Qin Shi Huang's National Road System

秦始皇之路网优化

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本文探讨了一个关于古代中国秦始皇时期道路系统的数学问题，通过寻找最优的次小生成树来最小化非魔法道路的总长度，并最大化受益人口与成本的比例。

次小生成树优先队列优化

Qin Shi Huang's National Road System

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5613 Accepted Submission(s): 1975

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

Sample Output

65.00
70.00

#include <cstdio>
#include <cstring>
#include <vector>
#include <cmath>
#include <queue>
using namespace std;
const int maxn=1<<30;
int n;
int coor[1010][2];
double pri[1010];
double G[1010][1010];
bool used[1010];
double lowc[1010];
double Max[1010][1010];
double get_dis(int x1,int y1,int x2,int y2){
    return sqrt((x1-x2)*(x1-x2)*1.0+(y1-y2)*(y1-y2)*1.0);
}
double cnt;
struct node{
    int v;
    double w;
    node(int vv,double ww):v(vv),w(ww){}
    bool operator<(const node &other)const{
        return w>other.w;
    }
};
void prim(){
    memset(Max,0,sizeof(Max));
    memset(used,0,sizeof(used));
    int pre[1010];
    cnt=0;
    for(int i=0;i<=n;i++){
        lowc[i]=G[1][i];
    }
    priority_queue<node> q;
    q.push(node(1,0));
    pre[1]=0;
    while(!q.empty()){
        node p=q.top();
        q.pop();
        int u=p.v;
        if(used[u])continue;
        used[u]=true;
        cnt+=p.w;
        for(int i=1;i<=n;i++){
            if(used[i]&&i!=u)
                Max[u][i]=Max[i][u]=max(Max[i][pre[u]],lowc[u]);
            if(used[i]||lowc[i]<G[u][i])continue;
            lowc[i]=G[u][i];
            q.push(node(i,lowc[i]));
            pre[i]=u;
        }
    }
}
int main(){
    int T;
    scanf("%d",&T);
    while(T--){
        memset(G,0,sizeof(G));
        scanf("%d",&n);
        for(int i=1;i<=n;i++)
            scanf("%d%d%lf",&coor[i][0],&coor[i][1],&pri[i]);
        for(int i=0;i<=n;i++)
            G[i][i]=0;
        for(int i=1;i<=n;i++)
            for(int j=i+1;j<=n;j++)
                G[i][j]=G[j][i]=get_dis(coor[i][0],coor[i][1],coor[j][0],coor[j][1]);
        prim();
        double ans=-1;
        for(int i=1;i<=n;i++)
        for(int j=i+1;j<=n;j++){
            if((pri[i]+pri[j])/(cnt-Max[i][j])>ans)
                    ans=(pri[i]+pri[j])/(cnt-Max[i][j]);
        }
        printf("%.2lf\n",ans);
    }
}

求出了次小生成树中需要的Max数组，修改下即可求得次小生成树