POJ 3925 Minimal Ratio Tree 最小生成树

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

#include<cstdio>
#include<cmath>
using namespace std;
#define M 30
#define MAX 900000000
int w[M],g[M][M];
int n,m;
int ans[M];
int node[M];
double minans;
void prim()
{
    bool used[M];
    int dis[M];
    int i,j,nd=1;
    double totale,totaln;
    totale=totaln=0;
    for(i=0;i<m;i++)
    {
        used[i]=false;
        dis[i]=MAX;
    }
    used[0]=true;
    for(i=0;i<m;i++)
    {
        dis[i]=g[ node[0] ][ node[i] ];
    }
    while(nd<m)
    {
        int min=MAX;
        int mini=0;
        for(i=0;i<m;i++)
        {
            if(!used[i]&&dis[i]<min)
            {
                min=dis[i];
                mini=i;
            }
        }

        used[mini]=true;
        totale+=min;
        nd++;

        for(i=0;i<m;i++)
        {
            if(!used[i]&&g[ node[mini] ][ node[i] ] < dis[i])
            {
                dis[i] = g[ node[mini] ][ node[i] ];
            }
        }
    }
    for(i=0;i<m;i++)
    {
        totaln+=w[ node[i] ];
    }
    double now=totale/totaln;
    if(abs(now-minans)>0&&now<minans)
    {
        minans=now;
        for(i=0;i<m;i++)
        {
            ans[i]=node[i];
        }
    }
}
void fun(int ns,int num)
{
    if(ns==m)
    {
        prim();
        return ;
    }
    for(int i=num;i<n;i++)
    {
        node[ns]=i;
        fun(ns+1,i+1);
    }
}
int main()
{
    int i,j;
    while(~scanf("%d%d",&n,&m))
    {
        if(n==0&&m==0)
        {
            break;
        }
        for(i=0;i<n;i++)
        {
            scanf("%d",w+i);
        }
        for(i=0;i<n;i++)
        {
            for(j=0;j<n;j++)
            {
                scanf("%d",&g[i][j]);
            }
        }
        minans=MAX;
        fun(0,0);

        printf("%d",ans[0]+1);
        for(i=1;i<m;i++)
        {
            printf(" %d",ans[i]+1);
        }
        putchar(10);
    }
    return 0;
}