After days of study, he finally figured his plan out. He wanted the average cost of each mile of the channels to be minimized. In other words, the ratio of the overall cost of the channels to the total length must be minimized. He just needs to build the necessary channels to bring water to all the villages, which means there will be only one way to connect each village to the capital.
His engineers surveyed the country and recorded the position and altitude of each village. All the channels must go straight between two villages and be built horizontally. Since every two villages are at different altitudes, they concluded that each channel between two villages needed a vertical water lifter, which can lift water up or let water flow down. The length of the channel is the horizontal distance between the two villages. The cost of the channel is the height of the lifter. You should notice that each village is at a different altitude, and different channels can't share a lifter. Channels can intersect safely and no three villages are on the same line.
As King David's prime scientist and programmer, you are asked to find out the best solution to build the channels.
Input
Output
Sample Input
4
0 0 0
0 1 1
1 1 2
1 0 3
0
Sample Output
1.000 题意：在这么一个图中求一棵生成树,这棵树点权和边权之比最大是多少？
#include<cstdio>
#include<cstring>
#include<cmath>
#include<iostream>
#include<algorithm>
#include<queue>
#define N 1007
#define eps 0.000001
using namespace std; int n;
double dis[N][N],h[N][N],line[N],p[N][N];
bool vis[N];
struct Node
{
double x,y,h;
}a[N]; double get_dis(int x,int y)
{return sqrt((a[x].x-a[y].x)*(a[x].x-a[y].x)+(a[x].y-a[y].y)*(a[x].y-a[y].y));}
/*struct cmp
{
bool operator()(int x,int y)
{return line[x]>line[y];}
};*/
double prim(double num)
{
for (int i=;i<=n;i++)
for (int j=;j<=n;j++)
p[i][j]=h[i][j]-dis[i][j]*num;
//priority_queue<int,vector<int>,cmp>q;
//while(!q.empty()) q.pop();
memset(vis,,sizeof(vis));
memset(line,,sizeof(line));
line[]=;
//for (int i=1;i<=n;i++) q.push(i);
for (int i=;i<=n;i++)
{
int u=;
for (int j=;j<=n;j++)
if (!vis[j]&&line[j]<line[u]) u=j;
vis[u]=;
for (int j=;j<=n;j++)
if (!vis[j]) line[j]=min(line[j],p[u][j]);
}
double sum=;
for (int i=;i<=n;i++)
sum+=line[i];
return sum;
}
int main()
{
while(~scanf("%d",&n)&&n)
{
for (int i=;i<=n;i++)
scanf("%lf%lf%lf",&a[i].x,&a[i].y,&a[i].h);
for (int i=;i<=n;i++)
for (int j=;j<=n;j++)
{
dis[i][j]=get_dis(i,j);
h[i][j]=fabs(a[i].h-a[j].h);
}
double l=0.0,r=100.0;
while(r-l>=eps)
{
double mid=(l+r)*1.0/;
if (prim(mid)>=) l=mid;
else r=mid;
}
printf("%.3f\n",l);
}
}