Desert King(POJ2728+最优比率生成树+二分）

题目链接：http://poj.org/problem?id=2728

题目：

题意：求一颗生成树，使得费用与距离的比值最小，其中距离等于两点之间的平面欧拉距离，费用为z坐标之差。

思路：

　　由上图我们可以得知，我们只需对x进行二分（最大化平均值），以cost[i]-len[i]*x为边权跑prime即可。

代码实现如下：

 #include <set>
 #include <map>
 #include <queue>
 #include <stack>
 #include <cmath>
 #include <bitset>
 #include <cstdio>
 #include <string>
 #include <vector>
 #include <cstdlib>
 #include <cstring>
 #include <iostream>
 #include <algorithm>
 using namespace std;

 typedef long long ll;
 typedef pair<ll, ll> pll;
 typedef pair<ll, int> pli;
 typedef pair<int, ll> pil;;
 typedef pair<int, int> pii;
 typedef unsigned long long ull;

 #define lson i<<1
 #define rson i<<1|1
 #define bug printf("*********\n");
 #define FIN freopen("D://code//in.txt", "r", stdin);
 #define debug(x) cout<<"["<<x<<"]" <<endl;
 #define IO ios::sync_with_stdio(false),cin.tie(0);

 const double eps = 1e-;
 const int mod = ;
 const int maxn =  + ;
 const double pi = acos(-);
 const int inf = 0x3f3f3f3f;
 const ll INF = 0x3f3f3f3f3f3f3f;

 int n, x, y, z;
 double ans;
 int vis[maxn];
 double mp[maxn][maxn], dis[maxn];

 struct node {
     int x, y ,z;
 }p[maxn];

 double dist(node& a, node& b) {
     return sqrt((a.x - b.x) * (a.x - b.x) + (a.y - b.y) * (a.y - b.y));
 }

 bool prime(double x) {
     for(int i = ; i <= n; i++) {
         dis[i] = inf;
         vis[i] = ;
     }
     ans = ;
     dis[] = ;
     for(int i = ; ; i++) {
         double mx = inf;
         int t = -;
         for(int j = ; j <= n; j++) {
             if(!vis[j] & mx > dis[j]) {
                 mx = dis[j];
                 t = j;
             }
         }
         if(t == -) break;
         ans += mx;
         vis[t] = ;
         for(int j = ; j <= n; j++) {
             if(!vis[j] && dis[j] > fabs(p[t].z - p[j].z) - mp[t][j] * x) {
                 dis[j] = fabs(p[t].z - p[j].z) - mp[t][j] * x;
             }
         }
     }
     return ans >= ;
 }

 int main() {
     //FIN;
     while(~scanf("%d", &n) && n) {
         for(int i = ; i <= n; i++) {
             scanf("%d%d%d", &p[i].x, &p[i].y, &p[i].z);
         }
         for(int i = ; i <= n; i++) {
             for(int j = ; j <= n; j++) {
                 mp[i][j] = dist(p[i], p[j]);
             }
         }
         double ub = , lb = , mid;
         while(ub - lb > eps) {
             mid = (ub + lb) / ;
             if(prime(mid)) lb = mid;
             else ub = mid;
         }
         printf("%.3f\n", lb);
     }
     return ;
 }