UVALive - 5713 Qin Shi Huang's National Road System-优快云博客

本文介绍了一种使用Prim算法结合邻接矩阵来求解最小生成树中两点间最大距离的方法，并通过枚举所有点对来找出特定条件下最优解的过程。代码实现了从输入点坐标到输出结果的完整流程。

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prim + 邻接矩阵储存两点间最大距离，然后枚举即可。

#include<iostream>
#include<string>
#include<cstdio>
#include<set>
#include<map>
#include<stack>
#include<list>
#include<vector>
#include<queue>
#include<algorithm>
#include<cstring>
#include<cmath>
#include<fstream>

using namespace std;
typedef long long ll;

const int maxn = 1010,INF = 0x3f3f3f3f;

int fa[maxn],n;
vector<int> son[maxn];
int x[maxn],y[maxn],pop[maxn];
double maxr[maxn][maxn];//最小生成树中两点间的最大边
double sum,ans;
double e[maxn][maxn];

struct Edge{
    int to,from;
    double val;
    bool operator < (const Edge &x) const{
        return val < x.val;
    }
};

int getfa(int x){
    if (fa[x] == x) return x;
    else return fa[x] = getfa(fa[x]);
}
inline double pf(double x){
    return x * x;
}
inline double getdis(int u,int v){
    return sqrt(pf(x[u] - x[v]) + pf(y[u] - y[v]));
}

void buildEdge(){
    for(int i = 1;i <= n;++i)
        for(int j = i+1;j <= n;++j){
            double dis = getdis(i, j);
            e[i][j] = e[j][i] = dis;
        }
}

void prim()
{
    bool vis[maxn] = {};
    double dist[maxn] = {};
    for(int i = 1;i <= n;++i) {
        dist[i] = INF;
    }
    vector<int> v;
    int now = 1;
    int pre[maxn] = {};
    sum = 0;
    vis[1] = true;
    dist[1] = 0;
    v.push_back(1);
    for(int i = 1;i < n;++i)
    {
        for(int j = 1;j <= n;++j)
        {
            if(!vis[j] && e[now][j] < dist[j]){
                pre[j] = now;
                dist[j] = e[now][j];
            }
        }
        double mi = INF;
        int k = i;
        for(int j = 1;j <= n;++j)
        {
            if(!vis[j] && dist[j] < mi)
            {
                mi = dist[j];
                k = j;
            }
        }
        sum += dist[k];
        now = k;
        vis[k] = 1;
        for(int j = 0;j < v.size();++j)
        {
            maxr[v[j]][k] = maxr[k][v[j]] = max(maxr[v[j]][pre[k]],dist[k]);
        }
        v.push_back(k);
    }
}

void solve(){
    double ans = 0;
    for(int i = 1;i <= n;++i)
        for(int j = i + 1;j <= n;++j){
            double A = pop[i] + pop[j];
            double B = sum - maxr[i][j];
            ans = max(ans,A/B);
        }
    printf("%.2f\n",ans);
}

void init(){
    cin >> n;
    for(int i = 1;i <= n;++i){
        cin >> x[i] >> y[i] >> pop[i];
    }
    memset(maxr,0,sizeof maxr);
    for(int i = 1;i <= n;++i) {
        for(int j = i + 1;j <= n;++j)
            e[i][j] = e[j][i] = INF;
    }
    buildEdge();
    prim();
    solve();
}

int main(){
    int t;
    cin >> t;
    while(t--){
        init();
    }
}