uva1494 最小生成树--例题

这题说的是n个城市 建路 使他们联通然后 ， 可以使用一条超级的路这条路不计入总长，此时路长度为B， 这条路链接的两个城市人口与和为A+B, 然后计算出最大的A/B

 解题

  先生成一颗最小生成树，然后 计算出这颗树上每两个节点之间要经过的最长的那条路，然后枚举每两个节点u 个v 求出答案

#include <iostream>
#include <algorithm>
#include <vector>
#include <cstdio>
#include <string.h>
#include <cmath>
using namespace std;
const int maxn =1000+5;
struct Edge{
  int u,v;
  double dist;
  bool operator <(const Edge &rhs)const{
     return dist<rhs.dist;
  }
};
vector<Edge>E;
struct point{
   int x,y;
}LOC[maxn];
int P[maxn],fa[maxn];
double maxcost[maxn][maxn];
struct ed{
    int to; double dist;
};
vector<ed>G[ maxn ];
int fid(int u){
   return fa[u]==u? u :( fa[u] = fid( fa[u] ) );
}
vector<int>use;
void dfs(int u, int per,double cost){
   for(int i =0; i < (int )use.size(); i++){
         int v = use[i];
         maxcost[u][v] = maxcost[v][u] = max( maxcost[per][v], cost);
   }
   use.push_back(u);
   for(int i =0; i < (int)G[u].size() ; i++ ){
           ed e = G[u][i];
           if(e.to!=per) dfs(e.to,u,e.dist);
   }
}
double distends(double x1, double y1, double x2, double y2){
     return sqrt((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2));
}
int main()
{
      int cas;
      scanf("%d",&cas);
      for(int cc =1; cc <= cas; ++cc ){
          int n;
          scanf("%d",&n);
          for(int i=0; i<n; i++){
              scanf("%d%d%d",&LOC[i].x,&LOC[i].y,&P[i]);
              G[i].clear(); fa[i] = i;
          }
          E.clear();
          for(int i=0; i < n; i++)
          for(int j = i+1; j < n; j++ ){
              double d = distends(LOC[i].x,LOC[i].y, LOC[j].x, LOC[j].y);
               E.push_back( (Edge){ i,j,d } );
          }
          sort(E.begin() , E.end());
          double sum_dist=0;
          int ge=n;
          for(int i =0; i<(int)E.size(); i++ ){
              int u = E[i].u, v = E[i].v;
              double dist = E[i].dist;
              int fu = fid(u),fv =fid(v);
              if(fu != fv){
                  G[u].push_back( (ed){v,dist} ); G[v].push_back( (ed){u,dist} );
                  fa[ fu ] = fv;
                  sum_dist+=dist;
                 ge--; if(ge==1) break;
              }
          }
          memset(maxcost,0,sizeof(maxcost));
          use.clear();
          dfs(0,-1,0.0);
          double ans =0;
          for(int i =0; i<n; i++)
          for(int j =i+1 ; j<n; j++ ){
             double  c = 1.0*(P[i]+P[j])/(sum_dist-maxcost[i][j]);
               if(ans<c){
                   ans=c;
               }
          }
            printf("%.2f
",ans);
      }
    return 0;
}