复习最小生成树问题

<span style="font-family: Arial, Helvetica, sans-serif; background-color: rgb(255, 255, 255);">忘得差不多了，回忆下kruskal求最小生成树算法，首先得到边信息，对边按权值大小排序，每次选择最小且不在一个联通分量的边，直到连成一棵最小生成树。其中要用上并查集的知识。</span>

Eddy begins to like painting pictures recently ,he is sure of himself to become a painter.Every day Eddy draws pictures in his small room, and he usually puts out his newest pictures to let his friends appreciate. but the result it can be imagined, the friends are not interested in his picture.Eddy feels very puzzled,in order to change all friends 's view to his technical of painting pictures ,so Eddy creates a problem for the his friends of you.
Problem descriptions as follows: Given you some coordinates pionts on a drawing paper, every point links with the ink with the straight line, causes all points finally to link in the same place. How many distants does your duty discover the shortest length which the ink draws?

Input

The first line contains 0 < n <= 100, the number of point. For each point, a line follows; each following line contains two real numbers indicating the (x,y) coordinates of the point.

Input contains multiple test cases. Process to the end of file.

Output

Your program prints a single real number to two decimal places: the minimum total length of ink lines that can connect all the points.

Sample Input

31.0 1.02.0 2.02.0 4.0

Sample Output

3.41

<span style="font-size:24px;">#include <cstdio>#include <cmath>#include <cstring>#include <iostream>#include <algorithm>using namespace std;const int maxn=105;const int maxm=maxn*maxn/2;struct Point{    double x,y;};struct Edge{    int u,v;    double w;};Point p[maxn];Edge e[maxm];int uset[maxn];int cmp(Edge x,Edge y){    return x.w<y.w;}int Find(int x) //并查集的查找函数{    if(x==uset[x]) return x;    else return uset[x]=Find(uset[x]);}int main(){    int n;    while(scanf("%d",&n)!=EOF)    {        for(int i=0;i<n;i++)        {            scanf("%lf%lf",&p[i].x,&p[i].y);        }        int nEdge=0;        for(int i=0;i<n-1;i++)        {            for(int j=i+1;j<n;j++)            {                e[nEdge].u=i;                e[nEdge].v=j;                double temp=(p[i].x-p[j].x)*(p[i].x-p[j].x)+(p[i].y-p[j].y)*(p[i].y-p[j].y);                e[nEdge++].w=sqrt(temp);            }        }        sort(e,e+nEdge,cmp); //对边排序        double sum=0;        for(int i=0;i<n;i++) uset[i]=i;        for(int i=0;i<nEdge;i++)        {            int u=Find(e[i].u);            int v=Find(e[i].v);            double w=e[i].w;            if(u!=v) //如果不在一个连通分量，合并            {                uset[u]=v;                sum+=w;            }        }        printf("%.2lf\n",sum);    }    return 0;}</span>