Minimal Ratio Tree prime()&&枚举点

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

***************************************************************************************************************************

从n个点抽出m个点，组成最小生成树

***************************************************************************************************************************

#include<iostream>
#include<string>
#include<cstring>
#include<cmath>
#include<cstdio>
using namespace std;
int vis1[1001],vis2[1001],ans[1001];
double e[1001][1001],p[1001],dis[1001],minRatio;
int n,m,i,j;
double prime()//求最小生成树
{
    int u;
    memset(vis1,0,sizeof(vis1));
    int it,jt;
    u=0;
    while(!vis2[u]&&u<=n)u++;
    vis1[u]=1;
    double psum=0.0,esum=0.0;
    for(it=1;it<=n;it++)
    {
        if(vis2[it])
        {
            dis[it]=e[u][it];
            psum+=p[it];
        }
    }
    for(it=1;it<m;it++)
    {
        u=-1;
        for(jt=1;jt<=n;jt++)
        {
            if(vis2[jt]&&!vis1[jt])
                if(u==-1||dis[jt]<dis[u])
                {
                   u=jt;
                }

        }
        esum+=dis[u];
        vis1[u]=1;
        for(jt=1;jt<=n;jt++)
        {
            if(!vis1[jt]&&vis2[jt])
              if(dis[jt]>e[u][jt])
                dis[jt]=e[u][jt];
        }
    }
    return esum/psum;
}
void dfs(int u,int num)//枚举
{
    if(num>m)return;
    if(u==n+1)
    {
        if(num!=m)return;
        double t=prime();
        //cout<<"T:: "<<t<<endl;
        if(t<minRatio)
        {
                minRatio=t;
              memcpy(ans,vis2,sizeof(vis2));//记录
        }
        return;
    }
    vis2[u]=1;
    dfs(u+1,num+1);
    vis2[u]=0;
    dfs(u+1,num);

}
int main()
{
    while(scanf("%d %d",&n,&m)!=EOF)
    {
        if(n==0&&m==0)
            break;
        for(i=1;i<=n;i++)
        {
            scanf("%lf",&p[i]);
            //printf("p::%lf
",p[i]);
        }
        for(i=1;i<=n;i++)
         for(j=1;j<=n;j++)
          {
            scanf("%lf",&e[i][j]);
          }
          minRatio=10000000.0;
          memset(vis2,0,sizeof(vis2));
          //memset(ans,0,sizeof(ans));
          dfs(1,0);
        bool gs=0;
          for(i=1;i<=n;i++)
          {
              if(ans[i])
               {
                   if(gs==0)
                   {
                       printf("%d",i);
                       gs=1;
                    }
                    else
                    {
                        printf(" %d",i);
                    }
               }
          }
         puts("");
    }
    return 0;
}