POJ 2420 A Star not a Tree?(計算幾何 多邊形的費馬點 類比退火)

A Star not a Tree? Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 2991   Accepted: 1533 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 Source Waterloo Local 2002.01.26

這個題目也是wa的我好傷心啊，原因就是類比退火的步長和精度沒控制好，嘗試好多次才AC

不過覺得這個題目用到的類比退火的思想很好，可以設定精度讓我們的結果無限接近真正的答案

思想就是這樣，開始選擇一個初始點作為開始，選擇一個步長度，然後走四個方向，選擇一個

離正確答案最近的點，同時步長減半，接著類比，直到步長小到一定程度就好了！

這個思想很精妙，也容易想到和實現，不過沒有OJ測試精度不好保證啊！

#include <iostream>#include <stdio.h>#include <string.h>#include <cmath>#include <algorithm>using namespace std;#define inf 1e-5struct point{double x;double y;}po[150];int n;double dis(point &a,point &b){return sqrt((a.x-b.x)*(a.x-b.x) + (a.y-b.y)*(a.y-b.y));}int find_ans(){point pp,ppp;point rec;pp=po[0];double step=1000000;int i,j,k;double d=0;double ans;for(i=0;i<n;i++)ans+=dis(pp,po[i]);rec=pp;while(step > 0.2){pp=rec;d=0;ppp=pp;ppp.x+=step;for(i=0;i<n;i++){d+=dis(ppp,po[i]);}if(d < ans){ans=d;rec=ppp;}ppp=pp;d=0;ppp.x-=step;for(i=0;i<n;i++){d+=dis(ppp,po[i]);}if(d < ans){ans=d;rec=ppp;}ppp=pp;d=0;ppp.y-=step;for(i=0;i<n;i++){d+=dis(ppp,po[i]);}if(d < ans){ans=d;rec=ppp;}ppp=pp;d=0;ppp.y+=step;for(i=0;i<n;i++){d+=dis(ppp,po[i]);}if(d < ans){ans=d;rec=ppp;}step/=2;}if(ans-int(ans) >=0.5)printf("%d\n",int(ans)+1);elseprintf("%d\n",int(ans));return 0;}int main(){int i,j,k;while(scanf("%d",&n)!=EOF){for(i=0;i<n;i++)scanf("%lf%lf",&po[i].x,&po[i].y);find_ans();}return 0;}