程序師世界是廣大編程愛好者互助、分享、學習的平台,程序師世界有你更精彩!
首頁
編程語言
C語言|JAVA編程
Python編程
網頁編程
ASP編程|PHP編程
JSP編程
數據庫知識
MYSQL數據庫|SqlServer數據庫
Oracle數據庫|DB2數據庫
 程式師世界 >> 編程語言 >> C語言 >> C++ >> C++入門知識 >> POJ 2420 模擬退火法

POJ 2420 模擬退火法

編輯:C++入門知識

A Star not a Tree? Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 3272 Accepted: 1664

Description

Luke wants to upgrade his home computer network from 10mbs to 100mbs. His existing network uses 10base2 (coaxial) cables that allow you to connect any number of computers together in a linear arrangement. Luke is particulary proud that he solved a nasty NP-complete problem in order to minimize the total cable length.
Unfortunately, Luke cannot use his existing cabling. The 100mbs system uses 100baseT (twisted pair) cables. Each 100baseT cable connects only two devices: either two network cards or a network card and a hub. (A hub is an electronic device that interconnects several cables.) Luke has a choice: He can buy 2N-2 network cards and connect his N computers together by inserting one or more cards into each computer and connecting them all together. Or he can buy N network cards and a hub and connect each of his N computers to the hub. The first approach would require that Luke configure his operating system to forward network traffic. However, with the installation of Winux 2007.2, Luke discovered that network forwarding no longer worked. He couldn't figure out how to re-enable forwarding, and he had never heard of Prim or Kruskal, so he settled on the second approach: N network cards and a hub.

Luke lives in a loft and so is prepared to run the cables and place the hub anywhere. But he won't move his computers. He wants to minimize the total length of cable he must buy.

Input

The first line of input contains a positive integer N <= 100, the number of computers. N lines follow; each gives the (x,y) coordinates (in mm.) of a computer within the room. All coordinates are integers between 0 and 10,000.

Output

Output consists of one number, the total length of the cable segments, rounded to the nearest mm.

Sample Input

4
0 0
0 10000
10000 10000
10000 0

Sample Output

28284


在多邊形內找一個點,使得到多邊形各個點的距離和最小,顯然是費馬點,模擬退火法

代碼:

/* ***********************************************
Author :_rabbit
Created Time :2014/5/3 16:37:48
File Name :8.cpp
************************************************ */
#pragma comment(linker, "/STACK:102400000,102400000")
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
using namespace std;
#define INF 0x3f3f3f3f
#define eps 1e-8
#define pi acos(-1.0)
typedef long long ll;
struct Point{
	double x,y;
	Point(double _x=0,double _y=0){
		x=_x;y=_y;
	}
};
int dcmp(double x){
	if(fabs(x)>n){
		for(int i=1;i<=n;i++)scanf("%lf%lf",&pp[i].x,&pp[i].y);
		double ans=INF,step=0;
		Point u,v;
		u.x=u.y=v.x=v.y=0;
		for(int i=1;i<=n;i++){
			step+=fabs(pp[i].x)+fabs(pp[i].y);
			u.x+=pp[i].x;
			u.y+=pp[i].y;
		}
		u=u/n;
		for(;step>eps;step/=2){
			for(int i=-4;i<=4;i++)
				for(int j=-4;j<=4;j++){
					v.x=u.x+step*i;
					v.y=u.y+step*j;
					double cnt=0;
					for(int p=1;p<=n;p++)cnt+=Length(v-pp[p]);
					if(cnt

  1. 上一頁:
  2. 下一頁:
Copyright © 程式師世界 All Rights Reserved