Minimal Ratio Tree
Time Limit : 2000/1000ms (Java/Other) Memory Limit : 32768/32768K (Java/Other)
Total Submission(s) : 12 Accepted Submission(s) : 7
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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
#include <iostream>
#include<cstdio>
#include<cstring>
#include<climits>
using namespace std; int a[],f[],p[];
int mp[][];
bool vis[];
int i,j,n,m;
double ans;
void prim() //最小生成树
{
bool vis[];
int dis[];
int sumnode,sumedge=,k;
memset(vis,,sizeof(vis));
vis[]=;
sumnode=p[a[]];
for(int i=;i<=m;i++) dis[i]=mp[a[]][a[i]];
for(int i=;i<m;i++)
{
int minn=INT_MAX;
for(int j=;j<=m;j++)
{
if (!vis[j] && dis[j]<minn)
{
minn=dis[j];
k=j;
}
}
vis[k]=;
sumedge+=minn;
sumnode+=p[a[k]];
for(int j=;j<=m;j++)
if (!vis[j] && dis[j]>mp[a[k]][a[j]])
dis[j]=mp[a[k]][a[j]];
}
double w=sumedge*1.0/sumnode;
if (w<ans) //把最优解存放在f数组中
{
ans=w;
for(int i=;i<=m;i++)
f[i]=a[i];
}
return;
}
void dfs(int k,int num)//dfs暴力枚举m个节点是哪几个存在a数组中
{
if (num==m)
{
prim();
return;
}
if (k>n) return; if (!vis[k])
{
vis[k]=;
a[num+]=k;
dfs(k+,num+);
vis[k]=;
}
dfs(k+,num);
return;
}
int main()
{
while(scanf("%d%d",&n,&m))
{
if (n== && m==) break;
for(i=;i<=n;i++) scanf("%d",&p[i]);
for(i=;i<=n;i++)
for(j=;j<=n;j++)
scanf("%d",&mp[i][j]);
memset(vis,,sizeof(vis));
ans=INT_MAX*1.0;
dfs(,);
for(i=;i<m;i++) printf("%d ",f[i]);
printf("%d\n",f[m]);
}
return ;
}