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 =+;
 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 =; 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 =; 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 =; cc <= cas; ++cc ){
           int n;
           scanf("%d",&n);
           for(int i=; 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=; i < n; i++)
           for(int j = i+; 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=;
           int ge=n;
           for(int i =; 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==) break;
               }
           }
           memset(maxcost,,sizeof(maxcost));
           use.clear();
           dfs(,-,0.0);
           double ans =;
           for(int i =; i<n; i++)
           for(int j =i+ ; j<n; j++ ){
              double  c = 1.0*(P[i]+P[j])/(sum_dist-maxcost[i][j]);
                if(ans<c){
                    ans=c;
                }
           }
             printf("%.2f\n",ans);
       }
     return ;
 }