poj 2728 最优比率生成树

思路：设sum(cost[i])/sum(dis[i])=r;那么要使r最小，也就是minsum(cost[i]-r*dis[i]);那么就以cost[i]-r*dis[i]为边权重新建边。当求和使得最小生成树的

sum(cost[i]-r*dis[i])==0时，这个r就是最优的。这个证明是01分数规划。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#define Maxn 1010
#define Maxm Maxn*Maxn
#define inf 1e16
#define eps 1e-6
using namespace std;
int vi[Maxn],n;
double dis[Maxn][Maxn],cost[Maxn][Maxn],benefit[Maxn][Maxn],far[Maxn];
struct Point{
    double x,y,z;
}p[Maxn];
void init()
{
    memset(dis,0,sizeof(dis));
    memset(vi,0,sizeof(vi));
    memset(cost,0,sizeof(cost));
    memset(far,0,sizeof(far));
}
double Dis(Point a,Point b)
{
    return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
}
double prime(double r)
{
    int i,j,temp;
    double ans,Max;
    memset(vi,0,sizeof(vi));
    ans=0;
    for(i=2;i<=n;i++)
        far[i]=inf;
    far[1]=0;
    for(i=1;i<=n;i++)
    {
        Max=inf;
        for(j=1;j<=n;j++)
        {
            if(!vi[j]&&far[j]<Max)
            {
                Max=far[j];
                temp=j;
            }
        }
        vi[temp]=1;
        ans+=Max;
       // dis[temp][j]-cost[temp][j]
        for(j=1;j<=n;j++)
        {
            if(!vi[j]&&far[j]>cost[temp][j]-r*dis[temp][j])
                far[j]=cost[temp][j]-r*dis[temp][j];
        }
    }
    return ans;
}
int main()
{
    int i,j,a,b,c;
    double Max;
    while(scanf("%d",&n)!=EOF,n)
    {
        init();
        Max=-inf;
        for(i=1;i<=n;i++)
        {
            scanf("%lf%lf%lf",&p[i].x,&p[i].y,&p[i].z);
        }
        for(i=1;i<n;i++)
        {
            for(j=i+1;j<=n;j++)
            {
                dis[j][i]=dis[i][j]=Dis(p[i],p[j]);
                Max=max(dis[i][j],Max);
                cost[i][j]=cost[j][i]=fabs(p[i].z-p[j].z);
            }
        }
        double l,r,mid;
        l=0,r=100000;
        double temp;
        while(r-l>eps)
        {
            mid=(l+r)/2;
            temp=prime(mid);
            if(temp>0)
                l=mid;
            else
                r=mid;
        }
        printf("%.3lf
",l);
    }
    return 0;
}