Time Limit: 3000MS
Memory Limit: 65536K

Total Submissions: 27109
Accepted: 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 to every village. Villages which are connected to his capital village will be watered. As the dominate ruler and the symbol of wisdom in the country, he needs to build the channels in a most elegant way.
After days of study, he finally figured his plan out. He wanted the average cost of each mile of the channels to be minimized. In other words, the ratio of the overall cost of the channels to the total length must be minimized. He just needs to build the necessary channels to bring water to all the villages, which means there will be only one way to connect each village to the capital.
His engineers surveyed the country and recorded the position and altitude of each village. All the channels must go straight between two villages and be built horizontally. Since every two villages are at different altitudes, they concluded that each channel between two villages needed a vertical water lifter, which can lift water up or let water flow down. The length of the channel is the horizontal distance between the two villages. The cost of the channel is the height of the lifter. You should notice that each village is at a different altitude, and different channels can't share a lifter. Channels can intersect safely and no three villages are on the same line.
As King David's prime scientist and programmer, you are asked to find out the best solution to build the channels.

Input

There are several test cases. Each test case starts with a line containing a number N (2 <= N <= 1000), which is the number of villages. Each of the following N lines contains three integers, x, y and z (0 <= x, y < 10000, 0 <= z < 10000000). (x, y) is the position of the village and z is the altitude. The first village is the capital. A test case with N = 0 ends the input, and should not be processed.

Output

For each test case, output one line containing a decimal number, which is the minimum ratio of overall cost of the channels to the total length. This number should be rounded three digits after the decimal point.

Sample Input

4
0 0 0
0 1 1
1 1 2
1 0 3
0

Sample Output

1.000

Source

Beijing 2005

【翻译】题目大意:有n个村庄要连在一起,村与村之间的长度为他们之间的水平距离,连在一起的花费是两村的高度差。现求所花费和与长度和之比最小。

分析:

      ①Sigma的比值可以想到使用01分数规划。

      ②01分数规划有两种实现方式:

               (1)dichotomy[有点慢] (2)Dinkelbach[超级快]

      ③所以整个程序就是01分数规划然后使用Prim求出最小生成树。

      ④Prim很快,代码又好写。

#include<math.h>
#include<stdio.h>
#define go(i,a,b) for(int i=a;i<=b;i++)
#define G(i,j) (g[i][j].cost-r1*g[i][j].len)
const int N=1010;int vis[N];
double Sqr(double a){return a*a;};
double Abs(double a){return a<0?-a:a;}
struct E{double cost,len;}g[N][N],e[N];
int n,x[N],y[N],h[N],T=-1e9;double d[N],r1,r2;
double Dis(int i,int j){return sqrt(Sqr(x[i]-x[j])+Sqr(y[i]-y[j]));} void Prim()
{
double C=0,D=0;d[1]=1e15;int v;++T; go(u,2,n)d[u]=G(1,u),e[u]=g[1][u];go(i,2,n){v=1;
go(u,2,n)if(vis[u]!=T&&d[u]<d[v])v=u;vis[v]=T;C+=e[v].cost;
go(u,2,n)if(vis[u]!=T&&G(v,u)<d[u])d[u]=G(v,u),e[u]=g[v][u];D+=e[v].len;} r2=C/D;
} int main()
{
while(scanf("%d",&n),n)
{
go(i,1,n)scanf("%d%d%d",x+i,y+i,h+i);
go(i,1,n)go(j,1,n)g[i][j]=(E){Abs(h[i]-h[j]),Dis(i,j)};
r1=r2=0;while(Prim(),Abs(r1-r2)>1e-5)r1=r2;printf("%.3f\n",r1);
}
return 0;
}//Paul_Guderian

And every young mammal has multitudinous opportunities.————Judy·Hopps

【POJ 2728 Desert King】的更多相关文章

  1. poj 2728 Desert King (最小比例生成树)

    http://poj.org/problem?id=2728 Desert King Time Limit: 3000MS   Memory Limit: 65536K Total Submissio ...

  2. POJ 2728 Desert King 最优比率生成树

    Desert King Time Limit: 3000MS   Memory Limit: 65536K Total Submissions: 20978   Accepted: 5898 [Des ...

  3. poj 2728 Desert King (最优比率生成树)

    Desert King http://poj.org/problem?id=2728 Time Limit: 3000MS   Memory Limit: 65536K       Descripti ...

  4. POJ 2728 Desert King(最优比例生成树 二分 | Dinkelbach迭代法)

    Desert King Time Limit: 3000MS   Memory Limit: 65536K Total Submissions: 25310   Accepted: 7022 Desc ...

  5. POJ 2728 Desert King (01分数规划)

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

  6. POJ 2728 Desert King ★(01分数规划介绍 && 应用の最优比率生成树)

    [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 ...

  7. POJ 2728 Desert King

    Description David the Great has just become the king of a desert country. To win the respect of his ...

  8. POJ 2728 Desert King(最优比率生成树 01分数规划)

    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 ...

  9. POJ 2728 Desert King | 01分数规划

    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> ...

随机推荐

  1. 基于 win7下虚拟机的 GNSS-SDR安装过程

    最近在安装 GNSS-SDR软件时,遇到了很多问题,这里回顾了我的安装过程,罗列了所遇到的问题和解决办法.希望后来者不要再踩这些坑了! 首先,在官方文档中看到,GNSS-SDR目前并不支持直接在 Wi ...

  2. 连接MYSQL 错误代码2003

    问题是服务里面mysql没有启动或者mysql服务丢失 解决办法: 开始->运行->cmd,进到mysql安装的bin目录(以我的为例,我的安装在D盘)D:\MySQL\bin>my ...

  3. 多线程(threading module)

    一.线程与进程 线程定义:线程是操作系统能够进行运算调度的最小单位.它被包含在进程之中,是进程中的实际运作单位.一条线程指的是进程中一个单一顺序的控制流,一个进程中可以并发多个线程,每条线程并行执行不 ...

  4. (转)在图像处理中,散度 div 具体的作用是什么?

    出处http://www.zhihu.com/question/24591127 按:今天看到这篇文章,有点感慨,散度这个概念我初次接触到至少应该是在1998年,时隔这么多年后看到这篇文章,真的 佩服 ...

  5. 使用PHP生成分享图片

    小程序导航 wq.xmaht.top 假设代码中用到的资源文件夹在当前code_png目录下: /** * 分享图片生成 * @param $gData 商品数据,array * @param $co ...

  6. PPT入门学习笔记1:待修改

    一直被比人忽悠实在是累了,我可以接受自己的失误,但我接受不了别人一次又一次的坑我! 做PPT的原则是什么? 1.一个目标: "一个PPT只为一类人服务,针对不同的听众制作不同层次内容的PPT ...

  7. [Hdu4825]Xor Sum(01字典树)

    Description Zeus 和 Prometheus 做了一个游戏,Prometheus 给 Zeus 一个集合,集合中包含了N个正整数,随后 Prometheus 将向 Zeus 发起M次询问 ...

  8. sql查询题目

    --1.查询在1981年入职的员工信息select * from emp where hiredate between '01-1月-1981'and '31-12月-1981'; select * ...

  9. 一、MySQL数据库之简介和安装

    一.基础部分 1.数据库是简介     之前所学,数据要永久保存,比如用户注册的用户信息,都是保存于文件中,而文件只能存在于某一台机器上. 如果我们不考虑从文件中读取数据的效率问题,并且假设我们的程序 ...

  10. 11.1,nginx集群概念

    集群介绍 为什么要用集群