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…

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…

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…

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…

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: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>…