A - Treehouses Kattis - treehouses--分块并查集

In a rainforest there are n treehouses high in the forest canopy on different trees (numbered from 1 to n). The i-th tree’s location is at (xi,yi). The first e of them in the list are close enough to neighboring open land around the rainforest so that transportation between all of them is easy by foot. Some treehouses may already be connected by direct straight cables through the air that can allow transport between them.

Residents want easy transportation between all the treehouses and the open land, by some combination of walking (between those near the open land), and using one or more cables between treehouses. This may require the addition of more cables. Since the cables are expensive, they would like to add the smallest possible length of cable.

The height of a cable up two trees can be set so cables can criss-cross other cables, and not allow any snags or crashes. It is not safe to try to switch between two criss-crossed cables in mid-air!

Input
The input will start with the three integers n (1≤n≤1000), e (1≤e≤n), and p (0≤p≤1000), where p is the number of cables in place already.

Next come n lines, each with two real numbers x and y (|x|,|y|≤10000) giving the location of a treehouse. The i-th coordinate pair is for the treehouse with ID i. All coordinate pairs are unique. Real numbers are stated as integers or include one digit after a decimal point.

Next come p lines, each with two integers a, b, where 1≤a

Output
The output is the minimum total length of new cable that achieves the connection goal, expressed with absolute or relative error less than 0.001.

Sample Input 1 Sample Output 1
3 1 0
0.0 0.0
2.0 0.0
1.0 2.0
4.236067
Sample Input 2 Sample Output 2
3 1 1
0.0 0.0
0.5 2.0
2.5 2.0
1 2
2.000000
Sample Input 3 Sample Output 3
3 2 0
0.0 0.0
2.0 0.0
1.0 2.0
2.236067

题目意思：
给你一些点的坐标，已知前e个点不用连，已知中间有p对点不用连，求用最少的距离的线，将全部的点连起来；

#include
using namespace std;
double x[1005];
double y[1005];
int fa[1005];
int getf(int x)
{
     
    if(x==fa[x]) return x;
    return fa[x]=getf(fa[x]);
}
struct node{
     
    int u,v;
    double w;
}e[1000021];
int edn;
bool cmp(node x,node y)
{
     
    return x.w<y.w;
}
int main()
{
     
    int n,ee,p;
    cin>>n>>ee>>p;
    for(int i=1;i<=n;i++)
    {
     
        fa[i]=i;
        cin>>x[i]>>y[i];
    }
    for(int i=2;i<=ee;i++)
    {
     
        fa[getf(i)]=getf(1);
    }
    for(int i=1,u,v;i<=p;i++)
    {
     
        cin>>u>>v;
        fa[getf(u)]=getf(v);
    }
    for(int i=1;i<=n;i++)//将任意两点的距离存下来
    {
     
        for(int j=1+i;j<=n;j++)
        {
     
            e[++edn]=(node){
     getf(i),getf(j),sqrt(pow(x[i]-x[j],2)+pow(y[i]-y[j],2))};
        }
    }
    sort(e+1,e+1+edn,cmp);
    double ans=0;
    for(int i=1;i<=edn;i++)
    {
     
        int u=e[i].u;
        int v=e[i].v;
        if(getf(u)==getf(v)) continue;
        ans+=e[i].w;
        fa[getf(u)]=getf(v);
    }
    printf("%.6f\n",ans);
    return 0;
}

。