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[Mathimatics-Numerical algorithms] TK DL : 0: 旅行家问题一个旅行家想驾驶汽车以最少的费yi 用从一个城市到另一个城市（假设出发时油箱是空的）。给定两个城市之间的距离为D1、汽车油箱的容量为C(以升为单位），每升汽油能行驶的距离为 D2，出发点每升汽油价格P和沿途油站数N（N可以为零），油站i离出发点距离Di,每升汽油价格Pi(i=1,2...N)。计算结果四舍五入至小数点后两位。如果无法到达目的地，则输出“No Solution\"。-issue a travel home to travel home to drive a car with the minimum of cost yi used from one city to another city (assuming starting at the fuel tank was empty). Given the two cities for the distance between D1, car fuel tank capacity of the C (in liters) per liter petrol traveling distance to the D2, the starting point liter gasoline prices P and several petrol stations along the N (N can be zero), PFS i distance from the starting point Di per liter Steam oil prices Pi (i = 1,2 ... N). Calculation results rounded to two decimal places. If unable to reach their destination, the export of "No Solution."
Date : 2008-10-13 Size : 881byte User : 信仰
[Mathimatics-Numerical algorithms] TK DL : 0: 旅行家问题一个旅行家想驾驶汽车以最少的费yi 用从一个城市到另一个城市（假设出发时油箱是空的）。给定两个城市之间的距离为D1、汽车油箱的容量为C(以升为单位），每升汽油能行驶的距离为 D2，出发点每升汽油价格P和沿途油站数N（N可以为零），油站i离出发点距离Di,每升汽油价格Pi(i=1,2...N)。计算结果四舍五入至小数点后两位。如果无法到达目的地，则输出“No Solution"。-issue a travel home to travel home to drive a car with the minimum of cost yi used from one city to another city (assuming starting at the fuel tank was empty). Given the two cities for the distance between D1, car fuel tank capacity of the C (in liters) per liter petrol traveling distance to the D2, the starting point liter gasoline prices P and several petrol stations along the N (N can be zero), PFS i distance from the starting point Di per liter Steam oil prices Pi (i = 1,2 ... N). Calculation results rounded to two decimal places. If unable to reach their destination, the export of "No Solution."
Date : 2025-12-25 Size : 1kb User : 信仰
[Mathimatics-Numerical algorithms] guff DL : 0: 在一个操场的四周摆放着n 堆石子。现要将石子有次序地合并成一堆。规定在合并过程中最多可以有m(k)次选k 堆石子合并成新的一堆，2≤k≤n，合并的费用为新的一堆的石子数。试设计一个算法，计算出将n 堆石子合并成一堆的最小总费用。-In a playground around the stones are placed n heap. Stone is to have the order to merge into a pile. Provisions in the merger process can have up to m (k) election k heap stones into a pile of new, 2 ≤ k ≤ n, the combined cost for the new pile of stones a few. Try to design an algorithm to calculate the n heap a pile of stones into the smallest total cost.
Date : 2025-12-25 Size : 2kb User : chenxueping
[Mathimatics-Numerical algorithms] car DL : 0: 使用动态规划算法求解汽车加油问题，求得费用的最小值，-The use of dynamic programming algorithm to solve the issue of motor vehicle fuel to obtain the minimum cost,
Date : 2025-12-25 Size : 2kb User : lands
[Mathimatics-Numerical algorithms] jf DL : 0: 缔结特拉斯算法的c++实现，简单而实用，可扩充性强，抑郁修改- Fuzzy comprehensive evaluation, the right level change, time priority, the cost of priority
Date : 2025-12-25 Size : 116kb User : jf
[Mathimatics-Numerical algorithms] automatic-classification-cluster DL : 0: 一、问题描述若要在n个城市之间建役通信网络，只福要架设n-1条级路即可．如何以最低的经济代价建设这个通信网，是一个网的最小生成树问题。二、基本要求（1）利用克鲁斯卡尔算法求图的最小生成树。（2）能实现教科书6.5节中定义的抽象数据类型MFSet.以此表示构造生成树过程中的连通分量。 (3 ) 以文本形式输出生成树中各条边以及他们的权值．三、需求分析 1、构造图结构。 2、利用克鲁斯卡尔算法求图的最小生成树。 3、完成生成树的输出。 -I. Description of the problem to the n cities built between the service communication networks, Fuk only n-1 to set up the way to class Article. How to minimize the economic cost of building the communications network is a network of minimum spanning tree problem. Second, the basic requirements of (1) the use of Kruskal Algorithm for Minimum Spanning Tree. (2) to achieve 6.5 in the textbook definition of abstract data types MFSet. Spanning tree structure as that in the process of component connectivity. (3) to output text in the spanning tree edges and their weights. Third, a needs analysis, structural map structure. 2, the use of Kruskal Algorithm for Minimum Spanning Tree. 3, complete the spanning tree output.
