Hdu4081qin Shi Huang ' s national Road System sub-niche into a tree

Given the coordinates of each point and the amount of its population, find a spanning tree, which has a side cost of 0//ask A/b for the maximum value, a the sum of the two points connected by this edge//b the sum of the lengths of the edges other than this edge//first find the smallest spanning tree, then enumerate all the edges, If this edge is the edge of the smallest spanning tree,//directly calculates a A/(Sum-edge), if not, then this edge adds a//ring on the smallest spanning tree, removing the longest edge in the ring remains a tree,//with dp[u][v] represents the size of the longest side that passes through the unique path from U to V in the smallest spanning tree//dp[u][V[i]] = dp[V[i]][u]= Max (map[F[u]][u], dp[F[u]][v[i]]);#include <cstdio>#include <cstring>#include <iostream>#include <cmath>using namespace std; const int MAXN = 1010; const int inf = 0x3f3f3f3f;d ouble dp[MAXN][MAXN];d ouble map[MAXN][MAXN];d ouble X[MAXN], Y[MAXN], P[MAXN]; int F[MAXN]; int VIS[MAXN]; int V[MAXN];d ouble DIS[MAXN]; int n;d ouble Prim () {int len = 0;memset (Vis, 0, sizeof (VIS));memset (DP, 0, sizeof (DP));for (int i = 2;i <=n; i++)dis[i] = inf;dis[1] = 0; F[1] = 1;V[++len] = 1;double ans = 0;While (1)    {double mi = inf; int u;for (int i = 1;i <= n;i++)if (!vis[i]&&dis[i] < mi)mi = dis[u = i];if (mi = = inf) break;Vis[u] = 1;ans + = mi;for (int i = 1;i <= len; i++)Dp[u][v[i]] = dp[v[i]][u]= max (Map[f[u]][u], dp[f[u]][v[i]);V[++len] = u;for (int i = 1;i <= n;i++)if (!vis[i] && map[u][i] < dis[i])        {Dis[i] = map[u][i];f[i] = u;        }    }return ans;}int Main () {//freopen ("In.txt", "R", stdin);int t;scanf ("%d", &t);While (t--)    {scanf ("%d", &n);for (int i = 1;i <= n;i++)        {scanf ("%lf%lf%lf", &x[i], &y[i], &p[i]);For (int j = 1;j <= i;j++)Map[i][j] = map[j][i] = sqrt ((x[i]-x[j]) * (X[i]-x[j]) + (Y[i]-y[j]) * (Y[i]-y[j ]));        }Double sum = Prim ();double ans = 0;for (int i = 1;i <= n;i++)For (int j = 1; j < I; j + +)ans = max (ans, (p[i] + p[j])/(Sum-dp[i][j]));printf ("%.2lf\n", ans);    }return 0;}

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Hdu4081qin Shi Huang ' s national Road System sub-niche into a tree