LA5713 Qin Shi Huang's National Road System

题目大意：秦始皇要在n个城市之间修筑一条道路使得任意两个城市均可连通。有个道士可以用法力帮忙修一条路。秦始皇希望其他的道路总长B最短且用法术连接的两个城市的人口之和A尽量大，因此下令寻找一个A / B的最大方案。（转自http://blog.csdn.net/murmured/article/details/18865721，侵删）

题解：考虑修哪一条路，此时A确定，断掉这条路后，形成两个连通块，不难发现两个连通块都应是MST才能让B最短。此时B为两个连通块的MST权值和
这个过程相当于选了一条路u->v后，在整张图的MST上，把MST上u->v路径上最大边删掉。

其实不用枚举边，枚举u和v即可

空间开小了，RE了无数发。。。

 #include <iostream>
 #include <cstdio>
 #include <cstring>
 #include <cstdlib>
 #include <algorithm>
 #include <queue>
 #include <vector>
 #include <map>
 #include <string>
 #include <cmath>
 #define min(a, b) ((a) < (b) ? (a) : (b))
 #define max(a, b) ((a) > (b) ? (a) : (b))
 #define abs(a) ((a) < 0 ? (-1 * (a)) : (a))
 template<class T>
 inline void swap(T &a, T &b)
 {
     T tmp = a;a = b;b = tmp;
 }
 inline void read(int &x)
 {
     x = ;char ch = getchar(), c = ch;
     while(ch < '' || ch > '') c = ch, ch = getchar();
     while(ch <= '' && ch >= '') x = x *  + ch - '', ch = getchar();
     if(c == '-') x = -x;
 }

 const int INF = 0x3f3f3f3f;
 const int MAXN =  + ;
 const int MAXM =  + ;

 int x[MAXM], y[MAXM], node[MAXM], n, m, t, u[MAXM], v[MAXM], cnt[MAXM], fa[MAXM], vis[MAXN];
 double ma[MAXN][MAXN], w[MAXM];
 struct Edge
 {
     int u,v,nxt;
     double w;
     Edge(int _u, int _v, double _w, int _nxt){u = _u;v = _v;w = _w;nxt = _nxt;}
     Edge(){}
 }edge[MAXM << ];
 int head[MAXN], cntt;
 inline void insert(int a, int b, double c)
 {
     edge[++cntt] = Edge(a,b,c,head[a]);
     head[a] = cntt;
 }
 int find(int x){return x == fa[x] ? x : fa[x] = find(fa[x]);}
 int cmp(int x, int y){return w[x] < w[y];}

 int tiaoshi;
 void dfs(int x, int pre)
 {
     vis[x] = ;
     for(int pos = head[x];pos;pos = edge[pos].nxt)
     {
         int v = edge[pos].v;
         if(v == pre) continue;
         for(int i = ;i <= n;++ i)
             if(i == v || !vis[i]) continue;
             else ma[i][v] =ma[v][i] = max(ma[i][v], max(ma[v][i], max(ma[x][i], max(ma[i][x], edge[pos].w))));
         dfs(v, x);
     }
 }
 int main()
 {
     read(t);
     for(;t;--t)
     {
         tiaoshi = ;
         read(n);memset(ma, , sizeof(ma)), memset(vis, , sizeof(vis)), memset(head, , sizeof(head)), cntt = , m = ;
         for(int i = ;i <= n;++ i) read(x[i]), read(y[i]), read(node[i]);
         for(int i = ;i <= n;++ i)
             for(int j = i + ;j <= n;++ j)
                 ++ m, u[m] = i, v[m] = j, w[m] = sqrt((x[i] - x[j]) * (x[i] - x[j]) + (y[i] - y[j]) * (y[i] - y[j])), cnt[m] = m, fa[m] = m;
         std::sort(cnt + , cnt +  + m, cmp);
         int tmp = ;
         double ans = ;
         double ans2 = ;
         for(int i = ;i <= m;++ i)
         {
             int f1 = find(u[cnt[i]]), f2 = find(v[cnt[i]]);
             if(f1 == f2) continue;
             fa[f1] = f2;
             insert(u[cnt[i]], v[cnt[i]], w[cnt[i]]), insert(v[cnt[i]], u[cnt[i]], w[cnt[i]]);
             ans += w[cnt[i]];
             ++ tmp;if(tmp == n - ) break;
         }
         dfs(, -);
         for(int i = ;i <= n;++ i)
             for(int j = i + ;j <= n;++j)
             {
                 if(ans == ma[i][j]) continue;
                 ans2 = max(ans2, (double)(node[i] + node[j]) / (ans - ma[i][j]));
             }
         printf("%.2lf\n", ans2);
     }
     return ;
 }

LA5713