poj 2031 && zoj 1718 Building a Space Station

题意：将给定的所有空间站用最小代价连接起来，不过空间站可以重合或相交。

思路：对于不相交的情况，两个空间站的距离就是 圆心距离-两个空间站的半径和，相交或重合的情况就把距离看作0处理。把空间站作为图中的顶点，两个空间站的距离作为边，求一次最小生成树即可得到结果。

注意：在poj上g++输出double要用%f 不然会wrong.

代码：

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cmath>

using namespace std;

const int maxn = 105;
const int maxm = 5005;

struct Station{
	double x,y,z;
	double r;
};

struct Edge{
	int u,v;
	double w;
	
	bool operator < (const Edge &b) const{
		return w < b.w;
	}
};

int n,m;
Station s[maxn];
Edge a[maxm];
int p[maxn];

void UFset(){
	for(int i = 0; i < n; i++){
		p[i] = -1;
	}
}

int Find(int x){
	return p[x] >= 0 ? p[x] = Find(p[x]) : x;
}

void Union(int r1, int r2){
	r1 = Find(r1);
	r2 = Find(r2);
	int t = p[r1] + p[r2];
	if(p[r1] > p[r2]){
		p[r1] = r2;
		p[r2] = t;
	}
	else{
		p[r2] = r1;
		p[r1] = t;
	}
}

bool Input(){
	cin>>n;
	if(n == 0)
		return false;
	m = 0;
	for(int i = 0; i < n; i++){
		cin>>s[i].x>>s[i].y>>s[i].z;
		cin>>s[i].r;
	}
	return true;
}

double Dis(Station &a, Station &b){
	return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y)+(a.z-b.z)*(a.z-b.z))-a.r-b.r;
}


void Convert(){
	for(int i = 0; i < n-1; i++){
		for(int j = i+1; j < n; j++){
			a[m].u = i;
			a[m].v = j;
			double d = Dis(s[i],s[j]);
			a[m].w = (d <= 0 ? 0.0 : d);
			m++;
		}
	}
}

double Kruskal(){
	double sum = 0;
	int num = 0;
	UFset();
	sort(a,a+m);
	for(int i = 0; i < m; i++){
		int u = a[i].u, v = a[i].v;
		if(Find(u) != Find(v)){
			Union(u,v);
			sum += a[i].w;
			num++;
		}
		if(num >= n-1)
			break;
	}
	return sum;
}

int main(){
	
	while(Input()){
		Convert();
		printf("%.3lf\n",Kruskal());
	}
	return 0;
}