hdu 携程全球数据中心建设 （球面距离 + 最小生成树）

题目

 #include <iostream>
 #include <cstdio>
 #include <cstring>
 #include <cstdlib>
 #include <cmath>
 #include <algorithm>
 using namespace std;
 double G[][], sum;
 const double pi = acos(-);
 const int INF = (<<);

 struct node
 {
     double x, y;    //x是经度， y是纬度。
 }p[];

 double angle(double lng1, double lat1, double lng2, double lat2)
 {
     double dlng = fabs(lng1-lng2)*pi/;
     while(dlng >= pi+pi)
         dlng -= (pi+pi);
     if(dlng > pi)
         dlng = pi+pi-dlng;
     lat1 *= pi/;
     lat2 *= pi/;
     return acos(cos(lat1)*cos(lat2)*cos(dlng)+sin(lat1)*sin(lat2));
 }

 double sphere_dist(double r, double lng1, double lat1, double lng2, double lat2)
 {
     return r*angle(lng1, lat1, lng2, lat2);
 }

 void prime(int n)
 {
     int i, j, pos;
     int Min, vis[];
     double d[];
     memset(vis, , sizeof(vis));
     memset(d, , sizeof(d));

     for(i = ; i <= n; i++)
     d[i] = G[][i];
     vis[] = ;
     for(i = ; i <= n; i++)
     {
         Min = INF;
         for(j = ; j <= n; j++)
         {
             if(!vis[j] && Min>d[j])
             {
                 Min = d[j];
                 pos = j;
             }
         }
         sum += Min;
         vis[pos] = ;
         for(j = ; j <= n; j++)
         {
             if(!vis[j])
             {
                 if(d[j] > G[pos][j])
                     d[j] = G[pos][j];
             }
         }
     }
 }
 int main()
 {
     int n, t, c, i, j;
     double d, l, r;
     cin>>t;
     while(t--)
     {
         sum = ;
         cin>>d>>l>>c;
         r = d/*1.0;
         n = c;
         for(i = ; i <= n; i++)
         {
             for(j = ; j <= n; j++)
                 G[i][j] = INF;
             G[i][i] = ;
         }
         for(i = ; i <= n; i++)
             cin>>p[i].x>>p[i].y;

         for(i = ; i <= n; i++)
             for(j = i+; j <= n; j++)
         {
             G[i][j] = sphere_dist(r, p[i].y, p[i].x, p[j].y, p[j].x);
             G[j][i] = G[i][j];
         }
         prime(n);
         if(sum - l> )
             cout<<"N"<<endl;
         else
         cout<<"Y"<<endl;
     }
     return ;
 }