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：

 #include<string.h>
 #include<stdlib.h>
 #include<stdio.h>
 #include<math.h>
 #define max 0x3f3f3f3f
 #define maxn 17
 int node_weight[maxn];
 int edge_weight[maxn][maxn];
 int depath[maxn];      //以这些点形成一颗最小生成树
 int  m , n ;
 double res;
 int stu[maxn];
 int sub_map[maxn][maxn];
 void Prime()
 {
   int vis[maxn]={};
   int lowc[maxn];
   int i,j,k,minc;
   double ans=;
   for(i=;i<=m;i++)  //从n中挑出m个点形成一个子图
   {
     for(j=;j<=m;j++)
     {
       if(edge_weight[depath[i]][depath[j]]==)
          sub_map[i][j]=max;
       else
          sub_map[i][j]=edge_weight[depath[i]][depath[j]];
     }
   }
   vis[]=;
   for(i=;i<=m;i++)
   {
     lowc[i]=sub_map[][i];
   }
   for(i=;i<=m;i++)
   {
      minc=max;
      k=;
     for(j=;j<=m;j++)
     {
       if(vis[j]==&&minc>lowc[j])
         {
          minc=lowc[j];
          k=j;
         }
     }
     if(minc==max) return ;  //表示没有联通
     ans+=minc;
     vis[k]=;
     for(j= ; j<=m;j++)
     {
         if(vis[j]==&&lowc[j]>sub_map[k][j])
               lowc[j]=sub_map[k][j];
     }
   }
   int  sum=;
    for(i=;i<=m;i++)  //统计点权值的和
       sum+=node_weight[depath[i]];
       ans/=sum;
   if(res+0.00000001>=ans)
    {
        if((res>=ans&&res<=ans+0.000001)||(res<=ans&&res+0.000001>=ans+0.000001))
        {
            for(i=;i<=m;i++)
            {
               if(stu[i]<depath[i]) return;
            }
        }
        res=ans;
        memcpy(stu,depath,sizeof(depath));
    }
 }
 void C_n_m(int st ,int count)
 {
     if(count==m)
     {
         Prime();
         return ;
     }
     for(int i=st ;i<=n;i++ )
     {
         depath[count+]=i;
        C_n_m(i+,count+);
     }
 }
 int main()
 {
     int i,j;
     while(scanf("%d%d",&n,&m),m+n)
     {
         for(i=;i<=n;i++)
            scanf("%d",node_weight+i);    //记录节点权值
         for(i=;i<=n;i++)                //记录边权值
             for(j=;j<=n;j++)
               scanf("%d",&edge_weight[i][j]);
       // C(n,m)
         res=max;
         C_n_m(,);
         for(i=;i<=m;i++)
         {
            printf("%d",stu[i]);
            if(i!=m)printf(" ");
         }
            putchar();
     }
     return ;
 }