poj 2420(模拟退火)

A Star not a Tree?

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 6066		Accepted: 2853

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

在平面内找到一个点,到给出的n个点的距离最短.
题解:利用模拟退火找费马点。

#include <stdio.h>
#include <stdlib.h>
#include <math.h>
#include <iostream>
#define Lim 0.999999
using namespace std;
const double eps = 1e-; ///温度下限
const double delta = 0.98;
const double T = ; ///初始温度
const double INF = ;
const int N = ;
struct Point{
    double x,y;
}p[N];
int n;
int dir[][] = {{-,},{,},{,},{,-}};
double dis(Point a,Point b) ///距离
{
    return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
}
double getSum(Point p[],Point s){
    double ans = ;
    for(int i=;i<n;i++){
        ans += dis(p[i],s);
    }
    return ans;
}
double Search(Point p[]){
    Point s = p[]; ///随机一个点
    double res = INF;
    double t = T;
    while(t>eps){
        bool flag = true;
        while(flag){
            flag = false;
            for(int i=;i<;i++){
                Point next;
                next.x = s.x+dir[i][]*t;
                next.y = s.y+dir[i][]*t;
                double ans = getSum(p,next);
                if(ans<res){
                    res = ans;
                    s = next;
                    flag = true;
                }
            }
        }
        t = t*delta;
    }
    return res;
}
int main(){
    while(scanf("%d",&n)!=EOF){
        for(int i=;i<n;i++){
            scanf("%lf%lf",&p[i].x,&p[i].y);
        }
        printf("%.0lf\n",Search(p));
    }
}