KY40 Freckles

最小生成树板子
虽然不知道我写的那一版为什么错
需要注意的是，递归调用层数过多，也会显示内存超限
ti

#include<bits/stdc++.h>
using namespace std;
struct Node{
    double x, y;
};

struct opop{
	int xx, yy;
	double dd;
	bool operator <(const opop &t) const{
		return dd < t.dd;
	}
};
vector<Node>no;
vector<opop>oop;
int h[101];
int f[101];
bool op[101];

void init(int n){
	for(int i = 0; i <= n; i ++ ){
		f[i] = i;
		h[i] = 0;
	} 
	oop.clear();
	no.clear();
	return ;
}

int find(int x){
	if(x != f[x]){
		f[x] = find(f(x));
	}
	return f[x];
}

int main()
{
    int n;
    while(cin>>n){
		init(n);
	    Node q;
	    opop p;
	    for(int i = 0; i < n; i ++ ){
	    	cin>>q.x>>q.y;
	    	no.push_back(q);
	    }
	
		for(int i = 0; i < no.size() - 1; i ++ ){
			for(int j = i + 1; j < no.size(); j ++ ){
				double l = sqrt((no[i].x - no[j].x) * (no[i].x - no[j].x) + (no[i].y - no[j].y) * (no[i].y - no[j].y));  //计算任意两点间距离 
				p.xx = i, p.yy = j, p.dd = l;
				oop.push_back(p);
			}
		}
		
		sort(oop.begin(), oop.end());
		
	    double ans = 0;  //记录最小生成树边长 
		
	    for(int i = 0; i < oop.size(); i ++ ){
	    	
	        int xt = find(oop[i].xx);
	        int yt = find(oop[i].yy);
	        double d = oop[i].dd;
	        
	        if(f[xt] != f[yt]){
				ans += d;
				if(h[xt] > h[yt]){
					f[yt] = xt;
				}
				else if(h[yt] > h[xt]){
					f[xt] = yt;
				}
				else {
					f[xt] = yt;
					h[yt] ++ ;
				}
			}
	    }
	    
	    printf("%.2f", ans);
	}
    
    return 0;
}