uvaLive5713 次小生成树

  1 #include <stdio.h>
  2 #include <string.h>
  3 #include <stdlib.h>
  4 #include <algorithm>
  5 #include <iostream>
  6 #include <queue>
  7 #include <stack>
  8 #include <vector>
  9 #include <map>
 10 #include <set>
 11 #include <string>
 12 #include <math.h>
 13 using namespace std;
 14 #pragma warning(disable:4996)
 15 #pragma comment(linker, "/STACK:1024000000,1024000000")
 16 typedef long long LL;                   
 17 const int INF = 1<<30;
 18 /*
 19 
 20 */
 21 const int N = 1000 + 10;
 22 struct Edge
 23 {
 24     int form, to, p;
 25     double dist;
 26     bool operator<(const Edge&rhs)const
 27     {
 28         return dist < rhs.dist;
 29     }
 30 }a[N*N];
 31 int x[N], y[N], p[N];
 32 int father[N];
 33 double maxCost[N][N];
 34 int p2[N][N];
 35 vector<Edge> g[N];
 36 vector<int> st;
 37 int find(int x)
 38 {
 39     if (x == father[x])
 40         return x;
 41     return father[x] = find(father[x]);
 42 }
 43 double kruskal(int n)
 44 {
 45     double sum = 0;
 46     for (int i = 0; i < n; ++i)
 47     {
 48         int fx = find(a[i].form);
 49         int fy = find(a[i].to);
 50         if (fx != fy)
 51         {
 52             //记录生成树，用来等下dfs
 53             g[a[i].form].push_back(a[i]);
 54             swap(a[i].form, a[i].to);
 55             g[a[i].form].push_back(a[i]);
 56             sum += a[i].dist;
 57             father[fx] = fy;
 58         }
 59     }
 60     return sum;
 61 }
 62 
 63 /*
 64 这个dfs其实就是给定一棵树，然后求任意两点之间的最大边权， 因为是任意两点， 那么时间复杂度无疑就是O(n*n)
 65 */
 66 void dfs(int u, int fa, double dis)
 67 {
 68     if (fa!=-1)
 69     for (int i = 0; i < st.size(); ++i)
 70     {
 71         int x = st[i];
 72         maxCost[x][u] = maxCost[u][x] = max(maxCost[x][fa], dis);
 73     }
 74     st.push_back(u);
 75     for (int i = 0; i < g[u].size(); ++i)
 76     {
 77         int v = g[u][i].to;
 78         if (v == fa) continue;
 79         dfs(v, u, g[u][i].dist);
 80     }
 81 }
 82 int main()
 83 {
 84     int t, n, cnt;
 85     double B;
 86     scanf("%d", &t);
 87     while (t--)
 88     {
 89         scanf("%d", &n);
 90         cnt = 0;
 91         st.clear();
 92         for (int i = 0; i < n; ++i)
 93         {
 94             father[i] = i;
 95             g[i].clear();
 96             scanf("%d%d%d", &x[i], &y[i], &p[i]);
 97         }
 98         for (int i = 0; i < n; ++i)
 99         {
100             for (int j = i + 1; j < n; ++j)
101             {
102                 p2[i][j] = p2[j][i] = p[i] + p[j];
103                 //边集数组
104                 a[cnt].form = i;
105                 a[cnt].to = j;
106                 a[cnt].p = p[i] + p[j];
107                 a[cnt++].dist = sqrt((x[i] - x[j])*(x[i] - x[j]) + (y[i] - y[j])*(y[i] - y[j]));
108             }
109         }
110         sort(a, a + cnt);
111         B = kruskal(cnt);
112         dfs(0, -1, 0);
113         double ans = 0;
114 
115         //枚举在哪两个之间建免费的道路
116         for (int i = 0; i < n; ++i)
117             for (int j = i + 1; j < n; ++j)
118                 ans = max(ans, p2[i][j] / (B - maxCost[i][j]));
119         printf("%.2lf\n", ans);
120     }
121     return 0;
122 }