poj2728 Desert King【最优比率生成树】【Prim】【0/1分数规划】
Time Limit: 3000MS | Memory Limit: 65536K | |
Total Submissions:31622 | Accepted: 8670 |
Description
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
Output
Sample Input
4
0 0 0
0 1 1
1 1 2
1 0 3
0
Sample Output
1.000
Source
题意:
有$n$个城市,每个城市有对应的坐标和海拔。两个城市之间的距离是坐标的距离,路的花费是海拔之差。
现在要建$n-1$条路,使得花费之和比距离之和最小。
思路:
一道0/1分数规划的题目。
0/1分数规划模型指:给定整数$a_1,a_2,...,a_n$,以及$b_1,b_2,...,b_n$,求一组解$x_i(1\leq i \leq n,x_i=0或1)$,
使得$\frac{\sum_{i=1}^{n} a_i*x_i}{\sum_{i=1}^{n}b_i*x_i}$最大化。
这种类型的题目我们可以通过二分答案来做。
因为如果对于$mid$,存在一组解,使得$\sum_{i=1}^{n}(a_i - mid * b_i) * x_i \geq 0$那么我们可以变形得到
存在一组解,使得$\frac{\sum_{i=1}^{n} a_i*x_i}{\sum_{i=1}^{n}b_i*x_i} \geq mid$
也就是说,$mid$比我们实际的答案要小。
反之,如果对于任意一组解,都有$\sum_{i=1}^{n}(a_i - mid * b_i) * x_i <0$那么我们可以得到
任意的解都有$\frac{\sum_{i=1}^{n} a_i*x_i}{\sum_{i=1}^{n}b_i*x_i} < mid$
也就是说,$mid$比我们实际的答案要大。
所以我们每次只需要计算$\sum_{i=1}^{n}(a_i - mid * b_i)*x_i$的最大值,如果最大值非负,令$st = mid$,否则$ed = mid$
对于这道题,也是类似。我们二分最终的答案。
每一次重新建图,城市和城市之间的边变为$cost - mid * length$
然后在新图上跑最小生成树。
如果最小生成树的值非负,说明实际答案比$mid$要大,令$st = mid$
由于这道题是完全图,而点的个数只有$1000$所以用prim比较好
WA了好久,后来发现是对d赋初值的问题。
对于double的数组赋初值$+\infty$不能用$0x3f$而应该用$0x7f$,就改了这里就过了。
#include<iostream>
//#include<bits/stdc++.h>
#include<cstdio>
#include<cmath>
//#include<cstdlib>
#include<cstring>
#include<algorithm>
//#include<queue>
#include<vector>
//#include<set>
//#include<climits>
//#include<map>
using namespace std;
typedef long long LL;
#define N 100010
#define pi 3.1415926535
#define inf 0x3f3f3f3f const int maxn = ;
const double eps = 1e-;
int n;
struct city{
double x, y, height;
}c[maxn];
double dist[maxn][maxn], cost[maxn][maxn], g[maxn][maxn]; double get_dist(int i, int j)
{
return sqrt((c[i].x - c[j].x) * (c[i].x - c[j].x) + (c[i].y - c[j].y) * (c[i].y - c[j].y));
} double d[maxn];
bool vis[maxn];
void prim()
{
memset(d, 0x7f, sizeof(d));
memset(vis, , sizeof(vis));
/*for(int i = 1; i <= n; i++){
d[i] = (double)inf;
vis[i] = 0;
}*/
d[] = ;
for(int i = ; i <= n; i++){
int x = ;
for(int j = ; j <= n; j++){
if(!vis[j] && (x == || d[j] < d[x]))x = j;
}
vis[x] = ;
for(int y = ; y <= n; y++){
if(!vis[y])d[y] = min(d[y], g[x][y]);
}
}
} void getgraph(double mid)
{
for(int i = ; i <= n; i++){
for(int j = i; j <= n; j++){
g[i][j] = g[j][i] = cost[i][j] - mid * dist[i][j];
}
}
} bool check(double mid)
{
getgraph(mid);
prim();
double res = ;
for(int i = ; i <= n; i++){
res += d[i];
}
return res >= ;
} int main()
{
while(scanf("%d", &n) != EOF && n){
for(int i = ; i <= n; i++){
scanf("%lf%lf%lf", &c[i].x, &c[i].y, &c[i].height);
} for(int i = ; i <= n; i++){
for(int j = i; j <= n; j++){
dist[i][j] = dist[j][i] = get_dist(i, j);
cost[i][j] = cost[j][i] = fabs(c[i].height - c[j].height);
}
}
double st = 0.0, ed = 100.0;
while(ed - st >= eps){
double mid = (st + ed) / ;
if(check(mid))st = mid;
else ed = mid;
}
printf("%.3f\n", ed);
} return ;
}
poj2728 Desert King【最优比率生成树】【Prim】【0/1分数规划】的更多相关文章
- POJ2728 Desert King 最优比率生成树
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/ ...
