HDU 2489 Minimal Ratio Tree (DFS枚举+最小生成树Prim)

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$lt;=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

Sample Output

1 3
1 2

题目大意：

有一个n个点的图，然后给出n*n的邻接矩阵图，要求这个图的m个结点的子图，使得这个子图所有边之和与所有点之和的商值最小。

分析与总结：

直接dfs枚举出n个点所有的m个点的组合，然后对m个点求最小生成树，便可得出答案。

dfs枚举n个点的m个点组合，对于每个点，要么属于这个组合，要么是不属于，所以复杂度为2^n, n最大为15，再加上减枝，时间足足矣。

#include<cstdio>
#include<cstring>
#define N 20
int n,m,vis[N], ans[N], pre[N], hash[N];
double G[N][N], weight[N], minCost[N], minRatio;
double prim(){
    int u;
   memset(hash,0,sizeof(hash));
   for(int i=1;i<=n;i++){
       if(vis[i]){
        u=i;
        break;
       }
   }

   hash[u]=1;
   double weightsum=0,edgesum=0;
   for(int i=1;i<=n;i++)
   if(vis[i]){
    minCost[i]=G[u][i];
    pre[i]=u;
    weightsum+=weight[i];
   }

   for(int i=1;i<m;i++){
    u=-1;
      for(int j=1;j<=n;j++)
      if(vis[j]&&!hash[j]){
        if(minCost[u]>minCost[j]||u==-1)
            u=j;
      }

      edgesum+=G[pre[u]][u];
      hash[u]=1;
      for(int j=1;j<=n;j++)

      if(vis[j]&&!hash[j]){
        if(minCost[j]>G[u][j]){
            minCost[j]=G[u][j];
            pre[j]=u;
        }
      }

   }


   return edgesum/weightsum;
}


void dfs(int u,int num){
   if(num>m)
    return;
   if(u==n+1){
       if(num!=m)
        return ;
        double t=prim();
        if(t<minRatio){
            minRatio=t;
            memcpy(ans,vis,sizeof(vis));
        }
       return ;
   }
   vis[u]=1;
   dfs(u+1,num+1);
   vis[u]=0;
   dfs(u+1,num);
}


int main(){
    while(~scanf("%d%d",&n,&m)){
        if(m==0&&n==0) break;
        memset(G,0,sizeof(G));
        memset(weight,0,sizeof(weight));
        memset(vis,0,sizeof(vis));

        for(int i=1; i<=n; ++i)
            scanf("%lf",&weight[i]);
        for(int i=1; i<=n; ++i)
            for(int j=1; j<=n; ++j)
                scanf("%lf",&G[i][j]);

        minRatio = 100000000;
        dfs(1, 0);
        bool flag=false;
        for(int i=1; i<=n; ++i)if(ans[i]){
                if(flag) printf(" %d", i);
                else{
                    printf("%d",i);
                    flag=true;
                }
            }
           puts("");
    }
    return 0;
}