Qin Shi Huang‘s National Road System

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Problem Description

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.

Input

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.

Output

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.

Sample Input

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

Sample Output

65.00
70.00

题意描述：

秦始皇想要修路使得n个城市连通，同时使得这些路尽量短。这时徐福说他可以在2个城市之间不花费人力物力建一条路，但他只能建1条这样的路。秦始皇想要使除了徐福造的那条路以外的路的总长度尽量短，而徐福想要造福更多的百姓，他想要使自己用法术建的那条路的两端的城市的总人口数最多。最终二人达成协议。假设A=用法术建的那条路的两端的城市的总人口数，B=除了徐福造的那条路以外的路的总长度，建造的路要使得A/B最大。输出A/B。

解题思路：

我们可以先建一棵最小生成树，然后再从这n个城市中选择2个城市，把二者之间的路看做用法术造的路。如果这条路在最小生成树上，即去掉这条边后最小生成树会变成两棵独立的树，那么A/B就是（二者城市的人口数总和/最小生成树的值减去二者之间的长度）；如果这条路不在生成树上，那么为了使A/B最大，我们要在最小生成树上删除一条权值最大的边使得生成树分为两棵树，同时这2个城市不在同一颗树上，然后就可以把这两个城市之间的路视为用法术造出的路，A/B就是

（二者城市的人口数总和/最小生成树的值减去二者之间某条使二者连通的边的最大长度）。

综上：我们可以使用used[i][j]表示i与j之间的边是否在最小生成树上，dp[i][j]表示i到j之间的使得i与j连通的权值最大的边的权值。在prim内更新dp数组。

参考代码：

#include<stack>
#include<queue>
#include<cmath>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#pragma commment(linker,"/STACK: 102400000 102400000")
using namespace std;
const double eps=1e-6;
const int INF=0x3f3f3f3f;
const int MAXN=1e3+20;
struct city
{
    int x,y,p;
};

city c[MAXN];
int V,par[MAXN];//par数组记录生成树上i的上一个点
bool used[MAXN][MAXN],visit[MAXN];
double mincost[MAXN],cost[MAXN][MAXN],dp[MAXN][MAXN];

double prim()
{
    memset(visit,false,sizeof(visit));
    memset(dp,0,sizeof(dp));
    memset(used,false,sizeof(used));
    for(int i=1; i<=V; i++)
    {
        mincost[i]=cost[1][i];
        par[i]=1;
    }
    visit[1]=true;
    double res=0;
    for(int i=1; i<V; i++)
    {
        int v=-1;
        for(int j=1; j<=V; j++)
            if(!visit[j]&&(v==-1||mincost[j]<mincost[v]))
                v=j;
        if(v==-1)
            break;
        visit[v]=true;
        used[v][par[v]]=used[par[v]][v]=true;
        res+=mincost[v];
        for(int j=1; j<=V; j++)
        {
            if(!visit[j]&&cost[v][j]<mincost[j])
            {
                mincost[j]=cost[v][j];
                par[j]=v;
            }
            if(visit[j]&&j!=v)
            {
                dp[v][j]=dp[j][v]=max(dp[j][par[v]],mincost[v]);
            }
        }
    }
    return res;
}

int main()
{
#ifndef ONLINE_JUDGE
    freopen("in.txt","r",stdin);
#endif // ONLINE_JUDGE
    int tcase;
    scanf("%d",&tcase);
    while(tcase--)
    {
        scanf("%d",&V);
        for(int i=1; i<=V; i++)
            scanf("%d%d%d",&c[i].x,&c[i].y,&c[i].p);
        for(int i=1; i<=V; i++)
            for(int j=1; j<=V; j++)
                cost[i][j]=cost[j][i]=sqrt((c[i].x-c[j].x)*(c[i].x-c[j].x)+(c[i].y-c[j].y)*(c[i].y-c[j].y));
        double mst=prim();
        double ans=-1;
        for(int i=1; i<=V; i++)
        {
            for(int j=1; j<=V; j++)
            {
                if(i==j)
                    continue;
                if(used[i][j])//i和j之间的边是否在最小生成树上
                    ans=max(ans,(c[i].p+c[j].p+0.0)/(mst-cost[i][j]));
                else
                    ans=max(ans,(c[i].p+c[j].p+0.0)/(mst-dp[i][j]));
            }
        }
        printf("%.2lf\n",ans);
    }
    return 0;
}

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HDU 4081 Qin Shi Huang's National Road System（prim）