- POJ2728 Desert King —— 最优比率生成树 二分法
题目链接:http://poj.org/problem?id=2728 Desert King Time Limit: 3000MS Memory Limit: 65536K Total Subm ...
- POJ.2728.Desert King(最优比率生成树 Prim 01分数规划 二分/Dinkelbach迭代)
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 ...
- 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 ...
- 【POJ2728】Desert King 最优比率生成树
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 ...
- Desert King(最优比率生成树)
Desert King Time Limit: 3000MS Memory Limit: 65536K Total Submissions: 22717 Accepted: 6374 Desc ...
- POJ 2728 Desert King(最优比率生成树, 01分数规划)
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 ...
- POJ 2728 Desert King (最优比率树)
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 ...
- poj-2728Desert King(最优比率生成树)
David the Great has just become the king of a desert country. To win the respect of his people, he d ...
- POJ 2728 Desert King (最优比例生成树)
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- ...
随机推荐
- ASP.NET MVC ViewBag/ViewData/TempData区别
ViewBag/ViewData public dynamic ViewBag { get; } public ViewDataDictionary ViewData { get; set; } Vi ...
- 3D建模软件的选择(UG,Solidworks,ProE)
转自:3D建模软件的选择(UG,Solidworks,ProE) 自述 咱是一个码农,和web.软件.控制台打交道太多了,很想玩玩炫的东西,于是学了点点PS,结果发现完全没有美术细胞TT.最近有碰到对 ...
- IOS开发系列之阿堂教程:玩转IPhone客户端和Web服务端交互(客户端)实践
说到ios的应用开发,我们不能不提到web server服务端,如果没有服务端的支持,ios应用开发就没有多大意义了,因为从事过手机开发的朋友都知道(Android也一样),大量复杂业务的处理和数据库 ...
- ESXi创建磁盘命令
[root@esx421 SAN]# vmkfstools -d thick -a lsilogic -c 10G lun00.vmdk Incorrect disk option "thi ...
- Python3多线程之间的执行顺序问题
[本文出自天外归云的博客园] 一个多线程的题:定义三个线程ID分别为ABC,每个线程打印10遍自己的线程ID,按ABCABC……的顺序进行打印输出. 我的解法: from threading impo ...
- PentesterLab渗透演练平台
转载自: https://www.blackh4t.org/archives/1143.html http://www.91ri.org/5958.html 1. 什么是WebApp Pen ...
- ADO.NET 数据库备份等操作
public class SqlServerBackup { private string database; private string server; private string uid; p ...
- cp显示进度条
cp显示进度条 alias cp='rsync -av --progress'
- Oracle分析函数-排序排列(rank、dense_rank、row_number、ntile)
(1)rank函数返回一个唯一的值,除非遇到相同的数据时,此时所有相同数据的排名是一样的,同时会在最后一条相同记录和下一条不同记录的排名之间空出排名. (2)dense_rank函数返回一个唯一的值, ...
- c++ 动态判断基类指针指向的子类类型(typeid)
我们在程序中定义了一个基类,该基类有n个子类,为了方便,我们经常定义一个基类的指针数组,数组中的每一项指向都指向一个子类,那么在程序中我们如何判断这些基类指针是指向哪个子类呢? 本文提供了两种方法 ( ...