POJ 2728 Desert King （最优比率生成树 01分数规划问题）

题意：
在这么一个图中求一棵生成树,这棵树的单位长度的花费最小是多少？
 
思路：
 
最小生成树的表达式可以这样写
∑x[i]*dis[i]-minsum>=0;(x[i]为0或者1,要求为一棵生成树)
 
这个题目ans<=(∑cost[i]*x[i])/(∑dis[i]*x[i]).变形可得∑x[i]*(cost[i]-dis[i]*ans)-0>=0;cost[i]-dis[i]*ans就相当于最小生成树中的dis[i];
二分ans,check的时候跑一边prime算法看得到的最小生成树的和是否>=0,最后可得答案;

#include <cstdio>
#include <iostream>
#include <cstring>
#include <string>
#include <cstdlib>
#include <algorithm>
#include <cmath>
#include <vector>
#include <set>
#include <list>
#include <queue>
#include <map>
using namespace std;
#define L(i) i<<1
#define R(i) i<<1|1
#define INF  0x3f3f3f3f
#define pi acos(-1.0)
#define eps 1e-9
#define maxn 100010
#define MOD 1000000007
struct node
{
int x,y,z;
}a[1010];
int n;
double mp[1010][1010];
double dis[1010],lowcost[1010];
int nearvex[1010];

double get_dis(int i,int j)
{
return sqrt((a[i].x-a[j].x)*(a[i].x-a[j].x)+(a[i].y-a[j].y)*(a[i].y-a[j].y));
}
int prim(double mid)
{
int k = 0;
double sum = 0;
for(int i = 1; i <= n; i++)
{
lowcost[i] = abs(a[1].z - a[i].z) - mp[1][i] * mid;
nearvex[i] = 1;
}
nearvex[1] = -1;
for(int i = 1; i <= n; i++)
{
int v = -1;
double Min = INF;
for(int j = 1; j <= n; j++)
if(nearvex[j] != -1 && lowcost[j] < Min)
{
v = j;
Min = lowcost[j];
}
if(v != -1)
{
nearvex[v] = -1;
sum += lowcost[v];
for(int j = 1; j <= n; j++)
if(nearvex[j] != -1 && abs(a[v].z - a[j].z)-mp[v][j]*mid < lowcost[j])
{
lowcost[j] = abs(a[v].z - a[j].z)-mp[v][j]*mid;
nearvex[j] = v;
}
}
}
return sum < 0;
}
int main()
{
int t,C = 1;
//scanf("%d",&t);
while(scanf("%d",&n) && n)
{
for(int i = 1; i <= n; i++)
scanf("%d%d%d",&a[i].x,&a[i].y,&a[i].z);
for(int i = 1; i <= n; i++)
for(int j = i; j <= n; j++)
mp[i][j] = mp[j][i] = get_dis(i,j);
double l = 0,r = 100;
while(r - l > 0.00001)
{
double mid = (l + r) / 2;
if(prim(mid))
r = mid;
else
l = mid;
}
printf("%.3f\n",l);
}
return 0;
}