POJ 2728 Desert King (最优比例生成树)

POJ2728 无向图中对每条边i 有两个权值wi 和vi 求一个生成树使得 （w1+w2+...wn-1）/(v1+v2+...+vn-1)最小。

采用二分答案mid的思想。

将边的权值改为 wi-vi*mid.

对所有边求和后除以v 即为 （w1+w2+...wn-1）/(v1+v2+...+vn-1)-mid. 因此，若当前生成树的权值和为0，就找到了答案。否则更改二分上下界。

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
#include<vector>
#include<queue>
using namespace std;

const int maxn=1000;
int n;
double lenth[maxn],nowans,lef,righ,w[maxn][maxn],v[maxn][maxn];
bool intree[maxn];

double ab(double x)
{
if(x>0)return x;
else return x*(-1);
}

class point
{
public:
double x;double y;double z;
};

point po[maxn];

double dist(point a,point b);

double cost(point a,point b)
{
return ab((a.z-b.z));
}

class edge
{
public:
int  pa;int pb;
double dis;double cos;double div;
edge(int a,int b,double(*distance)(point,point),double (*cost)(point,point))
{
this->pa=a;this->pb=b;
this->dis=distance(po[a],po[b]);
this->cos=cost(po[a],po[b]);
this->div=cos/dis;
}
bool operator <(const edge b)const
{
return (this->cos-(this->dis*nowans))>(b.cos-(b.dis*nowans));//这个算法的核心
}
};

double dist(point a,point b)
{
return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
}

double min(double a,double b)
{
if(a<b)return a;
else return b;
}

int main()
{
ios::sync_with_stdio(false);
while(cin>>n)
{
if(n==0)return 0;
lef=0.0;righ=1e6;
for(int i=0;i<n;i++)
{
cin>>po[i].x>>po[i].y>>po[i].z;
}
for(int i=0;i<n;i++)
for(int j=0;j<n;j++)
{
w[i][j]=cost(po[i],po[j]);
v[i][j]=dist(po[i],po[j]);
}
double minnow;
while(true)
{
nowans=(lef+righ)/2;minnow=0;
memset(intree,0,sizeof(intree));
for(int i=1;i<n;i++)
{
lenth[i]=w[0][i]-v[0][i]*nowans;
}
lenth[0]=0.0;intree[0]=true;
for(int i=1;i<n;i++)
{
double temp=1e+30;int tj=0;
for(int j=1;j<n;j++)
if(!intree[j]&&lenth[j]<temp)
{
tj=j;
temp=lenth[j];
}
minnow+=lenth[tj];
intree[tj]=true;
for(int j=1;j<n;j++)
{
if(!intree[j])lenth[j]=min(lenth[j],w[tj][j]-v[tj][j]*nowans);
}
}
if(ab(minnow-0.0)<1e-5)break;
if(minnow>0)lef=nowans;
else righ=nowans;
}
printf("%.3lf\n",(lef+righ)/2);
}
return 0;
}

　　用二分答案思想解决的生成树问题还有单度限制最小生成树，参考CODEFORCES 125E.