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POJ 2420 模拟退火法

2024-07-02 17:43:12   235人阅读

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


在多边形内找一个点，使得到多边形各个点的距离和最小，显然是费马点，模拟退火法

代码：

/* ***********************************************
Author :_rabbit
Created Time :2014/5/3 16:37:48
File Name :8.cpp
************************************************ */
#pragma comment(linker, "/STACK:102400000,102400000")
#include <stdio.h>
#include <iostream>
#include <algorithm>
#include <sstream>
#include <stdlib.h>
#include <string.h>
#include <limits.h>
#include <string>
#include <time.h>
#include <math.h>
#include <queue>
#include <stack>
#include <set>
#include <map>
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)<eps)return 0;
	return x<0?-1:1;
}
Point operator + (Point a,Point b){
	return Point(a.x+b.x,a.y+b.y);
}
Point operator - (Point a,Point b){
	return Point(a.x-b.x,a.y-b.y);
}
Point operator * (Point a,double p){
	return Point(a.x*p,a.y*p);
}
Point operator / (Point a,double p){
	return Point(a.x/p,a.y/p);
}
bool operator <(const Point &a,const Point &b){
	return a.x<b.x||(a.x==b.x&&a.y<b.y);
}
bool operator == (const Point &a,const Point  &b){
	return dcmp(a.x-b.x)==0&&dcmp(a.y-b.y)==0;
}
double Dot(Point a,Point b){
	return a.x*b.x+a.y*b.y;
}
double Length(Point a){
	return sqrt(Dot(a,a));
}
double Angle(Point a,Point b){
	return acos(Dot(a,b)/Length(a)/Length(b));
}
double angle(Point a){
	return atan2(a.y,a.x);
}
double Cross(Point a,Point b){
	return a.x*b.y-a.y*b.x;
}
Point Rotate(Point a,double rad){
	return Point(a.x*cos(rad)-a.y*sin(rad),a.x*sin(rad)+a.y*cos(rad));
}
Point GetLineIntersection(Point p,Point v,Point q,Point w){
	Point u=p-q;
	double t=Cross(w,u)/Cross(v,w);
	return p+v*t;
}
Point perpencenter(Point a,Point b,Point c){
	Point A,B,C,D;
	A=c;
	B.x=c.x-a.y+b.y;
	B.y=c.y+a.x-b.x;
	C=b;
	D.x=b.x-a.y+c.y;
	D.y=b.y+a.x-c.x;
	return GetLineIntersection(A,B-A,C,D-C);
}
Point pp[110];
int main(){
	int n;
	while(cin>>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<ans){
						ans=cnt;
						u=v;
					}
				}
		}
		int hh;
		double ss=ans-(int)ans;
		if(ss<0.5)hh=(int)ans;else hh=(int)ans+1;
		printf("%d\n",hh);
	}
}



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