poj 2728 最优比例生成树（01分数规划）模板

/*
迭代法 ：204Ms
*/
#include<stdio.h>
#include<string.h>
#include<math.h>
#define N  1100
#define eps 1e-10
#define inf  0x3fffffff
struct node {
 int u,v,w;
}p[N];
double ma[N][N];
double distance(int i,int j) {
return sqrt(1.0*(p[i].u-p[j].u)*(p[i].u-p[j].u)+1.0*(p[i].v-p[j].v)*(p[i].v-p[j].v));
}
int n;
double prime(int u,double r) {
  int i,j,vis[N],pre[N];
  double dis[N],len=0,cost=0,total=0;
  for(i=1;i<=n;i++)  {
      dis[i]=fabs(1.0*p[u].w-1.0*p[i].w)-ma[u][i]*r;
      pre[i]=u;
  }
  memset(vis,0,sizeof(vis));
  vis[1]=1;
  for(i=1;i<n;i++) {
    double  minn=inf;
    int index=-1;
    for(j=1;j<=n;j++)
        if(!vis[j]&&minn>dis[j]) {
            minn=dis[j];
            index=j;
        }
    if(index!=-1) {
        vis[index]=1;
        len+=ma[pre[index]][index];
        cost=cost+fabs(1.0*p[pre[index]].w-1.0*p[index].w);
        //total+=dis[index];
        for(j=1;j<=n;j++) {
            double f=fabs(1.0*p[index].w-1.0*p[j].w)-ma[index][j]*r;
            if(!vis[j]&&f<dis[j]) {
                dis[j]=f;
                pre[j]=index;
            }
        }
    }
  }
 // return total;
 return cost/len;
}
int main() {
    int i,j;
    while(scanf("%d",&n),n) {
        for(i=1;i<=n;i++)
            scanf("%d%d%d",&p[i].u,&p[i].v,&p[i].w);
        for(i=1;i<=n-1;i++)
            for(j=i+1;j<=n;j++)
            ma[i][j]=ma[j][i]=distance(i,j);
          /*  double l=0.0,r=100.0;
            while(r-l>eps) {
                double mid=(l+r)/2;
                if(prime(1,mid)>=0)l=mid;
                else  r=mid;
            }*/
            double a=0,b;
            while(1) {
                b=prime(1,a);
                if(fabs(a-b)<eps)break;
                a=b;
            }
        //printf("%.3f\n",r);
        printf("%.3f\n",b);
    }
return 0;
}

/*
二分法：1766ms
*/
#include<stdio.h>
#include<string.h>
#include<math.h>
#define N  1100
#define eps 1e-10
#define inf  0x3fffffff
struct node {
 int u,v,w;
}p[N];
double ma[N][N];
double distance(int i,int j) {
return sqrt(1.0*(p[i].u-p[j].u)*(p[i].u-p[j].u)+1.0*(p[i].v-p[j].v)*(p[i].v-p[j].v));
}
int n;
double prime(int u,double r) {
  int i,j,vis[N],pre[N];
  double dis[N],len=0,cost=0,total=0;
  for(i=1;i<=n;i++)  {
      dis[i]=fabs(1.0*p[u].w-1.0*p[i].w)-ma[u][i]*r;
      pre[i]=u;
  }
  memset(vis,0,sizeof(vis));
  vis[1]=1;
  for(i=1;i<n;i++) {
    double  minn=inf;
    int index=-1;
    for(j=1;j<=n;j++)
        if(!vis[j]&&minn>dis[j]) {
            minn=dis[j];
            index=j;
        }
    if(index!=-1) {
        vis[index]=1;
       // len+=ma[pre[index]][index];
        //cost=cost+fabs(1.0*p[pre[index]].w-1.0*p[index].w);
        total+=dis[index];
        for(j=1;j<=n;j++) {
            double f=fabs(1.0*p[index].w-1.0*p[j].w)-ma[index][j]*r;
            if(!vis[j]&&f<dis[j]) {
                dis[j]=f;
                pre[j]=index;
            }
        }
    }
  }
  return total;
// return cost/len;
}
int main() {
    int i,j;
    while(scanf("%d",&n),n) {
        for(i=1;i<=n;i++)
            scanf("%d%d%d",&p[i].u,&p[i].v,&p[i].w);
        for(i=1;i<=n-1;i++)
            for(j=i+1;j<=n;j++)
            ma[i][j]=ma[j][i]=distance(i,j);
            double l=0.0,r=100.0;
            while(r-l>eps) {
                double mid=(l+r)/2;
                if(prime(1,mid)>=0)l=mid;
                else  r=mid;
            }
            /*double a=0,b;
            while(1) {
                b=prime(1,a);
                if(fabs(a-b)<eps)break;
                a=b;
            }*/
        printf("%.3f\n",r);
        //printf("%.3f\n",b);
    }
return 0;
}