[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\({\sum C\over \sum D}<L\)…

[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,…

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…

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>…

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…

数据结构/可并堆 啊……换换脑子就看了看数据结构……看了一下左偏树和斜堆,鉴于左偏树不像斜堆可能退化就写了个左偏树. 左偏树介绍:http://www.cnblogs.com/crazyac/articles/1970176.html 体会:合并操作是可并堆的核心操作(就像LCT里的access),进堆和弹堆顶都是直接调用合并操作实现的. 而合并的实现是一个递归的过程:将小堆与大堆的右儿子合并(这里的大小指的是堆顶元素的大小),直到某个为0.(是不是有点启发式合并的感觉……) 在合并的过程中要维…

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>…

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…

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…

[算法]状态压缩型DP [题解]http://www.cnblogs.com/xtx1999/p/4620227.html (orz) https://www.cnblogs.com/zbtrs/p/6189240.html dp[i][j][k]为前i行已经放了j个国王并且第i行的状态为k(二进制)的方案数. 状态左移右移预处理两行合法. dp[i][j][k] = Σdp[i-1][j - num[k]][p] #include<cstdio> #include<algorithm&…