解 f[n][n].s的前半部分一定是回文串的前半部分(想想就行了) 当s的长度为奇时要多输出一个(因为这样长度+1,并且字典序保证最小(如axyzb bzyxa,就是axb|||不全是回文串的原因是后半部分的字典序回文串可能不是最小,多…

UVA&&POJ离散概率与数学期望入门练习[4]

POJ3869 Headshot 题意:给出左轮手枪的子弹序列,打了一枪没子弹,要使下一枪也没子弹概率最大应该rotate还是shoot 条件概率,|00|/(|00|+|01|)和|0|/n谁大的问题 |00|+|01|=|0| 注意序列是环形 // // main.cpp // poj3869 // // Created by Candy on 25/10/2016. // Copyright © 2016 Candy. All rights reserved. // #include <i…

UVA计数方法练习[3]

UVA - 11538 Chess Queen 题意:n*m放置两个互相攻击的后的方案数分开讨论行列两条对角线一个求和式可以化简后计算 // // main.cpp // uva11538 // // Created by Candy on 24/10/2016. // Copyright © 2016 Candy. All rights reserved. // #include <iostream> #include <cstdio> #include <cst…

UVA数学入门训练Round1[6]

UVA - 11388 GCD LCM 题意:输入g和l,找到a和b,gcd(a,b)=g,lacm(a,b)=l,a<b且a最小 g不能整除l时无解,否则一定g,l最小 #include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #include <cmath> using namespace std; typedef long long ll…

UVA - 1625 Color Length[序列DP 代价计算技巧]

UVA - 1625 Color Length 白书很明显f[i][j]表示第一个取到i第二个取到j的代价问题在于代价的计算,并不知道每种颜色的开始和结束和模拟赛那道环形DP很想,计算这次转移会给其他的元素带来的代价,也就是转移前已经出现但还没结束的元素都会代价+1 预处理每种颜色在两个序列中出现的位置bg[i][0/1]和ed[i][0/1], 计算f[i][j]时同时计算w[i][j]为(i,j)这个状态已经出现还没结束的个数注意bg要先初始化为INF,计算w: if(b…

UVA - 10375 Choose and divide[唯一分解定理]

UVA - 10375 Choose and divide Choose and divide Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 4053 Accepted: 1318 Description The binomial coefficient C(m,n) is defined as m! C(m,n) = -------- n!(m-n)! Given four natural numbers p, q…

UVA - 11584 Partitioning by Palindromes[序列DP]

UVA - 11584 Partitioning by Palindromes We say a sequence of char- acters is a palindrome if it is the same written forwards and backwards. For example, ‘racecar’ is a palindrome, but ‘fastcar’ is not. A partition of a sequence of characters is a lis…

UVA - 1025 A Spy in the Metro[DP DAG]

UVA - 1025 A Spy in the Metro Secret agent Maria was sent to Algorithms City to carry out an especially dangerous mission. After several thrilling events we find her in the first station of Algorithms City Metro, examining the time table. The Algorit…

UVA - 11134 Fabled Rooks[贪心问题分解]

UVA - 11134 Fabled Rooks We would like to place n rooks, 1 ≤ n ≤ 5000, on a n × n board subject to the following restrictions The i-th rook can only be placed within the rectan- gle given by its left-upper corner (xli,yli) and its right- lower corner…

UVA - 11987 Almost Union-Find[并查集删除]

UVA - 11987 Almost Union-Find I hope you know the beautiful Union-Find structure. In this problem, you’re to implement something similar, but not identical. The data structure you need to write is also a collection of disjoint sets, supporting 3 oper…

BZOJ 1260&UVa 4394 区间DP

题意: 给一段字符串成段染色,问染成目标串最少次数. SOL: 区间DP... DP[i][j]表示从i染到j最小代价转移:dp[i][j]=min(dp[i][j],dp[i+1][k]+dp[k+1][j]); CODE: BZ: /*================================================================= # Created time: 2016-03-28 21:10 # Filename: uva4394.cpp # Desc…

uva 147 Dollars

http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=83 147 - Dollars Time limit: 3.000 seconds Dollars New Zealand currency consists of $100, $50, $20, $10, and $5 notes and $2, $1, 50c, 20c, 10c and…

hduoj 1455 && uva 243 E - Sticks

http://acm.hdu.edu.cn/showproblem.php?pid=1455 http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=243 uva开头描述: 307 - Sticks Time limit: 3.000 seconds hduoj 开头描述: E - Sticks Time Limit:3000MS Memo…

Uva 1630 折叠串

题目链接:https://uva.onlinejudge.org/external/16/1630.pdf 题意:折叠串,给一个字符串,相同部分可以折叠,折叠可以嵌套.求最短长度的一种折叠方法.括号和数字的长度也要考虑进去. 刚看到这个题目,没有一点思路,还是大牛们厉害! 分析:一个串,可以转成两种形式,要么本身可以转成有重叠部分的串,要么分成两个部分,再转成有重叠部分的串. 本身是否是重叠串,利用kmp查,分成两个部分,遍历一遍所有情况,这样,dp顺序就出来了,最外层是每次查的长度,第二层就是…

Uva 10891 经典博弈区间DP

经典博弈区间DP 题目链接:https://uva.onlinejudge.org/external/108/p10891.pdf 题意: 给定n个数字,A和B可以从这串数字的两端任意选数字,一次只能从一端选取. 并且A B都尽力使自己选择的结果为最大的,可以理解成A B每一步走的都是最优的. 如果A先选择,则A B差值最大是多少. 分析: 总和是一定的,所以一个得分越高,另一个人的得分越低.当前状态总是最开始的状态的一个子状态. d(i,j): 先手取 i ~ j 最优策略下,得分最大值. d…