Building a Space Station POJ 2031

题目链接： http://poj.org/problem?id=2031

 

题意：现给定一些细胞的坐标以及它们的半径，求它们彼此联通的最短路径是多少。实则是最小生成树。

////特别心塞，G++提交就错，C++提交就A，害我找错好半天。。。

＊＊＊　百度说Ｇ＋＋　和Ｃ＋＋　的区别在于　％ｌｆ　和　％ｆ　由于这道题里面涉及到了　　所以才会ＷＡ　

 

 

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<math.h>
#include<algorithm>
#include<queue>
#include<iostream>
using namespace std;
#define maxn 10000
double maps[maxn][maxn];
double x[maxn], y[maxn], z[maxn], r[maxn], dist[maxn];
int v[maxn];
int n;

void Init()
{
    int i, j;
    for(i=1; i<=n; i++)
    for(j=1; j<=n; j++)
    if(i==j) maps[i][j] = 0;
    else maps[i][j]=maps[j][i]= 10000000 ;
}

double Dij()
{
    memset(dist, 0, sizeof(dist));
    memset(v, 0, sizeof(v));
    for(int i=1; i<=n; i++)
        dist[i] = maps[1][i];

    v[1] = 1;

    double sum = 0;

    for(int i=1; i<n; i++)
    {
        int index;
        double  mins=10000000;

        for(int j=1; j<=n; j++)
        {
            if(!v[j] && dist[j] < mins)
            {
                index = j;
                mins = dist[j];
            }
        }

    if(mins == 10000000)  return 0;
        sum += mins;
        v[index] = 1;

        for(int j=1; j<=n; j++)
        {
            if(!v[j] && dist[j] >= maps[index][j])
            {
                dist[j]= maps[index][j];
            }
        }
    }

    return sum;
}
int main()
{
    while(scanf("%d", &n), n)
    {
         for(int i=1; i<=n; i++)
            scanf("%lf %lf %lf %lf", &x[i], &y[i], &z[i], &r[i]);

         Init();

         for(int i=1; i<n; i++)
         {
             for(int j=i+1; j<=n; j++)
             {

                 double d=sqrt((x[j]-x[i])*(x[j]-x[i])+(y[j] - y[i])*(y[j] - y[i])+(z[j]-z[i])*(z[j]-z[i])) ;

                  double p = r[i] + r[j];

                  if(d <= p)
                    {
                        maps[i][j]=maps[j][i]=0;
                        continue;
                    }

                 maps[i][j] = maps[j][i] = min(maps[i][j], d-p);

             }
         }

        double ans = Dij();

        printf("%.3lf\n",ans);
    }
    return 0;
}