引用别人的解释: 题意:有n个村庄,村庄在不同坐标和海拔,现在要对所有村庄供水,只要两个村庄之间有一条路即可, 建造水管距离为坐标之间的欧几里德距离(好象是叫欧几里德距离吧),费用为海拔之差 现在要求方案使得费用与距离的比值最小 很显然,这个题目是要求一棵最优比率生成树, 概念 有带权图G, 对于图中每条边e[i], 都有benifit[i](收入)和cost[i](花费), 我们要求的是一棵生成树T, 它使得 ∑(benifit[i]) / ∑(cost[i]), i∈T 最大(或最小). 这…

Description David the Great has just become the king of a desert country. To win the respect of his people, he decided to build channels all over his country to bring water to every village. Villages which are connected to his capital village will be…

http://poj.org/problem?id=2728 Desert King Time Limit: 3000MS   Memory Limit: 65536K Total Submissions: 18595   Accepted: 5245 Description David the Great has just become the king of a desert country. To win the respect of his people, he decided to b…

用01分数规划 + prime + 二分 竟然2950MS惊险的过了QAQ 前提是在TLE了好几次下过的 = = 题目意思:有n个村庄,村庄在不同坐标和海拔,现在要对所有村庄供水,只要两个村庄之间有一条路即可,建造水管距离为坐标之间的欧几里德距离,费用为海拔之差,现在要求方案使得费用与距离的比值最小,很显然,这个题目是要求一棵最优比率生成树. 解题思路: 对答案进行二分,当把代进去的答案拿来算最小生成树的时候,一旦总路径长度为0,就是需要的答案. 0-1规划是啥? 概念有带权图G, 对于图中每条…

题目大意: 有n个村庄,村庄在不同坐标和海拔,现在要对所有村庄供水, 只要两个村庄之间有一条路即可,建造水管距离为坐标之间的欧几里德距离,费用为海拔之差, 现在要求方案使得费用与距离的比值最小,很显然,这个题目是要求一棵最优比率生成树. ———————————————————————————————————— 这是一道最优比率生成树的题目,是个很明显的0-1分数规划,设每条边代价为ci,距离为di 那么题目要求(∑(ci*xi))/(∑(di*xi))的最小值 xi∈{0,1} 我们进行一波转换…

题意:有M只猴子,他们的最大跳跃距离为Ai.树林中有N棵树露出了水面,给出了它们的坐标.问有多少只猴子能在这个地区露出水面的所有树冠上觅食. 解法:由于要尽量多的猴子能到达所有树冠,便用Kruskal求一次MST最小生成树,也就得到了最大边最小的最小瓶颈生成树.在用这个最大边权计算合法的猴子树. P.S.由于没有那些一下子输入几十万的数据,所以读优和不读优差别不大. 1 #include<cstdio> 2 #include<cstdlib> 3 #include<cstri…

poj2728 题意 给出 n 个点的坐标和它的高度,求一颗生成树使得树上所连边的两点高度差之和除以距离之和最小. 分析 01分数规划+最小生成树. 对于所有的边,在求最小生成树过程中有选或不选的问题, 首先根据01分数规划,我们要使 $ l = \frac{\sum_{i=1}^{n} height[i] * exist[i]}{\sum_{i=1}^{n} dis[i] * exist[i]}$ (exist[i]表示是否有这条边)最小, 设 \(F(l) = {\sum height[i]…

题意 n个点完全图,每个边有两个权值,求分数规划要求的东西的最小值. (n<=1000) 题解 心态炸了. 堆优化primT了. 普通的就过了. 我再也不写prim了!!!! 咳咳 最优比率生成树板子题. 公式不是很难推吧. #include<iostream> #include<cmath> #include<iomanip> #include<string.h> ; const int inf=0x7fffffff; using namespace…

