POJ 2728 Desert King (最优比例生成树)

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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.

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