POJ 2728（最优比率生成树+01规划）

                                                                                                Desert King

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 25729		Accepted: 7143

Description

David the Great has just become the king of a desert country. To win the respect of his people, he decided to build channels all over his country to bring water to every village. Villages which are connected to his capital village will be watered. As the dominate ruler and the symbol of wisdom in the country, he needs to build the channels in a most elegant way.

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

There
are several test cases. Each test case starts with a line containing a
number N (2 <= N <= 1000), which is the number of villages. Each
of the following N lines contains three integers, x, y and z (0 <= x,
y < 10000, 0 <= z < 10000000). (x, y) is the position of the
village and z is the altitude. The first village is the capital. A test
case with N = 0 ends the input, and should not be processed.

Output

For
each test case, output one line containing a decimal number, which is
the minimum ratio of overall cost of the channels to the total length.
This number should be rounded three digits after the decimal point.

Sample Input

4
0 0 0
0 1 1
1 1 2
1 0 3
0

Sample Output

1.000

Source

Beijing 2005

先贴个代码  明天找时间来补解释

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cctype>
#include<cmath>
#include<cstring>
#include<map>
#include<stack>
#include<set>
#include<vector>
#include<algorithm>
#include<string.h>
typedef long long ll;
typedef unsigned long long LL;
using namespace std;
const int INF=0x3f3f3f3f;
const double eps=0.0000000001;
const int N=+;
const int MAX=+;
int vis[N];
double x[N],y[N],z[N];
double w[MAX][MAX],v[MAX][MAX];
double low[N];
int n;
double fun(double a,double b,double c,double d){
    double ans=(a-c)*(a-c)+(b-d)*(b-d);
    return sqrt(ans);
}
int prime(double x){
    double sum=;
    memset(vis,,sizeof(vis));
    for(int i=;i<n;i++)low[i]=v[][i]-x*w[][i];
    vis[]=;
    for(int i=;i<n;i++){
        int k;
        double maxx=INF*1.0;
        for(int j=;j<n;j++)if(vis[j]==&&low[j]<maxx){
            maxx=low[j];
            k=j;
        }
        if(maxx==1.0*INF)break;
        sum=sum+maxx;
        vis[k]=;
        for(int j=;j<n;j++)
        if(vis[j]==&&low[j]>v[k][j]-x*w[k][j])
        low[j]=v[k][j]-x*w[k][j];
    }
    if(sum>)return ;
    else
        return ;
}
int main(){
    while(scanf("%d",&n)!=EOF){
        if(n==)break;
        double maxx=;
        double minn=INF*1.0;
        for(int i=;i<n;i++)scanf("%lf%lf%lf",&x[i],&y[i],&z[i]);
        for(int i=;i<n;i++)
        for(int j=i+;j<n;j++){
            double t=fun(x[i],y[i],x[j],y[j]);
            w[i][j]=w[j][i]=t;
            v[i][j]=v[j][i]=fabs(z[i]-z[j]);
            maxx=max(v[i][j],maxx);
            minn=min(minn,t);
        }
        double low=0.0;
        double high=maxx/minn;
        double ans;
        while(low+eps<high){
            double mid=(low+high)/;
            if(prime(mid)){
                ans=mid;
                low=mid;
            }
            else
                high=mid;
        }
        printf("%.3f\n",ans);
    }
}