P2872 - 不稳定岛屿

我的题解
```cpp
#include<iostream>
#include<cmath>
#include<algorithm>
#include<cstdio>
using namespace std;
const int maxn=1000001;
double ans;
int f[maxn],n,m;
struct loc{//坐标存储 
    int x;int y;
}loc[maxn];
struct node{//点距 
    int x;int y;double val;
}g[maxn];
bool comp(node a,node b){
    if(a.val==b.val){
        return a.x<b.x;
    }
    return a.val<b.val;
}
int find(int x){//找爹 
    if(f[x]==x) return x;
    else return f[x]=find(f[x]);
}
void merge(int x,int y)//合并
{
    f[find(x)]=find(y);    
}
int main(){
    cin>>n>>m;
    for(int i=1;i<=n;i++){//初始化 
        f[i]=i;
    }
    for(int i=1;i<=n;i++){//存坐标 
        int x,y;cin>>x>>y;
        loc[i].x=x;loc[i].y=y;
    }
    int k=1;//有距边数统 
    for(int i=1;i<=n;i++){//计算距离 
        for(int j=i+1;j<=n;j++){
            g[k].x=i;g[k].y=j;
            g[k].val=1.0*sqrt(1.0*(loc[i].x-loc[j].x)*(loc[i].x-loc[j].x)+1.0*(loc[i].y-loc[j].y)*(loc[i].y-loc[j].y));
            k++;
        }
    }
    
    for(int i=1;i<=m;i++){//已有边设零  !
        int a,b;
        cin>>a>>b;
        g[++k].x=a;g[k].y=b;g[k].val=0;
    }
    /*
    for(int i=1;i<=n*n;i++){//调试输出 
        cout<<"val:"<<g[i].val<<"lcox:"<<g[i].x<<"locy"<<g[i].y<<endl;
    }
    */
    sort(g+1,g+1+k,comp);
    int top=0;
    for(int i=1;i<=k;i++) {//模板，最小生成树
        if(find(g[++top].x)!=find(g[top].y)){
            ans+=g[top].val;
            merge(g[top].x,g[top].y);
        }
    }
    printf("%.2lf",ans);
}
```
老师的题解
```cpp
#include<iostream>
#include<algorithm>
#include<cmath>
#include<cstdio>
using namespace std;
const int maxn=1001;
int n,m,f[maxn],tot;
bool vis[1001][1001];
struct node
{
    int x,y;
}g[maxn];
struct tree
{
    int x,y;double d;
}tr[1000007];
double dis(int x1,int y1,int x2,int y2)
{
    return (double)sqrt((double)(x2-x1)*(x2-x1)+(double)(y2-y1)*(y2-y1));
}
double ans;
bool comp(tree &a,tree &b)
{
    if(a.d==b.d)
        return a.x<b.x;
    return a.d<b.d;
}
int find(int x)
{ 
    if(f[x]==x) return x;
    return f[x]=find(f[x]);
}
void krucal()
{
    for(int i=1;i<=tot;i++)
    {
        int x1=tr[i].x,y1=tr[i].y;
        int a1=find(x1),b1=find(y1);
        if(a1!=b1)
        {
            f[a1]=b1;
            ans+=tr[i].d;
        }
    }
}
int main()
{
    cin>>n>>m;
    for(int i=1;i<=n;i++) f[i]=i;
    for(int i=1;i<=n;i++)
    {
        cin>>g[i].x>>g[i].y;
    }
    for(int i=1;i<=m;i++)
    {
        int x,y;cin>>x>>y;vis[x][y]=1,vis[y][x]=1;
        int a=find(x),b=find(y);
        if(a!=b) f[a]=b;
    }
    for(int i=1;i<n;i++)
    {
        for(int j=i+1;j<=n;j++)
        {
            if(vis[i][j]==0)
            tr[++tot].x=i,tr[tot].y=j,tr[tot].d=dis(g[i].x,g[i].y,g[j].x,g[j].y);
        }
    }
    sort(tr+1,tr+1+tot,comp);
    krucal();
    printf("%.2lf",ans);
    return 0;
}
```

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https://shrinken.pw/crash-2021-07-08_48-fml.html