题目大意:在三维坐标中给出n个细胞的x,y,z坐标和半径r。如果两个点相交或相切则不用修路,否则修一条路连接两个细胞的表面,求最小生成树。

题目思路:最小生成树树模板过了,没啥说的

#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<iostream>
#include<algorithm>
#include<cstring>
#include<vector>
#include<queue>
#define INF 0x3f3f3f3f
#define MAX 1005 using namespace std; double x[MAX],y[MAX],z[MAX],r[MAX],Map[MAX][MAX],dist[MAX];
int n,vis[MAX]; double Dist(int a,int b)//计算两球心距离
{
double k1=x[a]-x[b];
double k2=y[a]-y[b];
double k3=z[a]-z[b];
return sqrt(k1*k1+k2*k2+k3*k3);
} void prm()
{
int i,j,k;
double minn,sum=;
vis[]=;
for(i=;i<=n;i++)
dist[i]=Map[][i];
for(i=;i<n;i++)
{
minn=INF;
for(j=;j<=n;j++)
{
if(!vis[j]&&dist[j]<minn)
{
minn=dist[j];
k=j;
}
}
if(minn==INF)
break;
sum+=dist[k];
vis[k]=;
for(j=;j<=n;j++)
{
if(Map[k][j]<dist[j] && !vis[j])
{
dist[j]=Map[k][j];
}
}
}
printf("%.3lf\n",sum);
} int main()
{
int i,j;
while(scanf("%d",&n),n)
{
for(i=;i<MAX;i++)
for(j=;j<MAX;j++)
Map[i][j]=INF;
memset(vis,,sizeof(vis));
memset(dist,,sizeof(dist));
for(i=;i<=n;i++)
{
scanf("%lf%lf%lf%lf",&x[i],&y[i],&z[i],&r[i]);
}
for(i=;i<=n;i++)
{
for(j=;j<=n;j++)
{
if(i!=j)
{
double d=Dist(i,j);
if(d>(r[i]+r[j]))
{
Map[i][j]=Map[j][i]=min(Map[i][j],(d-r[i]-r[j]));
}
else
{
Map[i][j]=Map[j][i]=;
}
}
}
}
prm();
}
return ;
}

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