POJ2728 最优比例生成树裸题

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <stack>
 
using namespace std;
#define N 1010
#define For(i,j,n) for(int i=j;i<=n;i++)
double dis[N][N] ,cost[N][N] ,n ,x[N] ,y[N] ,z[N] ,d[N] ,dist[N];
double eps = 1e-5;
bool vis[N];
template<class T> void read(T&num) {
    char CH; bool F=false;
    for(CH=getchar();CH<'0'||CH>'9';F= CH=='-',CH=getchar());
    for(num=0;CH>='0'&&CH<='9';num=num*10+CH-'0',CH=getchar());
    F && (num=-num);
}

double get_dis(int a, int b)
{
    return sqrt((x[b]-x[a])*(x[b]-x[a]) + (y[b]-y[a])*(y[b]-y[a]));
}

void clear()
{
    memset(vis , 0 ,sizeof(vis));

}

bool check(double a)
{
    clear();
    double sum = 0.0;
    For(i ,1 ,n)
        d[i] = 100000000;
    For(i ,1 ,n)
    {
        d[i] = cost[1][i] - dis[1][i]*a; 
    }
    double maxx;
    int k;
    vis[1] = 1;
    For(i ,2 ,n)
    {
        maxx = 100000000;
        For(j ,2 ,n)
            if(!vis[j]&&d[j] < maxx)
            {
                maxx = d[j];
                k = j;
            }
        vis[k] = 1;
        For(j ,2 ,n)
            if(!vis[j]&&cost[k][j]-a*dis[k][j]<d[j])
                d[j] = cost[k][j]-a*dis[k][j];
        sum+=d[k];
    }
    return sum>=0 ? true:false;
}

int main()
{
        while(1)
        {
            read(n);
            if(n==0)break;
            For(i,1,n)
            {
                scanf("%lf%lf%lf",&x[i],&y[i],&z[i]);
            }
            For(i,1,n)
            For(j,i+1,n)
            {
                dis[i][j]=dis[j][i]=get_dis(i,j);
                cost[i][j]=cost[j][i]=abs(z[i]-z[j]);
            }
            double l=0.0,r=100.0;
            while(r-l>=eps)
            {
                double mid=(l+r)/2;
                if(check(mid))l=mid;
                else r=mid;
            }
            printf("%.3f\n",r);
        }
        return 0;
}