图论训练2A--hdu4081

Qin Shi Huang's National Road System

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5986    Accepted Submission(s): 2085

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

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

 

Sample Output

65.0070.00

 

题意：

有N个城市，修n-1条路，输入信息为城市坐标以及该城市人口，n-1条路中，有一条是魔法路，输出A/B,A是魔法路两边城市人口和，B是非魔法路路径总长，求A/B的最大值，

解：

A/B，B越小越好，

先求出最小生成树，枚举每条路，如果是最小生成树中的边，就把这条边直接去掉，A/（D-cost[i][j]),如果不是，加上i-->j这条边，就会形成一个环，那么要去掉i-->j 这个环上的最长的路径，利用prime求次小生成树算法，MAX[ i ][ j ]表示最小生成树上i-->j的最大距离，操作A/(D-Max[i][j]),

#include <iostream>

#include <stdio.h>

#include <string.h>#include <stdlib.h>#include <math.h>using namespace std;const int maxn=1010;const double  inf=2147483647.0;bool vis[maxn];double lowc[maxn];int pre[maxn];double  Max[maxn][maxn];bool used[maxn][maxn];double cost[maxn][maxn];double max(double a,double b){    if(a>b)        return a;    return b;}double Prim(int n){    double  ans=0;    memset(vis,false,sizeof(vis));    memset(used,false,sizeof(used));    memset(Max,0,sizeof(Max));    vis[0]=true;    pre[0]=-1;    for(int i=1; i<n; ++i)    {        lowc[i] = cost[0][i];        pre[i] = 0;    }    for(int i=1; i<n; ++i)    {        int u=-1;        for(int j=0; j<n; ++j)            if(!vis[j])            {                if(u==-1 || lowc[j]<lowc[u])                    u = j;            }        used[u][pre[u]]=used[pre[u]][u] = true;        ans+= cost[pre[u]][u];        vis[u] = true;        for(int j=0; j<n; ++j)        {            if(vis[j]&&j!=u)            {                Max[u][j]=Max[j][u]=max(Max[j][pre[u]], lowc[u]);            }            if(!vis[j])            {                if(lowc[j]>cost[u][j])                {                    lowc[j] = cost[u][j];                    pre[j] = u;                }            }        }    }    return ans;}struct point{    double  x,y,cos;} d[maxn];double g(point a,point b){    return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));}int main(){    int t;    int n;    cin>>t;    while(t--)    {        scanf("%d",&n);        memset(cost,0,sizeof(cost));        for(int i=0; i<n; i++)            scanf("%lf%lf%lf",&d[i].x,&d[i].y,&d[i].cos);        for(int i=0; i<n; i++)            for(int j=0; j<n; j++)                if(i!=j)                    cost[i][j]=g(d[i],d[j]);        double B=Prim(n);        double ans=-1.0;        for(int i=0; i<n; i++)        {            for(int j=0; j<n; j++)            {                if(i!=j)                {                    if(used[i][j])                    {                        ans=max(ans,(d[i].cos+d[j].cos)/(B-cost[i][j]));                    }                    else                    {                        ans=max(ans,(d[i].cos+d[j].cos)/(B-Max[i][j]));                    }                }            }        }        printf("%.2lf\n",ans);    }    return 0;}