（最小生成树） Building a Space Station -- POJ -- 2031

链接：

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

代码：

#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <algorithm>
using namespace std;

const int N = 110;
const int INF = 0xfffffff;

struct node
{
    double x, y, z, r;
}s[N];

int n;
double J[N][N], dist[N];
bool vis[N];

double Prim()
{
    int i, j, index;
    double ans=0;

    memset(vis, 0, sizeof(vis));
    vis[1]=1;

    for(i=1; i<=n; i++)
        dist[i]=J[1][i];

    for(i=1; i<n; i++)
    {
        double MIN=INF;
        for(j=1; j<=n; j++)
        {
            if(!vis[j] && dist[j]<MIN)
            {
                index=j;
                MIN=dist[j];
            }
        }
        vis[index]=1;
        if(MIN<INF)
        ans += MIN;

        for(j=1; j<=n; j++)
        {
            if(!vis[j] && dist[j]>J[index][j] )
                dist[j]=J[index][j];
        }
    }
    return ans;
}

int main ()
{

    while(scanf("%d", &n),n)
    {
        int i, j;

        memset(s, 0, sizeof(s));
        memset(J, 0, sizeof(J));
        for(i=1; i<=n; i++)
            scanf("%lf%lf%lf%lf", &s[i].x, &s[i].y, &s[i].z, &s[i].r);

        for(i=1; i<=n; i++)
        for(j=1; j<i; j++)
        {
            double k=sqrt((s[i].x-s[j].x)*(s[i].x-s[j].x)+(s[i].y-s[j].y)*(s[i].y-s[j].y)+(s[i].z-s[j].z)*(s[i].z-s[j].z))-s[i].r-s[j].r;
            if(k<0)
                J[i][j]=J[j][i]=0;
            else
                J[i][j]=J[j][i]=k;
        }

        double ans=Prim();

        printf("%.3f\n", ans);

    }
    return 0;
}