[POJ 2728] Desert King

[题目链接]

        http://poj.org/problem?id=2728

[算法]

        0/1分数规划 + 最小生成树
[代码]

        在本题中，prim算法的时间复杂度优于kruskal算法，且实现较为容易，因此，笔者程序中使用的是prim算法

         

#include <algorithm>  
#include <bitset>  
#include <cctype>  
#include <cerrno>  
#include <clocale>  
#include <cmath>  
#include <complex>  
#include <cstdio>  
#include <cstdlib>  
#include <cstring>  
#include <ctime>  
#include <deque>  
#include <exception>  
#include <fstream>  
#include <functional>  
#include <limits>  
#include <list>  
#include <map>  
#include <iomanip>  
#include <ios>  
#include <iosfwd>  
#include <iostream>  
#include <istream>  
#include <ostream>  
#include <queue>  
#include <set>  
#include <sstream>  
#include <stdexcept>  
#include <streambuf>  
#include <string>  
#include <utility>  
#include <vector>  
#include <cwchar>  
#include <cwctype>  
#include <stack>  
#include <limits.h>
using namespace std;
#define MAXN 1010
const double eps = 1e-7;
const double INF = 1e50;

int i,j,n;
double l,r,mid,ans;
double dist[MAXN][MAXN],cost[MAXN][MAXN];
double x[MAXN],y[MAXN],h[MAXN];

inline double prim(double mid)
{
        int i,j,p;
        double mn,res = 0;
        static double d[MAXN];
        static bool visited[MAXN];
        for (i = 1; i <= n; i++) 
        {
                d[i] = INF;
                visited[i] = false;
        }
        d[1] = 0;
        for (i = 1; i < n; i++)
        {
                p = 0;
                mn = INF;
                for (j = 1; j <= n; j++)
                {
                        if (!visited[j] && d[j] < mn)
                        {
                                mn = d[j];
                                p = j;
                        }
                }
                visited[p] = true;
                for (j = 1; j <= n; j++)
                {
                        if (!visited[j])
                                d[j] = min(d[j],1.0 * cost[p][j] - mid * dist[p][j]);
                }
        }
        for (i = 1; i <= n; i++) res += d[i];
        return res;
}
int main() 
{
        
        while (scanf("%d",&n) && n)
        {
                for (i = 1; i <= n; i++) scanf("%lf%lf%lf",&x[i],&y[i],&h[i]);
                for (i = 1; i <= n; i++)
                {
                        for (j = i + 1; j <= n; j++)
                        {
                                dist[i][j] = dist[j][i] = 1.0 * sqrt(1.0 * (x[i] - x[j]) * (x[i] - x[j]) + (y[i] - y[j]) * (y[i] - y[j]));
                                cost[i][j] = cost[j][i] = 1.0 * abs(h[i] - h[j]);
                        }
                }
                l = 0.0; r = 100.00;
                while (r - l > eps)
                {
                        mid = (l + r) / 2.00;
                        if (prim(mid) >= 0) 
                        {
                                ans = mid;
                                l = mid;
                        } else r = mid;
                }
                printf("%.3f
",ans);
        }
        
        return 0;
    
}