[POJ2728] Desert King

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题目大意

有$n$个点的完全图，每条边有价值$c$与权值$w$，求$\frac{\sum c}{\sum r}$最小

试题分析

一道分数规划，我们二分$k$值，判断当每条边边权为$c-k\times w$时的最小生成树是否大于$0$，然后就会$T$掉，需要卡常，好像说正解是牛顿迭代法。

#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<algorithm>
#include<climits>
using namespace std;
inline int read(){
    int f=1,ans=0;char c=getchar();
    while(c<'0'||c>'9'){if(c=='-')f=-1;c=getchar();}
    while(c>='0'&&c<='9'){ans=ans*10+c-'0';c=getchar();}
    return f*ans;
}
const int N=1000001;const int M=1000001;
struct node{
    int u,v;
    double c,w;
}x[N];
struct Node{
    int u,v;double w;
}s[N];
double l,r,eps=1e-4;
int xx[N],yy[N],h[N],cnt,f[N],n;
bool cmp(Node x1,Node x2){return x1.w<x2.w;}
int find(int x){if(f[x]==x) return x;return f[x]=find(f[x]);}
double dis(int x1,int y1,int x2,int y2){return (double)sqrt((double)(x1-x2)*(x1-x2)*1.0+(double)(y1-y2)*(y1-y2)*1.0);}
bool check(double k){
    for(int i=1;i<=cnt;i++) s[i].u=x[i].u,s[i].v=x[i].v,s[i].w=x[i].c-k*x[i].w;
    for(int i=1;i<=n;i++) f[i]=i;
    sort(s+1,s+cnt+1,cmp);
    double ans=0;
    for(int i=1;i<=cnt;i++){
        int t1=find(s[i].u),t2=find(s[i].v);
        if(t1!=t2){
            ans+=s[i].w;
            f[t2]=t1;
        } 
    }return ans>=0;
}
double maxn=-INT_MAX;
signed main(){
//    freopen("3.in","r",stdin);
    while(1){
        n=read();
        maxn=-INT_MAX,cnt=0;
        if(!n) return 0;
    for(int i=1;i<=n;i++) xx[i]=read(),yy[i]=read(),h[i]=read();
    for(int i=1;i<=n;i++)
        for(int j=i+1;j<=n;j++) x[++cnt].u=i,x[cnt].v=j,x[cnt].w=dis(xx[i],yy[i],xx[j],yy[j]),x[cnt].c=abs(h[i]-h[j]);
    l=0,r=101;
    while(l<=r){
        double mid=(l+r)/2;
        if(check(mid)) maxn=max(maxn,mid),l=mid+eps;
        else r=mid-eps;
    }
    printf("%.3lf\n",maxn);
    }
    
}

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