题意:N个村庄,给出每个村庄的坐标和海拔,两个村庄之间的距离就用两点之间距离公式来求即可,两个村庄之间修路的费用为海拔之差,构建一棵生成树,使得费用之和与距离之和的比值最小?
分析:01分数规划.设每条边的权值为\(high[i][j]-mid*dist[i][j]\).因为是稠密图,所以只能Prim跑最小生成树.
//#include<bits/stdc++.h>
#include<iostream>
#include<cstdio>
#include<cmath>
#include<algorithm>
using namespace std;
inline int read(){
int s=0,w=1;char ch=getchar();
while(ch<'0'||ch>'9'){if(ch=='-')w=-1;ch=getchar();}
while(ch>='0'&&ch<='9'){s=s*10+ch-'0';ch=getchar();}
return s*w;
}
const int N=1005;
int n,a[N],b[N],c[N],vis[N],path[N],high[N][N];
double Map[N][N],dis[N];
double eps=1e-6,dist[N][N];
inline double Prim(double mid){
int i,j;
for(i=1;i<=n;i++)
for(j=i+1;j<=n;j++)
Map[i][j]=Map[j][i]=high[i][j]-mid*dist[i][j];
for(i=1;i<=n;i++){
dis[i]=Map[1][i];
vis[i]=0;
path[i]=1;
}
dis[1]=0;vis[1]=1;
int k;
for(i=1;i<n;i++){
double tmp=1e9;
for(j=1;j<=n;j++)
if(!vis[j]&&tmp>dis[j]){
k=j;
tmp=dis[j];
}
vis[k]=1;
for(j=1;j<=n;j++)
if(!vis[j]&&dis[j]>Map[k][j]){
dis[j]=Map[k][j];
path[j]=k;
}
}
double c=0,d=0;
for(i=1;i<=n;i++){
c+=high[i][path[i]];
d+=dist[i][path[i]];
}
return c/d;
}
int main(){
while(1){
n=read();if(!n)return 0;
for(int i=1;i<=n;i++)
a[i]=read(),b[i]=read(),c[i]=read();
for(int i=1;i<=n;i++){
for(int j=1;j<i;j++){
dist[i][j]=dist[j][i]=sqrt(1.0*(a[i]-a[j])*(a[i]-a[j])+1.0*(b[i]-b[j])*(b[i]-b[j]));
high[i][j]=high[j][i]=abs(c[i]-c[j]);
}
}
double l=0,r=0;
while(1){
r=Prim(l);
if(fabs(l-r)<=eps)break;
l=r;
}
printf("%.3f\n",l);
}
return 0;
}