题意:给指定数量的数字“1”,“2”,“3”……,“9”.用所有这些数字加上任意个0组成一个数,要求数能被11整除,且数的位数尽量小. 能被11整除的数有一个特点,奇数位数字之和与偶数位之和的差为11的倍数. 所以想到把所有数字分成两部分,即奇数位部分和偶数位部分,两部分的差相0即能被11整除(MOD 11). 所求可以化为,其中一部分%11的余数为所有数字之和%11的余数的一半. dp[k][r] := 能否找到 任意k个数之和 %11 == R #include<bits/stdc++.h>…

Codeforces Round #598 (Div. 3)- E. Yet Another Division Into Teams - 动态规划 [Problem Description] 给你\(n\)个数,将其划分为多组,对于每个组定义其\(d\)值为 组内的最大值减最小值,问如何划分使得最终所有组的\(d\)值之和最小.每个组至少要保证有\(3\)个数. [Solution] 将所有值从小到大排序,然后我们知道最多有\(5\)个人划分到同一组中,如果有\(6\)个人,那么划分为两组一定比…

题意: 有N个格子排成一排,在每个格子里填上1到N的数(每个只能填一次),分别代表每个格子的高度.现在给你两个数left和right,分别表示从左往右看,和从右往左看,能看到的格子数.问有多少种情况. 数据范围: N<5000: 思路: 首先枚举最高的一块,在最高的格子的后面的格子都一定会被挡住.所以,除了最高的那一格之外,从左边能看到的格子,从右边一定看不到:从右边能看到的也是一样. 因此,除了最高的那个格子,左边的是否能被看见,和右边的无关.所以,我们可以以最高的格子为界,把这一排格子分成左…

PHP的SPL标准库里面的堆(SplHeap)怎么使用 一.总结 1.因为SplHeap是抽象类,所以要先继承,实现里面的抽象方法compare后,才能new对象使用. 二.PHP的SPL标准库里面的堆(SplHeap)怎么使用 堆(Heap)就是为了实现优先队列而设计的一种数据结构,它是通过构造二叉堆(二叉树的一种)实现.根节点最大的堆叫做最大堆或大根堆,根节点最小的堆叫做最小堆或小根堆.二叉堆还常用于排序(堆排序).如下:最小堆(任意节点的优先级不小于它的子节点) 看看PHP SplHeap…

Accepted Papers by Session Research Session RT01: Social and Graphs 1Tuesday 10:20 am–12:00 pm | Level 3 – Ballroom AChair: Tanya Berger-Wolf Efficient Algorithms for Public-Private Social NetworksFlavio Chierichetti,Sapienza University of Rome; Ales…

POJ 3342 Party at Hali-Bula / HDU 2412 Party at Hali-Bula / UVAlive 3794 Party at Hali-Bula / UVA 1220 Party at Hali-Bula(树型动态规划) Description Dear Contestant, I'm going to have a party at my villa at Hali-Bula to celebrate my retirement from BCM. I w…

[CF1133E]K Balanced Teams(动态规划,单调队列) 题面 CF 让你把一堆数选一些出来分成不超过\(K\)组,每一组里面的最大值和最小值之差不超过\(5\),求最多有多少个人元素可以被分组. 题解 设\(f[i][j]\)表示把前\(i\)个数分成\(j\)组的最多人数. 然后单调队列转移一下完了. #include<iostream> #include<cstdio> #include<cstdlib> #include<cstring&g…

UVA 1394 And Then There Was One / Gym 101415A And Then There Was One / UVAlive 3882 And Then There Was One / POJ 3517 And Then There Was One / Aizu 1275 And Then There Was One (动态规划,思维题) Description Let's play a stone removing game. Initially, n ston…

UVAlive 4794 Sharing Chocolate 题目: http://acm.hust.edu.cn/vjudge/problem/viewProblem.action?id=12055 思路:   设d[S][r][c]表示形如r*c的矩形是否可以划分为S中的子集,10表示可否. 转移方程: d[S][r][c] = d[S0][r0][c] || d[S0][r][c0]  优化:    首先注意到S r c三者知二求一,所以将状态优化为d[S][x]表示有短边x的矩形是否可以…

 UVAlive 3983 Robotruck 题目: Robotruck   Time Limit: 3000MS   Memory Limit: Unknown   64bit IO Format: %lld & %llu Submit Status Description   Problem C - Robotruck Background This problem is about a robotic truck that distributes mail packages to sev…

题意:求出将两个字符串改成一样长度所能形成最大的相似度. 思路:这个可以说是编辑距离的一个变形,编辑距离最终状态时要两个字符串完全一致,这个就是要求长度一样,而且这个只允许插入“—”这一个字符.模仿编辑距离定义状态,dp[i][j]表示将第一个字符串的前i个字符与第二个字符串的前j个字符变为相同长度所能形成的最大相似度.设两个字母的相似度为g[i][j]; 那状态转移为 dp[i][j] = max( dp[i][j-1] + g[j][5], d[i-1][j] + g[i][5],dp[i-…

f(i,j)表示从A序列前面取i个,从B序列前面取j个时,能否拼成C序列.转移自行脑补. A train yard is a complex series of railroad tracks for storing, sorting, or loading/unloading railroad cars. In this problem, the railroad tracks are much simpler, and we are only interested in combining…

 Problem A: Football (aka Soccer)  The Problem Football the most popular sport in the world (americans insist to call it "Soccer", but we will call it "Football"). As everyone knows, Brasil is the country that have most World Cup title…

Check the difficulty of problems Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 4522   Accepted: 1993 Description Organizing a programming contest is not an easy job. To avoid making the problems too difficult, the organizer usually exp…

