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Time Limit: 20 Sec
Memory Limit: 256 MB
题目连接
http://acm.hust.edu.cn/vjudge/contest/view.action?cid=88808#problem/B
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
HINT
题意
平面上有100个点，让你找到一个点，使得这些点到这个点的距离和最小
就是让你找广义费马点
题解：
爬山算法
有一个蠢萌兔子，喝醉了，四处走，然后清醒了，就滚回去了，就是这样……
代码：
//qscqesze #include <cstdio> #include <cmath> #include <cstring> #include <ctime> #include <iostream> #include <algorithm> #include <set> #include <bitset> #include <vector> #include <sstream> #include <queue> #include <typeinfo> #include <fstream> #include <map> #include <stack> typedef long long ll; using namespace std; //freopen("D.in","r",stdin); //freopen("D.out","w",stdout); #define sspeed ios_base::sync_with_stdio(0);cin.tie(0) #define maxn 200051 #define mod 10007 #define eps 1e-9 int Num; //const int inf=0x7fffffff; //нчоч╢С const int inf=0x3f3f3f3f; inline ll read() { ll x=0,f=1;char ch=getchar(); while(ch<‘0‘||ch>‘9‘){if(ch==‘-‘)f=-1;ch=getchar();} while(ch>=‘0‘&&ch<=‘9‘){x=x*10+ch-‘0‘;ch=getchar();} return x*f; } //************************************************************************************** int n; struct node { double x,y; }; double sqr(double x) { return x*x; } double dis(double x,double y,node p) { return sqrt(sqr(x-p.x)+sqr(y-p.y)); } node p[105]; double get_sum(double x,double y) { double ans=0; for(int i=0;i<n;i++) { ans+=dis(x,y,p[i]); } return ans; } int main() { n=read(); for(int i=0;i<n;i++) scanf("%lf%lf",&p[i].x,&p[i].y); node ans; ans.x=0,ans.y=0; for(int i=0;i<n;i++) ans.x+=p[i].x,ans.y+=p[i].y; ans.x=(ans.x/(n*1.0)),ans.y=(ans.y/(n*1.0)); double Ans = get_sum(ans.x,ans.y); double t = 10000; double x,y,tmp; while(t>0.02) { x=0,y=0; for(int i=0;i<n;i++) { x+=(p[i].x-ans.x)/dis(ans.x,ans.y,p[i]); y+=(p[i].y-ans.y)/dis(ans.x,ans.y,p[i]); } tmp = get_sum(ans.x+x*t,ans.y+y*t); if(tmp<Ans) { Ans = tmp; ans.x+=x*t; ans.y+=y*t; } t*=0.9; } printf("%.0lf\n",Ans); }
POJ 2420 A Star not a Tree? 爬山算法
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原文地址：http://www.cnblogs.com/qscqesze/p/4761340.html
