这题数据量较大.普通的求MST是会超时的. d[i]=cost[i]-ans*dis[0][i] 据此二分. 但此题用Dinkelbach迭代更好 #include<cstdio> #include<cstring> #include<cmath> #include<iostream> #include<algorithm> using namespace std; #define N 1010 double mp[N][N],c[N][N],x…

题目:http://poj.org/problem?id=2728 第一道01分数规划题!(其实也蛮简单的) 这题也可以用迭代做(但是不会),这里用了二分: 由于比较裸,不作过多说明了. 代码如下: #include<iostream> #include<cstdio> #include<cstring> #include<cmath> #define eps 1e-6 using namespace std; int const inf=0x3f3f3f;…

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…

[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) &…

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…

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…

[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,…

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…

K Best Time Limit: 8000MS   Memory Limit: 65536K Total Submissions: 12812   Accepted: 3290 Case Time Limit: 2000MS   Special Judge Description Demy has n jewels. Each of her jewels has some value vi and weight wi. Since her husband John got broke aft…