題目:

http://poj.org/problem?id=2728

題解:

二分比率,然後每條邊邊權變成w-mid*dis,用prim跑最小生成樹就行

#include<cstdio>
#include<algorithm>
#include<cstring>
#include<cmath>
#define N 1005
using namespace std;
int n,tot;
double x[N],y[N],z[N],dis[N];
bool vis[N];
double mul(double x) {return x*x;}
double
 
 dist(int a,int b)
{
    return sqrt(mul(x[a]-x[b])+mul(y[a]-y[b]));
}
bool check(double mid)
{
    memset(vis,0,sizeof(vis));
    for (int i=2;i<=n;i++) dis[i]=fabs(z[1]-z[i])-mid*dist(1,i);
    vis[1]=1;
    int tot=n-1;
    int id=-1;
    double val=0.0,tmp=0.0;
    while(tot--)
    {
    id=-1;
    
 
for (int i=1;i<=n;i++)
    {
        if(!vis[i])
        {
        if(id==-1) id=i;
        else if(dis[id]>dis[i]) id=i;
        } 
    }
    tmp+=dis[id];
    vis[id]=1;
    for (int i=1;i<=n;i++)
    {
        if(!vis[i])
        {
        dis[i]=min(dis[i],fabs(z[i]-z[id])-mid*dist(i,id));
        } 
    }
    }
    
 
return tmp<=0.0;
} 
int main()
{
    while (scanf("%d",&n)!=EOF)
    {
    if (!n) break;
    for (int i=1;i<=n;i++) scanf("%lf%lf%lf",&x[i],&y[i],&z[i]);
    double l=0.0,r=0.0,mid;
    for (int i=2;i<=n;i++)
        r+=fabs(z[i]-z[1]);
    for(int i=1;i<=50;i++)
    {
        mid=(l+r)/2.0;
        if (check(mid)) r=mid;
        else l=mid;

    }
    printf("%.3lf\n",r);
    }
    return 0;
}

POJ 2728 Desert King | 01分數規劃