Building a Space Station(bfs)

http://poj.org/problem?id=2031

题意：给出n个球的圆心坐标x,y,z, 半径r，若任意两球不相交，则在两球间建桥。问需建桥的最短距离是多少。

思路：建图,以两球间相差的距离为权值，然后求最小生成树。

#include <stdio.h>
#include <math.h>
#include <string.h>
const int inf=1<<28;
const double eps=1e-8;
const int maxn=120;
double sum,Map[maxn][maxn],dis[maxn];
int vis[maxn];
struct point
{
    double x,y,z,r;
};
void prim(int n)
{
    for (int i = 1; i <= n; i++)
        dis[i]=inf;
    dis[1]=0;
    sum=0;
    for (int i = 1; i <= n; i++)
    {
        double Min=inf;
        int pos=0;
        for (int j = 1; j <= n; j++)
        {
            if (!vis[j]&&dis[j] < Min)
            {
                Min = dis[j];
                pos = j;
            }
        }
        if (Min==inf)
            return;
        vis[pos] = 1;
        sum+=Min;
        for (int j = 1; j <= n; j++)
        {
            if (!vis[j]&&dis[j] > Map[pos][j])
                dis[j] = Map[pos][j];
        }
    }
}
void init()
{
    sum = 0;
    memset(vis,0,sizeof(vis));
    memset(Map,0,sizeof(Map));
}
double dist(point &a,point &b)
{
    return (a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y)+(a.z-b.z)*(a.z-b.z);
}
int main()
{
    int n;
    while(~scanf("%d",&n)&&n)
    {
        point a[120];
        init();
        for (int i = 1; i <= n; i++)
        {
            scanf("%lf%lf%lf%lf",&a[i].x,&a[i].y,&a[i].z,&a[i].r);
        }
        for (int i = 1; i <= n; i++)
        {
            for (int j = 1; j <= n; j++)
            {
                double d = sqrt(dist(a[i],a[j]));
                if (d-(a[i].r+a[j].r)>eps)
                    Map[i][j] = d-a[i].r-a[j].r;
                else
                    Map[i][j] = 0;
            }
        }
        prim(n);
        printf("%.3f
",sum);
    }
    return 0;
}