最小生成树 + 01分数规划
好烦的卡精度题，调了一下午。。
题意大概就是给你一张稠密图，每条边都有收益和花费，要求一个花费之和与收益之和的比值和最小的生成树。
当然我们可以看成收益之和与花费之和的比值最大的生成树。。
这显然是一个01分数规划问题。然后二分枚举猜测的L值跑最大生成树就行了，若最大生成树大于等于0，代表我们的最大值大于等于L，于是mid = l，若最大生成树小于0，代表我们的最大值小于L， 于是mid = r。
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <cstring>
#include <iomanip>
#define INF 0x3f3f3f3f
#define full(a, b) memset(a, b, sizeof a)
using namespace std;
typedef long long ll;
inline int lowbit(int x){ return x & (-x); }
inline int read(){
int X = 0, w = 0; char ch = 0;
while(!isdigit(ch)) { w |= ch == '-'; ch = getchar(); }
while(isdigit(ch)) X = (X << 3) + (X << 1) + (ch ^ 48), ch = getchar();
return w ? -X : X;
}
inline int gcd(int a, int b){ return a % b ? gcd(b, a % b) : b; }
inline int lcm(int a, int b){ return a / gcd(a, b) * b; }
template<typename T>
inline T max(T x, T y, T z){ return max(max(x, y), z); }
template<typename T>
inline T min(T x, T y, T z){ return min(min(x, y), z); }
template<typename A, typename B, typename C>
inline A fpow(A x, B p, C lyd){
A ans = 1;
for(; p; p >>= 1, x = 1LL * x * x % lyd)if(p & 1)ans = 1LL * x * ans % lyd;
return ans;
}
const int N = 1005;
const double eps = 1e-7;
int n;
double x[N], y[N], z[N];
double dist[N][N], cost[N][N], cur[N][N], mst, d[N];
bool vis[N];
inline double getDis(double x1, double y1, double x2, double y2){
return sqrt((x1 - x2) * (x1 - x2) + (y1 - y2) * (y1 - y2));
}
inline void build(double mid){
for(int i = 1; i <= n; i ++){
for(int j = 1; j <= n; j ++) cur[i][j] = -INF;
}
for(int i = 1; i <= n; i ++){
for(int j = 1; j <= n; j ++){
if(i == j) continue;
cur[i][j] = dist[i][j] - mid * cost[i][j];
}
}
}
inline void prim(){
for(int i = 1; i <= n; i ++){
d[i] = -INF;
vis[i] = false;
}
d[1] = mst = 0.0;
for(int i = 1; i < n; i ++){
int x = 0;
for(int j = 1; j <= n; j ++){
if(!vis[j] && (x == 0 || d[j] > d[x])) x = j;
}
vis[x] = true;
for(int j = 1; j <= n; j ++){
if(!vis[j]) d[j] = max(d[j], cur[x][j]);
}
}
for(int i = 2; i <= n; i ++) mst += d[i];
}
inline bool check(){
prim();
return mst >= 0.0;
}
int main(){
ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
while(cin >> n && n){
for(int i = 1; i <= n; i ++)
cin >> x[i] >> y[i] >> z[i];
for(int i = 1; i <= n; i ++){
for(int j = i + 1; j <= n; j ++){
dist[i][j] = dist[j][i] = getDis(x[i], y[i], x[j], y[j]);
cost[i][j] = cost[j][i] = fabs(z[i] - z[j]);
}
}
double l = 0.0, r = 40.0;
while(r - l >= eps){
double mid = (l + r) / 2;
build(mid);
if(check()) l = mid;
else r = mid;
}
printf("%.3lf
", 1 / l);
}
return 0;
}