POJ 2420：A Star not a Tree?__PKU

原文連結：https://www.dreamwings.cn/poj2420/2838.html
A Star not a Tree?

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 5788		Accepted: 2730

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

40 00 1000010000 1000010000 0

Sample Output

28284

題意：

給出平面內一個點集，讓你求平面內一點距離這些點距離和的最小值~

好像退貨類比問題哎~

可惜我不會

只能用爬山演算法做了，爬山演算法講的是首先在平面內選一點，判斷它臨近的點所求得的距離和是否比它小，若小，更新點的座標

就這樣，不過它也存在很大的局限性，如在高峰處無法選擇下一點應該去哪兒，只是局部最優而已。

我們可以選擇自己控制迴圈次數，類似於隨機數之類的，然後根據大資料的判斷比較，然後得出最終答案

AC代碼：

#include<stdio.h>#include<math.h>#include<algorithm>#include<iostream>#define eps 1e-6using namespace std;struct point{    double x;    double y;    void input()                //輸入    {        cin>>x>>y;    }    double dis(point k)         //求兩點之間距離    {        return sqrt((x-k.x)*(x-k.x)+(y-k.y)*(y-k.y));    }} a[105];double getsum(point k,int n)    //求距離和{    double s=0;    for(int i=0; i<n; i++)        s+=k.dis(a[i]);    return s;}int main(){    int n;    while(cin>>n)    {        point one;        one.x=one.y=0;        for(int i=0; i<n; ++i)        {            a[i].input();            one.x+=a[i].x;            one.y+=a[i].y;        }        one.x/=n;        one.y/=n;           //選擇一個靠中的點        double ans=getsum(one,n);        double t=10000;        while(t>eps)        //迴圈次數        {            double x=0,y=0;            for(int i=0; i<n; ++i)  //選取臨近點            {                x+=(a[i].x-one.x)/one.dis(a[i]);                y+=(a[i].y-one.y)/one.dis(a[i]);            }            point ttm;            ttm.x=one.x+x*t;            ttm.y=one.y+y*t;            double tmp=getsum(ttm,n);            if(tmp<ans)         //判斷大小            {                ans=tmp;                one.x+=x*t;                one.y+=y*t;            }            t*=0.99;        }        printf("%.0f\n",ans);    }    return 0;}