HDU4081 Qin Shi Huang's National Road System 2017-05-10 23:16 41人阅读 评论(0) 收藏

Qin Shi Huang's National Road System

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

                                                                                             Total Submission(s): 7672    Accepted Submission(s): 2693

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

 

Source

2011 Asia Beijing Regional Contest

 

Recommend

lcy

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题目意思：秦始皇统一中国之后要在全国修公路连接各个城市，抠门皇帝只想修成最小生成树（距离最小，不考虑人力），一个道士说自己可以不花人力物力修一条路，经过两方妥协，选择max（两个城市人口/（生成树长度-这条路的长度））的路让他变，求这个比值最大值。

思路：先找出最小生成树，在枚举删除树上每一条边，删除边后会形成2个集合，找出两个集合的最大值，算出答案取最大值

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <string>
#include <set>
#include <map>
#include <queue>

using namespace std;
#define inf 0x3f3f3f3f
int n;
struct node
{
    int u,v;
    double w;
} p[1000006],mst[1006];
int pre[1006],cnt;
int a[1006];

bool cmp(node a,node b)
{
    return a.w<b.w;
}

void init()
{
    for(int i=0; i<1005; i++)
		pre[i]=i;
}
int fin(int x)
{
    return pre[x]==x?x:pre[x]=fin(pre[x]);
}

double kruskal()
{
    init();
    double cost=0;
    int ans=0;
    for(int i=0; i<cnt; i++)
    {
        int a=fin(p[i].u);
        int b=fin(p[i].v);
        if(a!=b)
        {
            pre[a]=b;
            cost+=p[i].w;
            mst[ans].u=p[i].u,mst[ans].v=p[i].v,mst[ans].w=p[i].w;
            ans++;
        }
        if(ans==n-1)
            break;
    }
    return cost;
}

int main()
{

    int x[1005],y[1005],T;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d",&n);
        for(int i=0; i<n; i++)
            scanf("%d%d%d",&x[i],&y[i],&a[i]);
        cnt=0;
        memset(p,0,sizeof p);
        memset(mst,0,sizeof mst);
        for(int i=0; i<n; i++)
            for(int j=i+1; j<n; j++)
            {
                p[cnt].u=i,p[cnt].v=j,p[cnt++].w=sqrt((double)((x[i]-x[j])*(x[i]-x[j])+(y[i]-y[j])*(y[i]-y[j])));
            }
        sort(p,p+cnt,cmp);
        double mincost=kruskal();
        double ans=-1;
        for(int i=0; i<n-1; i++)
        {
            init();
            for(int j=0; j<n-1; j++)
            {
                if(i==j)
                    continue;
                int aa=fin(mst[j].u);
                int bb=fin(mst[j].v);
                if(aa!=bb)
                {
                    pre[aa]=bb;
                }
            }
            int xx=fin(mst[i].u);
            int yy=fin(mst[i].v);
            int x1=-1,x2=-1;
            mincost-=mst[i].w;
            for(int j=0; j<n; j++)
                if(fin(j)!=xx)
                    x1=max(x1,a[j]);
            for(int j=0; j<n; j++)
                if(fin(j)!=yy)
                    x2=max(x2,a[j]);
            ans=max(ans,(x1+x2)*1.0/mincost);
            mincost+=mst[i].w;
        }
        printf("%.2f\n",ans);
    }
    return 0;
}