最优比率生成树 POJ 2728 迭代或者二分

别人解题报告的链接：

http://blog.sina.com.cn/s/blog_691190870101626q.html

说明一下关于精度的问题，当结果是精确到小数点后3为，你自然要把误差定为至少10^(-4),我定的是10^(-8)````这里多定点没事的···

然后对于POJ上的提交，如果是用C++提交，可以写printf("%.3lf\n",ans);

但是如果是用的G++提交，就得用printf("%.3f\n",ans);

当然ans定义的是double型的·····

补充的解题报告链接：

http://blog.csdn.net/sdj222555/article/details/7490797

 //#define debug
 #include <cstdio>
 #include <cmath>
 #include <cstring>
 using namespace std;
 #define N 1005
 #define eps 1e-8
 #define INF 1e300
 struct point
 {
     int x,y,z;
 } p[N];
 double dis(point a,point b)
 {
     double t = (a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y);
     return sqrt(t);
 }
 struct arc
 {
     double len,cost;
 } edge[N][N];
 double lowcost[N];
 int pre[N];
 bool vis[N];
 int n;
 double prim(double d)
 {
     double tl =,tc=,ra;
     memset(vis,,sizeof(vis));
     vis[] = ;
     for(int i=; i<n; ++i)
     {
         lowcost[i] = edge[][i].cost-d*edge[][i].len;
         pre[i] = ;
     }
     for(int k=; k<n; ++k)
     {
         double mi = INF;
         int v;
         for(int i=; i<n; ++i)
         {
             if(!vis[i] && lowcost[i] < mi)
             {
                 mi = lowcost[i];
                 v = i;
             }
         }
         vis[v] = ;
         tc += edge[v][pre[v]].cost;
         tl += edge[v][pre[v]].len;
         for(int i=; i<n; ++i)
         {
             if(!vis[i] && lowcost[i] > edge[v][i].cost-d*edge[v][i].len)
             {
                 lowcost[i] = edge[v][i].cost-d*edge[v][i].len;
                 pre[i] = v;
             }
         }
     }
     ra = tc/tl;
     return ra;
 }
 int main()
 {
 #ifdef debug
     freopen("in.c","r",stdin);
 #endif
     while(scanf("%d",&n),n)
     {
         for(int i=; i<n; ++i)
             scanf("%d%d%d",&p[i].x,&p[i].y,&p[i].z);
         for(int i=; i<n; ++i)
         {
             for(int j=i+; j<n; ++j)
             {
                 edge[i][j].len = edge[j][i].len = dis(p[i],p[j]);
                 edge[i][j].cost =  edge[j][i].cost = fabs(p[i].z-p[j].z);
             }
         }
         double r=;
         while(true)
         {
             double  t = prim(r);
             if(fabs(t-r) < eps) break;
             r = t;
         }
         printf("%.3f\n",r);
     }
     return ;
 }