UVA 104 Arbitrage

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uva 104 Arbitrage （DP + floyd）

uva 104 Arbitrage Description Download as PDF Background The use of computers in the finance industry has been marked with controversy lately as programmed trading – designed to take advantage of extremely small fluctuations in prices – has been outl…

UVa 104 - Arbitrage（Floyd动态规划）

题目来源:https://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=3&page=show_problem&problem=40 Arbitrage Background The use of computers in the finance industry has been marked with controversy lately as programmed tr…

动态规划类似FLOYD dp[i][j][k] 表示第i个点经过K次到达j点能获得的最大利润 #include <map> #include <set> #include <list> #include <cmath> #include <ctime> #include <deque> #include <stack> #include <queue> #include <cctype> #incl…

UVA 436 - Arbitrage (II)（floyd)

UVA 436 - Arbitrage (II) 题目链接题意:给定一些国家货币的汇率.问是否能通过不断换货币使钱得到增长思路:floyd,完事后推断一下有没有连到自己能大于1的情况代码: #include <cstdio> #include <cstring> #include <iostream> #include <string> #include <map> using namespace std; const int N = 35…

uva 104 Bandwidth

题意: 给一个图, 将其节点以任一序列排列. 1)计算每个节点距离相邻节点的最大距离 dis[i] 2)计算出当前序列中, 所有节点的dis[i], 并求出最大的dis[i] : max_dis 求最小的max_dis, 并且输出此序列. 节点数不超过8个思路: 节点数不超过八个, 那直接进行全排列, 求解最小值即可. 复杂度为O(n!) 坑爹的地方在读图, 模拟读图, 得处理好细节, 全排列的生成直接用dfs即可. 代码: #include <cmath> #include <cst…

UVA题目分类

题目 Volume 0. Getting Started 开始10055 - Hashmat the Brave Warrior 10071 - Back to High School Physics 10300 - Ecological Premium 458 - The Decoder 494 - Kindergarten Counting Game 414 - Machined Surfaces 490 - Rotating Sentences 445 - Marvelous Mazes…

UVA题解一

UVA 100 题目描述:经典3n+1问题在\(n \leq 10^6\)已经证明是可行的,现在记\(f[n]\)为从\(n\)开始需要多少步才能到\(1\),给出\(L, R\),问\(f[L], f[L+1], f[L+2], ... , f[R]\)中的最大值. solution 这题主要是坑... 1.\(L\)有可能大于\(R\) 2.数据范围应该是\(10^6\) 言归正传.通过打表发现,\(10^6\)以内的\(f[n]\)最大也只是几百,普遍都很小,所以不用记忆化也可以把\(f[…