HDU-2489 Minimal Ratio Tree(最小生成树)

Posted 2020-10-08 可惜没如果=_=

Minimal Ratio Tree

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4610    Accepted Submission(s): 1466

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

 

 

Recommend

 1 #include "bits/stdc++.h"
 2 using namespace std;
 3 typedef long long LL;
 4 const int MAX=20;
 5 int n,m;
 6 int a[MAX],b[MAX][MAX],c[MAX];
 7 int ans[MAX];double an;
 8 struct Edge{
 9     int u,v;
10     int w;
11     bool operator < (const Edge &tt) const {
12         return w<tt.w;
13     }
14 }edge[MAX*MAX];
15 int fa[MAX];
16 int getfather(int x){
17     if (fa[x]==x) return x;
18     return fa[x]=getfather(fa[x]);
19 }
20 void dfs(int no,int now){
21     int i,j;
22     if (now==m+1){
23         int x=0,y=0;
24         for (i=1;i<=m;i++){y+=a[c[i]];}
25         for (i=1;i<=n;i++){fa[i]=i;}
26         int len=0;
27         for (i=1;i<=m;i++){
28             for (j=i+1;j<=m;j++){
29                 edge[++len].u=c[i];
30                 edge[len].v=c[j];
31                 edge[len].w=b[c[i]][c[j]];
32             }
33         }
34         sort(edge+1,edge+len+1);
35         for (i=1;i<=len;i++){
36             int dx=getfather(edge[i].u);
37             int dy=getfather(edge[i].v);
38             if (dx!=dy){
39                 x+=edge[i].w;
40                 fa[dx]=dy;
41             }
42         }
43         double xy=(x*1.0)/(y*1.0);
44         if (xy<an){
45             for (i=1;i<=m;i++)
46                 ans[i]=c[i];
47             an=xy;
48         }
49     }
50     for (i=no+1;i<=n;i++){
51         c[now]=i;
52         dfs(i,now+1);
53     }
54 }
55 int main(){
56     int i,j;
57     while (1){
58     scanf("%d%d",&n,&m);if (n==0 && m==0) break;
59     an=100000000.0;
60     for (i=1;i<=n;i++){
61         scanf("%d",a+i);
62     }
63     for (i=1;i<=n;i++){
64         for (j=1;j<=n;j++){
65             scanf("%d",&b[i][j]);
66         }
67     }
68     dfs(0,1);
69     sort(ans+1,ans+m+1);
70     for (i=1;i<m;i++){
71         printf("%d ",ans[i]);
72     }
73     printf("%d\n",ans[m]);
74     }
75     return 0;
76 }