[POJ2728]Desert King(分数规划) 题面 vjudge 翻译: 有\(n\)个点,每个点有一个坐标和高度 两点之间的费用是高度之差的绝对值 两点之间的距离就是欧几里得距离 求一棵生成数,使得单位距离的费用最小 题解 使得\(\sum cost/\sum dis\)最小 这是分数规划问题 二分答案\(K\) 如果\(K\)满足,则有 \(\sum cost-K\sum dis\leq 0\) 定义生成树边权为\(cost-K·dis\) 做最小生成树检查答案即可. 因为是稠密图,…

poj2728:http://poj.org/problem?id=2728 题意:给你n的点,每一个点会有一个坐标(x,y),然后还有一个z值,现在上你求一棵生成树,是的这棵生成树的所有边的费用/所有边的距离最小,其中,边费用是指两点之z差值的绝对值,边距离是指两点之间的距离. 题解:这一题就是求最小比率生成树.采用的解法就是0-1分数规划. 其中设最后的比率是l 1,z(l)是单调递减的. 2,z(max(l))=0;这可以采用反证法进行证明. 3,因为是完全图,所以要采用prime求最小生…

http://poj.org/problem?id=2728 题目大意:求一棵生成树使得路费用和/路长之和最小(路的费用是两端点的高度差) 最小比率生成树. 我们还是01分数规划的思想将边权变为路费用-路长*枚举的答案,跑一遍最小生成树即可. 但是debug的三个小时的我要对出题人说一句. CNM无良卡常!K算法被卡正常,堆优化Prim都被卡你是什么东西????!!! #include<cstdio> #include<cmath> #include<cstring>…

以下内容均为转载 http://www.cnblogs.com/ftae/p/6947497.html poj2728(最小比率生成树)   poj2728 题意 给出 n 个点的坐标和它的高度,求一颗生成树使得树上所连边的两点高度差之和除以距离之和最小. 分析 01分数规划+最小生成树. 对于所有的边,在求最小生成树过程中有选或不选的问题, 首先根据01分数规划,我们要使 l=∑i=1nheight[i]∗exist[i]∑i=1ndis[i]∗exist[i] (exist[i]表示是否有这…

Desert King Time Limit: 3000MS   Memory Limit: 65536K Total Submissions: 20978   Accepted: 5898 [Description] David the Great has just become the king of a desert country. To win the respect of his people, he decided to build channels all over his co…

Desert King http://poj.org/problem?id=2728 Time Limit: 3000MS   Memory Limit: 65536K       Description David the Great has just become the king of a desert country. To win the respect of his people, he decided to build channels all over his country t…

POJ2728 Desert King Description David the Great has just become the king of a desert country. To win the respect of his people, he decided to build channels all over his country to bring water to every village. Villages which are connected to his cap…

题目描述: David the Great has just become the king of a desert country. To win the respect of his people, he decided to build channels all over his country to bring water to every village. Villages which are connected to his capital village will be water…

[题意]每条路径有一个 cost 和 dist,求图中 sigma(cost) / sigma(dist) 最小的生成树. 标准的最优比率生成树,楼教主当年开场随手1YES然后把别人带错方向的题Orz-- ♦01分数规划 参考Amber-胡伯涛神牛的论文<最小割模型在信息学竞赛中的应用> °定义 分数规划(fractional programming)的一般形式: Minimize  λ = f(x) = a(x) / b(x)   ( x∈S  && ∀x∈S, b(x) &…

Desert King Time Limit: 3000MS   Memory Limit: 65536K Total Submissions: 22717   Accepted: 6374 Description David the Great has just become the king of a desert country. To win the respect of his people, he decided to build channels all over his coun…

http://poj.org/problem?id=2728 题意: 在这么一个图中求一棵生成树,这棵树的单位长度的花费最小是多少? 思路: 最优比率生成树,也就是01分数规划,二分答案即可,题目很简单,因为这题是稠密图,所以用prim算法会好点. #include<iostream> #include<algorithm> #include<cstring> #include<cstdio> #include<vector> #include&…

