hdu 4081 Qin Shi Huang's National Road System

思路： 建立最小生成树的图，之后进行枚举边(两点)。然后去掉原来生成树上的边。 去掉一条遍之后，原图为两棵树。

A - 分别为两棵树中的最大人口和

B - 生成树边长之和减去一条生成树边的长度。

  1 #include<iostream>

  2 #include<cstdio>

  3 #include<cstdlib>

  4 #include<cstring>

  5 #include<string>

  6 #include<queue>

  7 #include<algorithm>

  8 #include<map>

  9 #include<iomanip>

 10 #include<climits>

 11 #include<string.h>

 12 #include<cmath>

 13 #include<stdlib.h>

 14 #include<vector>

 15 #include<stack>

 16 #include<set>

 17 #define INF 1e7

 18 #define MAXN 10010

 19 #define maxn 1000010

 20 #define Mod 1000007

 21 #define N 1010

 22 using namespace std;

 23 typedef long long LL;

 24 

 25 struct node{

 26     double x, y, num;

 27 }p[N];

 28 double G[N][N], dis[N], A, B;

 29 vector<int> vec[N];

 30 int vis[N];

 31 int T, n;

 32 int x, y;

 33 int pre[N];

 34 

 35 double getdis(node a, node b)

 36 {

 37     return sqrt((a.x - b.x)*(a.x - b.x) + (a.y - b.y)*(a.y - b.y));

 38 }

 39 

 40 void prim()            //对最小生成树建图

 41 {

 42     int now, pos;

 43     double min;

 44     for (int i = 1; i <= n; ++i) dis[i] = INF;

 45     now = 1;

 46     for (int i = 1; i < n; ++i) {

 47         min = INF;

 48         for (int j = 1; j <= n; ++j) {

 49             if (dis[j] > G[now][j])

 50                 dis[j] = G[now][j], pre[j] = now;

 51         }

 52         dis[now] = -1;

 53         for (int j = 1; j <= n; ++j) {

 54             if (dis[j] > 0 && min > dis[j])

 55                 min = dis[j], pos = j;

 56         }

 57         vec[pre[pos]].push_back(pos);

 58         vec[pos].push_back(pre[pos]);

 59         B += min;

 60         now = pos;

 61     }

 62 }

 63 

 64 int dfs(int u, int fa)

 65 {

 66     int max_num = p[u].num;

 67     int ret = u, tt;

 68     for (int i = 0; i < vec[u].size(); ++i) {

 69         int v = vec[u][i];

 70         if (v == fa) continue;

 71         tt = dfs(v, u);

 72         if (max_num < p[tt].num)

 73             max_num = p[tt].num, ret = tt;

 74     }

 75     return ret;

 76 }

 77 

 78 void run()

 79 {

 80     cin >> n;

 81     for (int i = 1; i <= n; ++i)

 82         cin >> p[i].x >> p[i].y >> p[i].num;

 83     for (int i = 1; i <= n; ++i)

 84         for (int j = 1; j <= n; ++j)

 85         {

 86             if (i == j)  continue;

 87             G[i][j] = G[j][i] = getdis(p[i], p[j]);

 88         }

 89     for (int i = 0; i <= n; ++i) vec[i].clear();

 90     B = 0;

 91     memset(pre,-1,sizeof(pre));

 92     prim();

 93     double ans = -1.0, res;

 94     int t1, t2;

 95     for (int i = 1; i <= n; ++i)

 96         for (int j = 0; j < vec[i].size(); ++j) {

 97             int v = vec[i][j];

 98             t1 = dfs(i, v);

 99             t2 = dfs(v, i);

100             res = (p[t1].num + p[t2].num)/(B - G[i][v]);

101             if (res > ans) ans = res;

102         }

103     printf("%.2lf\n",ans);

104 }

105 

106 int main()

107 {

108     cin >> T;

109     while (T--) {

110         run();

111     }

112     return 0;

113 }