POJ 2560 Freckles

Description
In an episode of the Dick Van Dyke show, little Richie connects the freckles on his Dad’s back to form a picture of the Liberty Bell. Alas, one of the freckles turns out to be a scar, so his Ripley’s engagement falls through.
Consider Dick’s back to be a plane with freckles at various (x,y) locations. Your job is to tell Richie how to connect the dots so as to minimize the amount of ink used. Richie connects the dots by drawing straight lines between pairs, possibly lifting the pen between lines. When Richie is done there must be a sequence of connected lines from any freckle to any other freckle.
Input
The first line contains 0 < n <= 100, the number of freckles on Dick’s back. For each freckle, a line follows; each following line contains two real numbers indicating the (x,y) coordinates of the freckle.
Output
Your program prints a single real number to two decimal places: the minimum total length of ink lines that can connect all the freckles.
Sample Input
3
1.0 1.0
2.0 2.0
2.0 4.0
Sample Output
3.41

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题意：给出坐标，然后用最短的直线连n个点，使得每两个点之间有连线、不就是最小生成树求整个网络的最短距离么、、

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#include<algorithm>
#include<iostream>
#include<cstdio>
#include<cmath>
using namespace std;
int n,m,f[105];
double ans,x[105],y[105];
struct pnt
{
    int l,r;
    double val;
}a[10005];
bool cmp(pnt c,pnt d)
{
    return c.val<d.val;
}
int fnd(int x)
{
    if(f[x]!=x)
        f[x]=fnd(f[x]);
    return f[x];
}
double dis(int i,int j)
{
    return (x[i]-x[j])*(x[i]-x[j])+(y[i]-y[j])*(y[i]-y[j]);
}
int main()
{
    scanf("%d",&n);
    for(int i=1;i<=n;i++)
        f[i]=i;
    for(int i=1;i<=n;i++)
        scanf("%lf%lf",&x[i],&y[i]);
    for(int i=1;i<=n;i++)
        for(int j=i+1;j<=n;j++)
        {
            m++;
            a[m].l=i;
            a[m].r=j;
            a[m].val=dis(i,j);
        }
    sort(a+1,a+m+1,cmp);
    for(int i=1;i<=m;i++)
        if(fnd(a[i].l)!=fnd(a[i].r))
        {
            f[fnd(a[i].l)]=fnd(a[i].r);
            ans+=sqrt(a[i].val);
        }
    printf("%.2f\n",ans);
    return 0;
}