Date : 2025-12-25 Size : 668kb User : 赵婧
[Mathimatics-Numerical algorithms] WorkAssignment DL : 0: acm算法设计设有n件工作分配给n个人。将工作i 分配给第j 个人所需的费用为Cij。试设计一个算法，为每一个人都分配1 件不同的工作，并使总费用达到最小。设计一个算法，对于给定的工作费用，计算最佳工作分配方案，使总费用达到最小。输入：第一行有1 个正整数n (1≤n≤30)。接下来的n行，每行n个数，表示工作费用。输出：最小总费用。例输入： 3 10 2 3 2 3 4 3 4 5 例输出：9-acm algorithm design with n pieces of work assigned to the n individuals. I will be allocated to the work of the first j for the cost of personal Cij. Try to design an algorithm for each individual a different allocation of effort and minimize the total cost. To design an algorithm for a given cost of the work to calculate the best distribution of work to minimize the total cost. Input: the first line there is a positive integer n (1 ≤ n ≤ 30). N the next line, n the number of each line, the cost of that work. Output: the minimum total cost. Cases of type: 31,023,234,345 cases output: 9
Date : 2025-12-25 Size : 347kb User : 张波
[Mathimatics-Numerical algorithms] zhipaisuanfa DL : 0: 指派问题的回溯算法，回溯算法解决此问题要找出问题所有的可行解，然后一次比较保留问题的最优解(即最少耗费的解)，并输出结果-The assignment problem, backtracking algorithm retrospective algorithm to solve the problem of finding problems all feasible solution, then a comparative retain the optimum solution of the minimum cost solution (namely) and output
Date : 2025-12-25 Size : 1kb User : 曹萌萌
[Mathimatics-Numerical algorithms] hebingguozi DL : 0: 在一个果园里，多多已经将所有的果子打了下来，而且按果子的不同种类分成了不同的堆。多多决定把所有的果子合成一堆。每一次合并，多多可以把两堆果子合并到一起，消耗的体力等于两堆果子的重量之和。可以看出，所有的果子经过n-1次合并之后，就只剩下一堆了。多多在合并果子时总共消耗的体力等于每次合并所耗体力之和。因为还要花大力气把这些果子搬回家，所以多多在合并果子时要尽可能地节省体力。假定每个果子重量都为1，并且已知果子的种类数和每种果子的数目，你的任务是设计出合并的次序方案，使多多耗费的体力最少，并输出这个最小的体力耗费值。例如有3种果子，数目依次为1，2，9。可以先将1、2堆合并，新堆数目为3，耗费体力为3。接着，将新堆与原先的第三堆合并，又得到新的堆，数目为12，耗费体力为12。所以多多总共耗费体力=3+12=15。可以证明15为最小的体力耗费值-In an orchard, a lot has shot down all the fruit, and of different types of fruit into a different pile. Decided to put all the fruit of a lot of synthetic pile. Each merger, a lot can be merged together two piles of fruit, equal to two piles of physical consumption of fruit and weight. It can be seen, all the fruit through the n-1 times after the merger, on the left a pile of. When a lot of fruit in the combined strength equal to the total consumption of each of the merger and the physical consumption. Because these also make great efforts to move the fruit home, so when a lot of fruit in the merger as much as possible to save energy. Assume that each fruit weight is 1, and the number of species known to fruit and the number of each type of fruit, your task is to design a combined order of programs, so a lot of physical least cost and physical output of the minimum cost value. For example, there are three kinds of fruit, followed by the number 1,2,9. 1,2 heap can be first merged, the
