不超过50个点,先给两个点s1,s2表示这两个点必须先连起来,接下来给出n个点的坐标,在先连接s1,s2的前提下求最小生成树。 N不到50..普利姆,克鲁斯卡尔随便都能搞吧..虽说完全图理论上普利姆快一点- -...
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <queue>
#include <stack>
#include <map>
#include <string>
#include <cstring>
#include <cmath>
using namespace std;
inline double dist(double x1,double y1,double x2,double y2)
{
return sqrt((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2));
}
struct node
{
double x,y;
}pt[100];
double g[100][100];
double dis[100];
bool ft[100];
int n,m,tt,s1,s2;
int main()
{
// freopen("in.txt","r",stdin);
while(scanf("%d",&n) && n)
{
memset(ft,0,sizeof ft);
scanf("%d%d",&s1,&s2);
for (int i=1; i<=n; i++)
{
scanf("%lf%lf",&pt[i].x,&pt[i].y);
}
for (int i=1; i<=n; i++)
for (int j=1; j<=n; j++)
g[i][j]=g[j][i]=dist(pt[i].x,pt[i].y,pt[j].x,pt[j].y);
for (int i=0; i<=n; i++) dis[i]=9999999999.999;
dis[s1]=dis[s2]=0;
ft[s1]=ft[s2]=true;
for (int i=1; i<=n; i++)
{
if (i!=s1 && i!=s2)
{
dis[i]=min(g[i][s1],g[i][s2]);
}
}
double minn=9999999999.999;
int k;
double ans=g[s1][s2];
for (int i=3; i<=n; i++)
{
minn=9999999999.999;
for (int j=1; j<=n; j++)
if (!ft[j] && minn>dis[j])
{
minn=dis[j];
k=j;
}
ans+=minn;
ft[k]=true;
for (int j=1; j<=n; j++)
{
if (!ft[j])
dis[j]=min(dis[j],g[k][j]);
}
}
printf("%.2lf\n",ans);
}
return 0;
}