最优比率生成树教程见http://blog.csdn.net/sdj222555/article/details/7490797 个人觉得很明白易懂,但他写的代码略囧. 模板题,但是必须Prim,不能用Kruscal,因为是完全图 Code: #include<cstdio> #include<cmath> #include<cstring> #include<algorithm> using namespace std; ]; int n; ]; ][],…

含[最小生成树Prim]模板. Prim复杂度为$O(n^2),适用于稠密图,特别是完全图的最小生成树的求解.   Desert King Time Limit: 3000MS   Memory Limit: 65536K Total Submissions:31622   Accepted: 8670 Description David the Great has just become the king of a desert country. To win the respect of h…

Language: Default Desert King Time Limit: 3000MS   Memory Limit: 65536K Total Submissions: 22113   Accepted: 6187 Description David the Great has just become the king of a desert country. To win the respect of his people, he decided to build channels…

题目链接:http://poj.org/problem?id=2728 Desert King Time Limit: 3000MS   Memory Limit: 65536K Total Submissions: 26878   Accepted: 7459 Description David the Great has just become the king of a desert country. To win the respect of his people, he decided…

Desert King Time Limit: 3000MS   Memory Limit: 65536K Total Submissions: 25310   Accepted: 7022 Description David the Great has just become the king of a desert country. To win the respect of his people, he decided to build channels all over his coun…

Desert King Time Limit: 3000MS   Memory Limit: 65536K Total Submissions:33847   Accepted: 9208 Description David the Great has just become the king of a desert country. To win the respect of his people, he decided to build channels all over his count…

题意:有n个村庄,村庄在不同坐标和海拔,现在要对所有村庄供水,只要两个村庄之间有一条路即可,建造水管距离为坐标之间的欧几里德距离,费用为海拔之差,现在要求方案使得费用与距离的比值最小,很显然,这个题目是要求一棵最优比率生成树. 析:也就是求 r = sigma(x[i] * d) / sigma(x[i] * dist)这个值最小,变形一下就可以得到 d * r - dist <= 0,当r 最小时,取到等号,也就是求最大生成树,然后进行判断,有两种方法,一种是二分,这个题时间长一点,另一种是迭…

分析:就是不断递归寻找靠近边界的最优解 学习博客(必须先看这个): 1:http://www.cnblogs.com/autsky-jadek/p/3959446.html 2:http://blog.csdn.net/u013849646/article/details/51524748 注:这里用的最小乘积生成树的思想,和dp结合 每次找满足条件的最优的点,只不过BZOJ裸题的满足条件是形成一棵树 这个题是大于m,生成树借用最小生成树进行求解最优,大于m用dp进行求解最优 #include…

01分数规划 前置技能 二分思想最短路算法一些数学脑细胞? 问题模型1 基本01分数规划问题 给定nn个二元组(valuei,costi)(valuei,costi),valueivaluei是选择此二元组获得的价值(非负),costicosti是选择此二元组付出的代价(非负),设xi(xi∈{0,1})xi(xi∈{0,1})代表第ii个二元组的选与不选,最大(小)化下式 maximize(or minimize)   r=∑valuei⋅xi∑costi⋅ximaximize(or mini…

[poj2728]Desert King 题意 最优比率生成树. http://blog.csdn.net/ophunter_lcm/article/details/10113817 分析 Dinkelbach算法,通过迭代的思想进行参数搜索. 每一次的参数搜索,求出来的最小生成树又会生成一个更优的参数,然后用新的参数继续搜...以此类推. 为什么生成的参数会更优? 这是因为我们要最小化\(\sum C\over \sum D\),我们设答案为\({\sum C\over \sum D}<L\)…

Desert King Time Limit: 3000MS   Memory Limit: 65536K Total Submissions: 28407   Accepted: 7863 Description David the Great has just become the king of a desert country. To win the respect of his people, he decided to build channels all over his coun…