HDUOJ----2489 Minimal Ratio Tree

Minimal Ratio Tree

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2180    Accepted Submission(s): 630

Problem Description

For a tree, which nodes and edges are all weighted, the ratio of it is calculated according to the following equation.




Given a complete graph of n nodes with all nodes and edges weighted, your task is to find a tree, which is a sub-graph of the original graph, with m nodes and whose ratio is the smallest among all the trees of m nodes in the graph.

 

 

Input

Input contains multiple test cases. The first line of each test case contains two integers n (2<=n<=15) and m (2<=m<=n), which stands for the number of nodes in the graph and the number of nodes in the minimal ratio tree. Two zeros end the input. The next line contains n numbers which stand for the weight of each node. The following n lines contain a diagonally symmetrical n×n connectivity matrix with each element shows the weight of the edge connecting one node with another. Of course, the diagonal will be all 0, since there is no edge connecting a node with itself.



All the weights of both nodes and edges (except for the ones on the diagonal of the matrix) are integers and in the range of [1, 100].

The figure below illustrates the first test case in sample input. Node 1 and Node 3 form the minimal ratio tree.  

 

 

Output

For each test case output one line contains a sequence of the m nodes which constructs the minimal ratio tree. Nodes should be arranged in ascending order. If there are several such sequences, pick the one which has the smallest node number; if there's a tie, look at the second smallest node number, etc. Please note that the nodes are numbered from 1 .

 

 

Sample Input

3 2

30 20 10

0 6 2

6 0 3

2 3 0

2 2

1 1

0 2

2 0

0 0

 

 

Sample Output

1 3

1 2

 

 

Source

2008 Asia Regional Beijing

 

 

 

这道题是2008年北京现场比赛的一道题，题意大致意思是给定n个节点的一个图，要你从中选出这小边的权值和除以节点权值和的最小的一个树

于是很好理解的为最小生成树，采用普利姆最小生成树....注意精度的问题，这里我wa了n次

哎，喵了个咪

代码：

  1 #include<string.h>

  2 #include<stdlib.h>

  3 #include<stdio.h>

  4 #include<math.h>

  5 #define max 0x3f3f3f3f

  6 #define maxn 17

  7 int node_weight[maxn];

  8 int edge_weight[maxn][maxn];

  9 int depath[maxn];      //以这些点形成一颗最小生成树

 10 int  m , n ;

 11 double res;

 12 int stu[maxn];

 13 int sub_map[maxn][maxn];

 14 void Prime()

 15 {

 16   int vis[maxn]={0};

 17   int lowc[maxn];

 18   int i,j,k,minc;

 19   double ans=0;

 20   for(i=1;i<=m;i++)  //从n中挑出m个点形成一个子图

 21   {

 22     for(j=1;j<=m;j++)

 23     {

 24       if(edge_weight[depath[i]][depath[j]]==0)

 25          sub_map[i][j]=max;

 26       else

 27          sub_map[i][j]=edge_weight[depath[i]][depath[j]];

 28     }

 29   }

 30   vis[1]=1;

 31   for(i=1;i<=m;i++)

 32   {

 33     lowc[i]=sub_map[1][i];

 34   }

 35   for(i=2;i<=m;i++)

 36   {

 37      minc=max;

 38      k=0;

 39     for(j=2;j<=m;j++)

 40     {

 41       if(vis[j]==0&&minc>lowc[j])

 42         {

 43          minc=lowc[j];

 44          k=j;

 45         }

 46     }

 47     if(minc==max) return ;  //表示没有联通

 48     ans+=minc;

 49     vis[k]=1;

 50     for(j=1 ; j<=m;j++)

 51     {

 52         if(vis[j]==0&&lowc[j]>sub_map[k][j])

 53               lowc[j]=sub_map[k][j];

 54     }

 55   }

 56   int  sum=0;

 57    for(i=1;i<=m;i++)  //统计点权值的和

 58       sum+=node_weight[depath[i]];

 59       ans/=sum;

 60   if(res+0.00000001>=ans)

 61    {

 62        if((res>=ans&&res<=ans+0.000001)||(res<=ans&&res+0.000001>=ans+0.000001))

 63        {

 64            for(i=1;i<=m;i++)

 65            {

 66               if(stu[i]<depath[i]) return;

 67            }

 68        }

 69        res=ans;

 70        memcpy(stu,depath,sizeof(depath));

 71    }

 72 }

 73 void C_n_m(int st ,int count)

 74 {

 75     if(count==m)

 76     {

 77         Prime();

 78         return ;

 79     }

 80     for(int i=st ;i<=n;i++ )

 81     {

 82         depath[count+1]=i;

 83        C_n_m(i+1,count+1);

 84     }

 85 }

 86 int main()

 87 {

 88     int i,j;

 89     while(scanf("%d%d",&n,&m),m+n)

 90     {

 91         for(i=1;i<=n;i++)

 92            scanf("%d",node_weight+i);    //记录节点权值

 93         for(i=1;i<=n;i++)                //记录边权值

 94             for(j=1;j<=n;j++)

 95               scanf("%d",&edge_weight[i][j]);

 96       // C(n,m)

 97         res=max;

 98         C_n_m(1,0);

 99         for(i=1;i<=m;i++)

100         {

101            printf("%d",stu[i]);

102            if(i!=m)printf(" ");

103         }

104            putchar(10);

105     }

106     return 0;

107 }