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…

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…

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…

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…

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: 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…

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&…

[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) &…

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…

Desert King Time Limit: 3000MS   Memory Limit: 65536K Total Submissions:29775   Accepted: 8192 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…

                                                                                                Desert King Time Limit: 3000MS   Memory Limit: 65536K Total Submissions: 25729   Accepted: 7143 Description David the Great has just become the king of a…

题目链接:http://poj.org/problem?id=2728 题意: 给你n个点(x,y,z),让你求一棵生成树,使得 k = ∑ |z[i]-z[j]| / ∑ dis(i,j)最小. |z[i]-z[j]|为一条边两端点的高度(z)之差,dis(i,j)为两端点在xy平面投影的欧几里得距离. 题解: 二分答案R. 如果当前的R还没有达到最小值ans,即R >= ans,则一定有一种方案使得 ∑ |z[i]-z[j]| / ∑ dis(i,j) <= R. 化简得: ∑ (|z[i…

题目链接 \(Description\) 将n个村庄连成一棵树,村之间的距离为两村的欧几里得距离,村之间的花费为海拔z的差,求花费和与长度和的最小比值 \(Solution\) 二分,假设mid为可行的某一生成树的解,则应有 \((∑cost)/(∑dis) = mid\) 变形得 \(\sum(cost-mid*dis) = 0\) 取cost-mid*dis为边权,Prim求最小生成树(即尽可能满足mid) 若\(\sum(cost-mid*dis) > 0\),说明怎么也满足不了mid,m…

一个完全图,每两个点之间的cost是海拔差距的绝对值,长度是平面欧式距离, 让你找到一棵生成树,使得树边的的cost的和/距离的和,比例最小 然后就是最优比例生成树,也就是01规划裸题 看这一发:http://blog.csdn.net/sdj222555/article/details/7490797 #include<stdio.h> #include<algorithm> #include<math.h> #include<queue> #includ…

题意: 给定n个村子的坐标(x,y)和高度z, 求出修n-1条路连通所有村子, 并且让 修路花费/修路长度 最少的值 两个村子修一条路, 修路花费 = abs(高度差), 修路长度 = 欧氏距离 分析: 01分数划分的题目, 构造出 d[i] = 修路花费 - L * 修路长度, 这个L值我们可以二分(这道题看数据范围的话二分上限其实挺大的, 但其实上限取到100就可以过), 也可以用Dinkelbach迭代出来. 二分(1422ms) #include <stdio.h> #include…

POJ2728 无向图中对每条边i 有两个权值wi 和vi 求一个生成树使得 (w1+w2+...wn-1)/(v1+v2+...+vn-1)最小. 采用二分答案mid的思想. 将边的权值改为 wi-vi*mid. 对所有边求和后除以v 即为 (w1+w2+...wn-1)/(v1+v2+...+vn-1)-mid. 因此,若当前生成树的权值和为0,就找到了答案.否则更改二分上下界. #include<iostream> #include<cstdio> #include<c…

#include <iostream> #include <cstdio> #include <cmath> #include <cstdlib> #include <cstring> #include <iomanip> using namespace std; const int maxn=1005; const double eps=1e-6; const double inf=0xffffffff; struct node{…

题目: http://poj.org/problem?id=2728 题解: 二分比率,然后每条边边权变成w-mid*dis,用prim跑最小生成树就行 #include<cstdio> #include<algorithm> #include<cstring> #include<cmath> #define N 1005 using namespace std; int n,tot; double x[N],y[N],z[N],dis[N]; bool v…

Time Limit: 3000MSMemory Limit: 65536K Total Submissions: 27109Accepted: 7527 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…

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…

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…

Network Time Limit: 1000MS   Memory Limit: 30000K Total Submissions: 15268   Accepted: 5987   Special Judge Description Andrew is working as system administrator and is planning to establish a new network in his company. There will be N hubs in the c…

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 watered. As…

http://poj.org/problem?id=2728 Time Limit: 3000MS   Memory Limit: 65536K Total Submissions: 27191   Accepted: 7557 Description David the Great has just become the king of a desert country. To win the respect of his people, he decided to build channel…

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…

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…