Date : 2025-12-25 Size : 1kb User : yanSS
[Mathimatics-Numerical algorithms] svm4 DL : 0: 　-s svm类型：SVM设置类型(默认0) 　　0 -- C-SVC 　　1 --v-SVC 　　2 – 一类SVM 　　3 -- e -SVR 　　4 -- v-SVR 　　-t 核函数类型：核函数设置类型(默认2) 　　0 – 线性：u v 　　1 – 多项式：(r*u v + coef0)^degree 　　2 – RBF函数：exp(-r|u-v|^2) 　　3 –sigmoid：tanh(r*u v + coef0) 　　-d degree：核函数中的degree设置(针对多项式核函数)(默认3) 　　-g r(gama)：核函数中的gamma函数设置(针对多项式/rbf/sigmoid核函数)(默认1/ k) 　　-r coef0：核函数中的coef0设置(针对多项式/sigmoid核函数)((默认0) 　　-c cost：设置C-SVC，e -SVR和v-SVR的参数(损失函数)(默认1) 　　-n nu：设置v-SVC，一类SVM和v- SVR的参数(默认0.5) 　　-p p：设置e -SVR 中损失函数p的值(默认0.1) 　　-m cachesize：设置cache内存大小，以MB为单位(默认40) 　　-e eps：设置允许的终止判据(默认0.001) 　　-h shrinking：是否使用启发式，0或1(默认1) 　　-wi weight：设置第几类的参数C为weight*C(C-SVC中的C)(默认1) 　　-v n: n-fold交互检验模式，n为fold的个数，必须大于等于2--s svm_type : set type of SVM (default 0) 0-- C-SVC 1-- nu-SVC 2-- one-class SVM 3-- epsilon-SVR 4-- nu-SVR -t kernel_type : set type of kernel function (default 2) 0-- linear: u *v 1-- polynomial: (gamma*u *v+ coef0)^degree 2-- radial basis function: exp(-gamma*|u-v|^2) 3-- sigmoid: tanh(gamma*u *v+ coef0) 4-- precomputed kernel (kernel values in training_instance_matrix) -d degree : set degree in kernel function (default 3) -g gamma : set gamma in kernel function (default 1/k) -r coef0 : set coef0 in kernel function (default 0) -c cost : set the parameter C of C-SVC, epsilon-SVR, and nu-SVR (default 1) -n nu : set the parameter nu of nu-SVC, one-class SVM, and nu-SVR (default 0.5) -p epsilon : set the epsilon in loss function of epsilon-SVR (default 0.1) -m cachesize : set cache memory size in MB (default 100) -e epsilon : set tolerance of termination criterion (default 0.001) -h shrinking: whether to use the shrinking heuristics, 0 or 1 (default 1) -b
Date : 2025-12-25 Size : 17kb User : little863
[Mathimatics-Numerical algorithms] 1-ETS DL : 0: 问题陈述：有一推销员，欲到n(n<=10)个城市推销产品。为了节省旅行费用，在出发前他查清了任意两个城市间的旅行费用，想找到一条旅行路线，仅经过每个城市一次，且使旅行费用最少。-Problem Statement: a salesman, n (n < = 10) cities want to sell products. Before departure in order to save travel expenses, he find out the cost of travel between any two cities, want to find a travel route, after each city only once, and the travel expenses to a minimum.
Date : 2025-12-25 Size : 275kb User : Kahn