Poj-2031 最小生成树+几何

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用最短距离使几个球联通，最小生成树，两点距离为几何知识。

#include<cstdio>#include<algorithm>#include<iostream>#include<cmath>using namespace std;struct edge{    int x,y;    double v;}e[20000];int fa[200],am[200];double ans;struct circle{     double x,y,z,dis;}c[200];int len,n;void init(){   for(int i=0;i<n;i++)   {    fa[i] = i;    am[i] = 1;   }}int cmp(edge a,edge b){    return a.v<b.v;}int toGetFarther(int a){    while(fa[a]!=a)        a = fa[a];    return a;}void unin(int a,int b,int z){   int faA = toGetFarther(a),faB = toGetFarther(b);   if(fa[faA]!=fa[faB])   {      if(am[faA]>am[faB]){am[faA]+=am[faB];fa[faB] = faA;}      else {am[faB]+=am[faA];fa[faA] = faB;}      ans+=e[z].v;   }}double Kruskal(){    ans = 0.0;  for(int i=0;i<len;i++)      unin(e[i].x,e[i].y,i);  return ans;}int main(){  while(scanf("%d",&n)&&n)  {      init();      len = 0;    for(int i=0;i<n;i++)      scanf("%lf%lf%lf%lf",&c[i].x,&c[i].y,&c[i].z,&c[i].dis);    for(int i=0;i<n;i++)      for(int j=i+1;j<n;j++)    {        e[len].x = i;        e[len].y = j;        double dist = sqrt(pow(c[i].x-c[j].x,2.0)+pow(c[i].y-c[j].y,2.0)+pow(c[i].z-c[j].z,2.0))-(c[i].dis+c[j].dis);        if(dist>=0)e[len].v = dist;        else e[len].v = 0.0;        len ++;    }    sort(e,e+len,cmp);    printf("%.3f\n",Kruskal());  }  return 0;}

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