【分数规划，最优比率生成树】POJ 2728 Desert King

题面在这里

分数规划经典题目啊……

考虑二分答案mid

则只需验证:

∑cost∑dist≤mid∑cost≤mid∗∑dist∑(costi−mid∗disti)

然后就直接最小生成树了

示例程序：

#include<cstdio>#include<cmath>#include<cstring>#include<algorithm>#define cl(x,y) memset(x,y,sizeof(x))#define sqr(x) ((x)*(x))#define _abs(x) ((x)>0?(x):-(x))using namespace std;inline char nc(){    static char buf[100000],*p1=buf,*p2=buf;    return p1==p2&&(p2=(p1=buf)+fread(buf,1,100000,stdin),p1==p2)?EOF:*p1++;}inline int red(){    int res=0,f=1;char ch=nc();    while (ch<'0'||'9'<ch) {if (ch=='-') f=-f;ch=nc();}    while ('0'<=ch&&ch<='9') res=(res<<3)+(res<<1)+ch-48,ch=nc();    return res*f;}const int maxn=1005;const double eps=1e-5;int n,x[maxn],y[maxn],z[maxn],cst[maxn][maxn];double dst[maxn][maxn],d[maxn];bool vis[maxn];inline double getdst(int x,int y,int xx,int yy){    return sqrt(sqr(x-xx)+sqr(y-yy));}bool check(double mid){    cl(vis,0);vis[1]=1;    for (int i=1;i<=n;i++) d[i]=cst[1][i]-mid*dst[1][i];    double res=0;    for (int i=1;i<n;i++){        int k;double Min=1e100;        for (int j=1;j<=n;j++)         if (!vis[j]&&Min>d[j]) Min=d[j],k=j;        vis[k]=1;res+=Min;        for (int j=1;j<=n;j++)         if (!vis[j]) d[j]=min(d[j],cst[k][j]-mid*dst[k][j]);    }    return res+1e-10<=0;}int main(){    n=red();    while (n){        for (int i=1;i<=n;i++)         x[i]=red(),y[i]=red(),z[i]=red();        double L=0,R=0,ans;        for (int i=1;i<=n;i++)         for (int j=1;j<=n;j++)          dst[i][j]=getdst(x[i],y[i],x[j],y[j]),          cst[i][j]=_abs(z[i]-z[j]),R=max(R,cst[i][j]/dst[i][j]);        while (R-L>=eps){            double mid=(L+R)/2;            if (check(mid)) ans=mid,R=mid;else L=mid;        }        printf("%.3f\n",ans);        n=red();    }    return 0;}