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

 

 

1 #include 
       
          2 #include 
        
           3 #include 
         
            4 #define inf 0x3f3f3f3f  5 using namespace std;  6 int n,m;  7 int num[20];  8 int map[20][20];  9 int ans[20][20]; 10 int dis[20]; 11 bool mark[20]; 12 int DJ(int pm) 13 { 14     memset(dis,inf,sizeof(dis)); 15     memset(mark,false,sizeof(mark)); 16     for(int i=1;i
          
           ans[temp][j]) 39             { 40                 dis[j]=ans[temp][j]; 41             } 42         } 43     } 44     return res; 45 } 46 int main() 47 { 48     while(~scanf("%d%d",&n,&m)) 49     { 50         if(n==0 && m==0) 51         { 52             break; 53         } 54         for(int i=0;i
           
            =(1<
           
          
         
        
       

 

数据量的允许，使用二进制压缩的方法，枚举各种可能，每次都求一次最小生成树；

注意的地方是double 判<0（ 小于0）使用高精度判断 例如判小于可以(a<b)：a-b<-(1e-8)

否则会出现WA的情况