[POJ2728 Desert King]

[关键字]：最优比率生成树

[题目大意]：对于一个图有n个点,m条边，每条边有一个ci,bi。求其生成树使得∑ci / ∑bi 最小（或最大）。为方便讨论这里统一记录求min ( ∑ci / ∑bi )

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[分析]：黑书上有详细证明，有些数学符号用电脑打也不太好打，实在懒得写了，这个博客上解答的很详细http://hi.baidu.com/zzningxp/blog/item/b2d1b4ec1f8bbc2262d09fc9.html。

[代码]：

View Code

 1 //迭代法
 2 #include<cstdio>
 3 #include<cstdlib>
 4 #include<cstring>
 5 #include<cmath>
 6 #include<iostream>
 7 #include<algorithm>
 8 using namespace std;
 9 
10 const double esp=1e-5;
11 const int MAXN=1010;
12 const double INF=1e9;
13 
14 struct point
15 {
16        int x,y,z;
17 } points[MAXN];
18 
19 int N;
20 bool v[MAXN];
21 double dis[MAXN];
22 int pre[MAXN];
23 
24 double cal(int x,int y)
25 {
26        return sqrt(1.*(points[x].x-points[y].x)*(points[x].x-points[y].x)+1.*(points[x].y-points[y].y)*(points[x].y-points[y].y));
27 }
28 
29 double prim(double x)
30 {
31        memset(v,0,sizeof(v));
32        for (int i=2;i<=N;i++)
33        {
34            dis[i]=abs(points[1].z-points[i].z)-x*cal(1,i);
35            pre[i]=1;
36        }
37        dis[1]=0;
38        v[1]=1;       
39        double cost=0,len=0;
40        for (int i=1;i<N;i++)
41        {
42            double min=INF;
43            int minj;
44            for (int j=2;j<=N;j++)
45                if (!v[j] && min>dis[j])
46                {
47                          min=dis[j];
48                          minj=j;
49                }
50            v[minj]=1;
51            cost+=abs(points[pre[minj]].z-points[minj].z);
52            len+=cal(pre[minj],minj);
53            for (int j=2;j<=N;j++)
54            {
55                double val=abs(points[j].z-points[minj].z)-x*cal(minj,j);
56                if (!v[j] && dis[j]>val)
57                {
58                          dis[j]=val;
59                          pre[j]=minj;
60                }
61            }
62        }
63        return cost/len;
64 }
65 
66 int main()
67 {
68     while (scanf("%d",&N),N)
69     {
70           for (int i=1;i<=N;i++)
71               scanf("%d%d%d",&points[i].x,&points[i].y,&points[i].z);
72           double a=0,b;
73           while (1)          
74           {                
75                 b=prim(a);
76                 if (fabs(b-a)<esp) break;
77                 a=b;
78           }
79           printf("%.3f\n",b);
80     }
81     system("pause");
82     return 0;
83 }