POJ2031 Building a Space Station 最小生成树 知道么 c++ 双精度 用%lf G++用 %f 我勒个去 不然会wa啊！！！

自己的方法

#include<stdio.h>
#include<string.h>
#include<math.h>
#include<algorithm>
using namespace std;

struct Island
{
double xi,yi,zi,r;
int father;
}island[101];

double dist(Island A,Island B)
{
return sqrt((A.xi-B.xi)*1.0*(A.xi-B.xi)+(A.yi-B.yi)*(A.yi-B.yi)+(A.zi-B.zi)*(A.zi-B.zi))-A.r-B.r;
}
struct Bridge
{
int a,b;
double price;
}bridge[10010];

int cmp(Bridge a,Bridge b)
{
return a.price < b.price;
}

int find(int x)
{
int t;
int child = x;
while(x != island[x].father)
{
x = island[x].father;
}
while(x != child)
{
t=island[child].father;
island[child].father = x;
child = t;
}
return x;
}

int merge(int x,int y)
{
int a =find(x);
int b =find(y);
if(a != b)
{
island[a].father = b;
return 1;
}
return 0;
}

int main()
{
int n,m,i,j,k,g,t; //g 是用来优化最后一步  如果当所有的岛都联通之后 停止 造桥
double ans;
while(scanf("%d",&n),n)
{
k = 1;
g =1;
ans = 0.000;
memset(bridge,0,sizeof(bridge));
memset(island,0,sizeof(island));
for(i = 0;i <= 100;i++ )
island[i].father = i;
for (i = 1;i <= n;i++)
{
scanf("%lf%lf%lf%lf",&island[i].xi,&island[i].yi,&island[i].zi,&island[i].r);
for( j = 1;j < i;j++)
{
if(dist(island[i],island[j]) > 0)
{
bridge[k].a = i;
bridge[k].price = dist(island[i],island[j]);
bridge[k++].b = j;
}
else
{
bridge[k].a = i;
bridge[k].price = 0.000;
bridge[k++].b = j;
}
}
}
sort(bridge+1,bridge+k,cmp);
for(i = 1;i < k;i++)
{
if (merge(bridge[i].a,bridge[i].b))
{
ans += bridge[i].price;
g++;
}
if(g == n)
break;
}
printf("%.3lf\n",ans);
}
return 0;
}

网上的方法1

这个真的是太简洁 了    好飘逸  ！！！  prim 算法  好好学！！！

#include<cstdio>
#include<iostream>
#include<cmath>
using namespace std;
const int mm=111;
double x[mm],y[mm],z[mm],r[mm],dis[mm];
bool p[mm];
int i,j,k,n;
double mn,ans;
double sq(double a)
{
return a*a;
}
double dd(int i,int j)
{
double ret=sqrt(sq(x[i]-x[j])+sq(y[i]-y[j])+sq(z[i]-z[j]))-r[i]-r[j];
return ret>0?ret:0;
}
int main()
{
while(scanf("%d",&n),n)
{
for(i=1;i<=n;++i)
scanf("%lf%lf%lf%lf",&x[i],&y[i],&z[i],&r[i]);
for(i=1;i<=n;++i)dis[i]=1e30,p[i]=1;
ans=dis[1]=0;
for(i=1;i<=n;++i)
{
mn=1e30;
for(j=1;j<=n;++j)
if(p[j]&&dis[j]<mn)
{
mn=dis[j];
k=j;
}
p[k]=0;
ans+=mn;
for(j=1;j<=n;++j)
if(p[j])dis[j]=min(dis[j],dd(k,j));
}
printf("%.3lf\n",ans);
}
return 0;
}

网上的方法2

#include<stdio.h>
#include<math.h>
#include<string.h>
#define N 99999999
struct node {
double x,y,z,r;
}lnode[150];
double map[150][150];
double d[150];
int flag[150],n;
double dist(int a,int b)
{
double sum;
sum=sqrt(pow((lnode[a].x-lnode[b].x),2)+pow((lnode[a].y-lnode[b].y),2)+pow((lnode[a].z-lnode[b].z),2))-lnode[a].r-lnode[b].r;
return sum;
}
double prim()
{
int i,j,z;
double min,sum=0;
memset(flag,0,sizeof(flag));
for(i=0;i<n;i++)
d[i]=map[0][i];
flag[0]=1;
for(i=1;i<n;i++)
{
min=N;
for(j=1;j<n;j++)
if(!flag[j]&&d[j]<min)
{
min=d[j];
z=j;
}
flag[z]=1;
//printf("%.3f\n",min);
sum+=min;
for(j=0;j<n;j++)
if(!flag[j]&&d[j]>map[j][z])
d[j]=map[z][j];
}
return sum;
}
int main()
{
int i,j;
while(scanf("%d",&n),n)
{
for(i=0;i<n;i++)
scanf("%lf%lf%lf%lf",&lnode[i].x,&lnode[i].y,&lnode[i].z,&lnode[i].r);
for(i=0;i<n;i++)
for(j=0;j<n;j++)
map[i][j]=N;
for(i=0;i<n;i++)
for(j=i+1;j<n;j++)
{
double v=dist(i,j);
//printf("%.3f\n",v);
if(v<=0)
{
map[i][j]=0;
map[j][i]=0;
}
else
{
map[i][j]=v;
map[j][i]=v;
}
}
printf("%.3f\n",prim());
}
return 0;
}