POJ 2728 JZYZOJ 1636 分数规划 最小生成树 二分 prim

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cstring>
 4 #include<algorithm>
 5 #include<cmath>
 6 #include<cstdlib>
 7 using namespace std;
 8 const int maxn=1100;
 9 const double eps=0.0001;
10 int n;
11 long long c[maxn][3]={};
12 double e[maxn][maxn]={};
13 double t[maxn][maxn]={};
14 double w[maxn][maxn]={},k[maxn]={};
15 bool vis[maxn]={};
16 inline long long mabs(long long x){ return x>0?x:-x; }
17 inline long long sqr(long long x){ return x*x; }
18 bool ke(double v){
19     memset(vis,0,sizeof(vis));
20     for(int i=1;i<=n;i++){
21         for(int j=i;j<=n;j++){
22             w[i][j]=w[j][i]=t[i][j]-v*e[i][j];
23         }
24     }vis[1]=1;double fla=0;
25     for(int i=1;i<=n;i++){
26         k[i]=w[1][i];
27     }
28     for(int i=1;i<n;++i){
29         int u=-1;
30         for(int j=1;j<=n;++j){
31             if(!vis[j])
32                 if(u==-1||k[j]<k[u])u=j;
33         }
34         fla+=k[u];vis[u]=1;
35         for(int j=1;j<=n;++j){
36             if(!vis[j])
37                 if(w[u][j]<k[j])k[j]=w[u][j];
38         }
39     }
40     return fla<=0;
41 }
42 double doit(double l,double r){
43     while(r-l>eps){
44         double mid=(l+r)/2;
45         if(ke(mid))r=mid;
46         else l=mid+eps;
47     }
48     return l;
49 }
50 int main(){
51     //freopen("wtf.in","r",stdin);
52     scanf("%d",&n);
53     for(int i=1;i<=n;i++){
54         scanf("%lld%lld%lld",&c[i][0],&c[i][1],&c[i][2]);
55     }
56     for(int i=1;i<=n;i++){
57         for(int j=i;j<=n;j++){
58             e[i][j]=sqrt((double)(sqr(c[i][0]-c[j][0])+sqr(c[i][1]-c[j][1])));
59             t[i][j]=(double)mabs(c[i][2]-c[j][2]);
60             e[j][i]=e[i][j];t[j][i]=t[i][j];
61         }
62     }printf("%.3f",doit(1,100));
63     return 0;
64 }