hdu 4081

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

241 1 201 2 30200 2 80200 1 10031 1 201 2 302 2 40

 

Sample Output

65.0070.00
题意：给n个位置的坐标以及每个坐标的人数，让所有的城市联通，并且我们还有一次施放魔法的机会（连接任意两个城市），让我们求A/B的最大值，A表示两个城市的人数 B表示剩余城市全部联通所需修建的路长
思路：直接暴力枚举施放魔法的路，但是如果每次都重新计算一遍最小生成树的话肯定会超时，那我们需要怎么办
首先我们分析，有没有一些特殊的路径我们可以简单计算去掉它之后的最小生成树，我们就会发现如果我们要去掉的边在最小生成树上，那么我们直接用最小生成树的值减去我们要去掉的路径的长度，若果不在的话我们可以像生成次小生成树那样处理，
ac代码：
#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <cmath>using namespace std;const int maxn =1005;struct node{    double x,y,v;}data[maxn];const double inf = 0x3f3f3f3f;double Map[maxn][maxn];double Max[maxn][maxn];bool used[maxn][maxn];int pre[maxn];double dis[maxn];double prim(int n){    double ans = 0;    bool vis[maxn];    memset(vis, false, sizeof(vis));    memset(used, false, sizeof(used));    memset(Max, 0, sizeof(Max));    for (int i = 2; i <= n; i++)    {        dis[i] = Map[1][i];        pre[i] = 1;    }    pre[1] = 0;    dis[1] = 0;    vis[1] = true;    for (int i = 2; i <= n; i++)    {        double min_dis = inf;        int k;        for (int j = 1; j <= n; j++)        {            if (!vis[j] && min_dis > dis[j])            {                min_dis = dis[j];                k = j;            }        }        if (min_dis == inf) return -1;        ans += min_dis;        vis[k] = true;        used[k][pre[k]] = used[pre[k]][k] = true;        for (int j = 1; j <= n; j++)        {            if (vis[j]&&j!=k) Max[j][k] = Max[k][j] = max(Max[j][pre[k]], dis[k]);            if (!vis[j] && dis[j] > Map[k][j])            {                dis[j] = Map[k][j];                pre[j] = k;            }        }    }    return ans;}double solve(int n, double min_ans){    double pep=-1,ds=1;    for(int i=1;i<=n;i++)    {        for(int j=i+1;j<=n;j++)        {            if(used[i][j]==true)            {               if(pep*(min_ans-Map[i][j])<=(data[i].v+data[j].v)*ds)               {                   pep=data[i].v+data[j].v;                   ds=min_ans-Map[i][j];               }            }            else            {               if(pep*(min_ans-Max[i][j])<=(data[i].v+data[j].v)*ds)               {                   pep=data[i].v+data[j].v;                   ds=min_ans-Max[i][j];               }            }        }    }    return pep/ds;}int main(){    int T, n, m;    scanf("%d", &T);    while (T--)    {        scanf("%d",&n);        for(int i=1;i<=n;i++)        scanf("%lf%lf%lf",&data[i].x,&data[i].y,&data[i].v);        for(int i=1;i<=n;i++)        {            for(int j=i+1;j<=n;j++)            {                double d=sqrt((data[i].x-data[j].x)*(data[i].x-data[j].x)+((data[i].y-data[j].y))*((data[i].y-data[j].y)));                Map[i][j]=d;                Map[j][i]=d;            }        }        double p=prim(n);        double q=solve(n,p);        printf("%.2lf\n",q);    }    return 0;}