POJ2728 Desert King 最小比率生成树

　　01规划问题，二分+prim。注意这里的图是稠密图，应该用prim而不是kruskal。堆优化的prim不适合稠密图，也不适合邻接矩阵建图，否则基本上没怎么优化O(n*(lgn+n))，因为每次操作的常数因子很大，所以lgn与n的差别不大了，尤其在n比较小的时候。我这题用堆优化的prim居然还超时了！

  1 //STATUS:C++_AC_1079MS_8124KB
  2 #include<stdio.h>
  3 #include<stdlib.h>
  4 #include<string.h>
  5 #include<math.h>
  6 #include<iostream>
  7 #include<string>
  8 #include<algorithm>
  9 #include<vector>
 10 #include<queue>
 11 #include<stack>
 12 using namespace std;
 13 #define LL __int64
 14 #define pdi pair<double,int>
 15 #define Max(a,b) ((a)>(b)?(a):(b))
 16 #define Min(a,b) ((a)<(b)?(a):(b))
 17 #define mem(a,b) memset(a,b,sizeof(a))
 18 #define lson l,mid,rt<<1
 19 #define rson mid+1,r,rt<<1|1
 20 const int N=1010,M=1000000,INF=0x3f3f3f3f,MOD=1999997;
 21 const LL LLNF=0x3f3f3f3f3f3f3f3fLL;
 22 const double DNF=100000000;
 23 
 24 struct Node{
 25     int x,y,z;
 26 }a[N];
 27 double w[N][N],d[N];
 28 int vis[N];
 29 int n;
 30 
 31 inline double dist(Node &a,Node &b)
 32 {
 33     return sqrt((double)((a.x-b.x)*(a.x-b.x)+
 34                 (a.y-b.y)*(a.y-b.y)));
 35 }
 36 /*
 37 double prim(int s,double mid)
 38 {
 39     int i,u,v;
 40     double sum=0,t;
 41     priority_queue<pdi,vector<pdi>,greater<pdi> > q;
 42     for(i=0;i<n;i++)d[i]=DNF;d[s]=0;
 43     q.push(make_pair(0,s));
 44     mem(vis,0);
 45     while(!q.empty()){
 46         u=q.top().second;q.pop();
 47         if(vis[u])continue;
 48         vis[u]=1;
 49         sum+=d[u];
 50         for(v=0;v<n;v++){
 51             if(!vis[v] && (t=abs(a[u].z-a[v].z)-mid*w[u][v])<d[v]){
 52                 d[v]=t;
 53                 q.push(make_pair(t,v));
 54             }
 55         }
 56     }
 57     return sum;
 58 }*/
 59 double prim(int s,double mid)
 60 {
 61     int i,u,v,x;
 62     double sum=0,t,m;
 63     mem(vis,0);
 64     for(i=0;i<n;i++)d[i]=DNF;d[s]=0;
 65     for(i=0;i<n;i++){
 66         m=DNF;
 67         for(x=0;x<n;x++)if(!vis[x] && d[x]<m)m=d[u=x];
 68         vis[u]=1;
 69         sum+=d[u];
 70         for(v=0;v<n;v++){
 71             if(!vis[v] && (t=abs(a[u].z-a[v].z)-mid*w[u][v])<d[v])
 72                 d[v]=t;
 73         }
 74     }
 75     return sum;
 76 }
 77 
 78 double binary()
 79 {
 80     double l=0,r=200000,mid;
 81     while(r-l>1e-4){
 82         mid=l+(r-l)/2;
 83         if(prim(1,mid)<0)r=mid;
 84         else l=mid;
 85     }
 86     return mid;
 87 }
 88 
 89 int main()
 90 {
 91   //  freopen("in.txt","r",stdin);
 92     int i,j;
 93     while(~scanf("%d",&n) && n)
 94     {
 95         for(i=0;i<n;i++)
 96             scanf("%d%d%d",&a[i].x,&a[i].y,&a[i].z);
 97         for(i=0;i<n;i++)
 98             for(j=i;j<n;j++)
 99                 w[i][j]=w[j][i]=dist(a[i],a[j]);
100 
101         printf("%.3lf\n",binary());
102     }
103     return 0;
104 }