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…

题目 http://poj.org/problem?id=2728 关键词:0/1分数规划,参数搜索,二分法,dinkelbach 参考资料:http://hi.baidu.com/zzningxp/item/28aa46e0fd86bdc2bbf37d03 http://hi.baidu.com/zheng6822/item/b31fbe9d5ae17536336eeb8f #include<stdio.h> #include<string.h> #include<iost…

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…

题目大意:给定一个 N 个点的无向完全图,边有两个不同性质的边权,求该无向图的一棵最优比例生成树,使得性质为 A 的边权和比性质为 B 的边权和最小. 题解:要求的答案可以看成是 0-1 分数规划问题,即:选定一个数 mid,每次重新构建边权为 \(a[i]-mid*b[i]\) 的图,再在图上跑一遍最小生成树(这里由于是完全图,应该采用 Prim 算法)判断最小值和给定判定的最小值的关系即可,这里为:若最小值大于 mid,则下界提高,否则上界下降. 代码如下 #include<cmath>…

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…

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…

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…

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…

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…

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…