hdu 4081 Qin Shi Huang's National Road System 树的基本性质 or 次小生成树思想 难度:1

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

思路:这个思路提示的很直接了,树的性质就是只要加边就会成环,减去这环上的任意边还是一棵联通的树,

跑出一棵最小生成树来,对任意两点试着加花费为0的边.取消掉花费最大的那条边,找到最优答案即可,因为不能每次建边都跑环,所以生成树时预处理

最小生成树的任意联通部分还是最小生成树

#include <cstdio>
#include <cstring>
#include  <cmath>
#include <algorithm>
#include <queue>
using namespace std;
const int maxn=1105;
int n;
struct P{
    int x,y,p;
}v[maxn];
double d[maxn][maxn];
bool vis[maxn];
double maxd[maxn][maxn];
typedef pair<int ,int > point;
typedef pair<double,point> pr;

priority_queue<pr,vector<pr>,greater<pr> >que;
double prim(){//最小生成树
    memset(vis,0,sizeof(vis));
    vis[0]=true;
    int num=1;
    while(!que.empty())que.pop();
    double ans=0;
    for(int i=1;i<n;i++){
        que.push(pr(d[0][i],point(i,0)));
    }
    while(num<n){
        double td=que.top().first;
        int  t=que.top().second.first;
        int f=que.top().second.second;
        que.pop();
      if(vis[t])continue;

        vis[t]=true;num++;ans+=td;

        maxd[t][f]=maxd[f][t]=td;

        for(int i=0;i<n;i++){
            if(!vis[i]){
                que.push(pr(d[t][i],point(i,t)));
            }
            else {
                if(i!=t){
                    maxd[t][i]=maxd[i][t]=max(maxd[f][i],td);//此边就是联系当前边到树中所有边的目前最大边
                }
            }
        }
    }

    return ans;
}
int main(){
    int t;
    scanf("%d",&t);
    while(t--){
        scanf("%d",&n);
        for(int i=0;i<n;i++){
            scanf("%d%d%d",&v[i].x,&v[i].y,&v[i].p);
        }
        for(int i=0;i<n;i++){//建图
            for(int j=0;j<=i;j++){
                d[i][j]=d[j][i]=sqrt((v[i].x-v[j].x)*(v[i].x-v[j].x)+(v[i].y-v[j].y)*(v[i].y-v[j].y));
            }
        }
        double allt=prim();
        double maxrate=-1;
        for(int i=0;i<n;i++){//求最优解
            for(int j=0;j<i;j++){
                double rate=(v[i].p+v[j].p)/(allt-maxd[i][j]);
                maxrate=max(maxrate,rate);
            }
        }
        printf("%.2f\n",maxrate);
    }
    return 0;
}