Qin Shi Huang's National Road System（次小生成树）

题目描述：

During the Warring States Period of ancient China(476 BC to 221 BC), there were seven kingdoms in China ---- they were Qi, Chu, Yan, Han, Zhao, Wei and Qin. Ying Zheng was the king of the kingdom Qin. Through 9 years of wars, he finally conquered all six other kingdoms and became the first emperor of a unified China in 221 BC. That was Qin dynasty ---- the first imperial dynasty of China(not to be confused with the Qing Dynasty, the last dynasty of China). So Ying Zheng named himself “Qin Shi Huang” because “Shi Huang” means “the first emperor” in Chinese.

Qin Shi Huang undertook gigantic projects, including the first version of the Great Wall of China, the now famous city-sized mausoleum guarded by a life-sized Terracotta Army, and a massive national road system. There is a story about the road system:
There were n cities in China and Qin Shi Huang wanted them all be connected by n-1 roads, in order that he could go to every city from the capital city Xianyang.
Although Qin Shi Huang was a tyrant, he wanted the total length of all roads to be minimum,so that the road system may not cost too many people’s life. A daoshi (some kind of monk) named Xu Fu told Qin Shi Huang that he could build a road by magic and that magic road would cost no money and no labor. But Xu Fu could only build ONE magic road for Qin Shi Huang. So Qin Shi Huang had to decide where to build the magic road. Qin Shi Huang wanted the total length of all none magic roads to be as small as possible, but Xu Fu wanted the magic road to benefit as many people as possible ---- So Qin Shi Huang decided that the value of A/B (the ratio of A to B) must be the maximum, which A is the total population of the two cites connected by the magic road, and B is the total length of none magic roads.
Would you help Qin Shi Huang?
A city can be considered as a point, and a road can be considered as a line segment connecting two points.

输入：

The first line contains an integer t meaning that there are t test cases(t <= 10).
For each test case:
The first line is an integer n meaning that there are n cities(2 < n <= 1000).
Then n lines follow. Each line contains three integers X, Y and P ( 0 <= X, Y <= 1000, 0 < P < 100000). (X, Y) is the coordinate of a city and P is the population of that city.
It is guaranteed that each city has a distinct location.

输出：

For each test case, print a line indicating the above mentioned maximum ratio A/B. The result should be rounded to 2 digits after decimal point.

样例输入：

2
4
1 1 20
1 2 30
200 2 80
200 1 100
3
1 1 20
1 2 30
2 2 40

样例输出：

65.00
70.00

题目大意：

给定n个点的点权及相互间的边权，求一棵树，其中一条边的边权变为0，树的比率值为该0值边所连的两点的点权和/剩下的树边和，求这个值最大是多少。

这题要用到次小生成树的思想，即找到最小生成树，然后添加一条边构成环，再删掉环中属于最小树的最大边，用这种方法遍历所有边以找到最终长度。

#include<iostream>
#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<math.h>
#include<vector>
using namespace std;
int father[1005];
int n,cnt;
vector<int> huang[1005];
double d[1005][1005];
double dis[1005][1005];
int visit[1005];
struct node{
	int start,over;
	double cost;
}jiang[1000050];
struct code{
	int x,y,p;
}feng[1005];
bool cmp(node a,node b)
{
	return a.cost<b.cost;
}
void init()
{
	for(int i=0;i<1005;i++)
	{
		father[i]=i;
		huang[i].clear();
	}
}//初始化
int findd(int a)
{
	if(father[a]==a)return a;
	else return father[a]=findd(father[a]);
}
void unit(int a,int b)
{
	int x=findd(a);
	int y=findd(b);
	if(x!=y)
	{
		father[x]=y;
	}
}
double juli(int a,int b,int x,int y)
{
	return sqrt((a-x)*(a-x)+(b-y)*(b-y));
}
void dfs(int root,int where)
{
	visit[where]=1;
	for(int i=0;i<huang[where].size();i++)
	{
		int what=huang[where][i];
		if(visit[what]==0)
		{
			dis[root][what]=max(d[where][what],dis[root][where]);
			dfs(root,what);
		}
	}
}
//krustra算法
void solve()
{
	init();
	memset(d,0,sizeof(d));
	memset(dis,0,sizeof(dis));
	int num=0;
	double sum=0;
	sort(jiang,jiang+cnt,cmp);
	for(int i=0;i<cnt;i++)
	{
		if(num==n-1)break;
		else if(findd(jiang[i].start)!=findd(jiang[i].over))
		{
			unit(jiang[i].start,jiang[i].over);
			num++;
			sum+=jiang[i].cost;
			huang[jiang[i].start].push_back(jiang[i].over);
			huang[jiang[i].over].push_back(jiang[i].start);
			d[jiang[i].start][jiang[i].over]=d[jiang[i].over][jiang[i].start]=jiang[i].cost;
		}
	}
	for(int i=1;i<=n;i++)
	{
		memset(visit,0,sizeof(visit));
		dfs(i,i);
	}//找到从i~j之间的最长的那一条边的长度
	double ans=0;
	for(int i=1;i<=n;i++)
	{
		for(int j=i+1;j<=n;j++)
		{
			double v=sum-dis[i][j];
			v=(feng[i].p+feng[j].p)*1.0/v;
			ans=max(ans,v);
		}
	}
	printf("%.2f\n",ans);//求最大值
}
int main()
{
	int ttt;
	scanf("%d",&ttt);
	while(ttt--)
	{
		scanf("%d",&n);
		cnt=0;
		for(int i=1;i<=n;i++)
		{
			scanf("%d%d%d",&feng[i].x,&feng[i].y,&feng[i].p);
		}
		for(int i=1;i<=n;i++)
		{
			for(int j=i+1;j<=n;j++)
			{
				jiang[cnt].start=i;
				jiang[cnt].over=j;
				jiang[cnt++].cost=juli(feng[i].x,feng[i].y,feng[j].x,feng[j].y);
			}
		}
		solve();
	}
	return 0;
}