Football the most popular sport in the world (americans insist to call it "Soccer", but we will call it "Football"). As everyone knows, Brasil is the country that have most World Cup titles (four of them: 1958, 1962, 1970 and 1994). As…

题目链接 http://acm.sdibt.edu.cn/vjudge/ojFiles/uvalive/pdf/61/6177.pdf 题意是  给定一个数n,代表着一共有n个人,且他们的身高从1到n. 要求让这n个人站成一行,使得身高的排列呈波浪形,比如低高低或者高低高. 注意:n = 1 ,  ans = 1;    n = 2 ,  ans = 2; 动态规划. 解题思路: 每次新加入的点k,可以看成将之前的序列分成前后两部分,并且因为 k是最大的,所以要求k前面数的趋势应该是高低,k后面…

A 1-1 tie at home was sufficient for Guangzhou Evergrande to clinch the Asian Champions League title, the first Chinese club to win the tournament in more than two decades. clinch:拥吻,钉牢 tournament:锦标赛,联赛 on aggregate:总共,总计 It was difficult to spot a…

/** 题目:Confusing Date Format UVALive 7711 链接:https://vjudge.net/contest/174844#problem/A 题意:给定mm-mm-mm格式的时间.年份(1900-1999)只给了后两位数,问有多少种合法的排列使时间正确. 思路: 第一次:快速读题,题意不清,没注意到这句话, To punish teams who did not read this problem statement carefully, we’ll add…

[SinGuLaRiTy-1026] Copyright (c) SinGuLaRiTy 2017. All Rights Reserved. [UVA 1025] A Spy in the Metro 题目描述 特工玛利亚被送到S市执行一个特别危险的任务.她需要利用地铁完成他的任务,S市的地铁只有一条线路运行,所以并不复杂. 玛利亚有一个任务,现在的时间为0,她要从第一个站出发,并在最后一站的间谍碰头.玛利亚知道有一个强大的组织正在追踪她,她知道如果一直呆在一个车站,她会有很大的被抓的风险,躲…

Tug of War Time Limit: 3000MS Memory Limit: 65536K Description A tug of war is to be arranged at the local office picnic. For the tug of war, the picnickers must be divided into two teams. Each person must be on one team or the other; the number of p…

Stickers Time Limit: 3000ms Memory Limit: 131072KB This problem will be judged on UVALive. Original ID: 651064-bit integer IO format: %lld      Java class name: Main 解题:动态规划 #include <bits/stdc++.h> using namespace std; ; ],a[][maxn],n; int main(){…

The King's Ups and Downs Time Limit: 3000ms Memory Limit: 131072KB This problem will be judged on UVALive. Original ID: 617764-bit integer IO format: %lld      Java class name: Main   The king has guards of all different heights. Rather than line the…

Salesmen Time Limit: 3000ms Memory Limit: 131072KB This problem will be judged on UVALive. Original ID: 425664-bit integer IO format: %lld      Java class name: Main   Traveling salesmen of nhn. (the prestigious Korean internet company) report their…

Dividing coins Time Limit: 3000ms Memory Limit: 131072KB This problem will be judged on UVALive. Original ID: 558364-bit integer IO format: %lld      Java class name: Main It's commonly known that the Dutch have invented copper-wire. Two Dutch men we…

time limit per test2 seconds memory limit per test256 megabytes inputstandard input outputstandard output The country Treeland consists of n cities connected with n - 1 bidirectional roads in such a way that it's possible to reach every city starting…

E. Yet Another Division Into Teams 首先要想明白一个东西,就是当一个小组达到六个人的时候,它一定可以拆分成两个更优的小组. 这个题可以用动态规划来写,用一个数组来保存状态,用一个队列来尝试新的状态,但是因为上面的这个特性,每一次只会有三个新的状态. 我们用sum来保存躲避选择的元素,举个例子: 分组情况为:1 2 3 | 5 6 8 11 | 20 21 22 (不一定满足题意,只是为了说明sum的意义) 则sum=(5-3)+(20-11)=11 那么这样分组…

上一篇我们已经说到了,增强学习的目的就是求解马尔可夫决策过程(MDP)的最优策略,使其在任意初始状态下,都能获得最大的Vπ值.(本文不考虑非马尔可夫环境和不完全可观测马尔可夫决策过程(POMDP)中的增强学习). 那么如何求解最优策略呢?基本的解法有三种: 动态规划法(dynamic programming methods) 蒙特卡罗方法(Monte Carlo methods) 时间差分法(temporal difference). 动态规划法是其中最基本的算法,也是理解后续算法的基础,因此本…

题目:House Robber You are a professional robber planning to rob houses along a street. Each house has a certain amount of money stashed, the only constraint stopping you from robbing each of them is that adjacent houses have security system connected a…

March 26, 2013 作者:Hawstein 出处:http://hawstein.com/posts/dp-novice-to-advanced.html 声明:本文采用以下协议进行授权: 自由转载-非商用-非衍生-保持署名|Creative Commons BY-NC-ND 3.0 ,转载请注明作者及出处. 前言 本文翻译自TopCoder上的一篇文章: Dynamic Programming: From novice to advanced ,并非严格逐字逐句翻译,其中加入了自己的…

转自:http://segmentfault.com/blog/exploring/ LCS 问题描述 定义: 一个数列 S,如果分别是两个或多个已知数列的子序列,且是所有符合此条件序列中最长的,则 S 称为已知序列的最长公共子序列. 例如:输入两个字符串 BDCABA 和 ABCBDAB,字符串 BCBA 和 BDAB 都是是它们的最长公共子序列,则输出它们的长度 4,并打印任意一个子序列. (Note: 不要求连续) 判断字符串相似度的方法之一 - LCS 最长公共子序列越长,越相似. Ju…