HDU 2489 Minimal Ratio Tree 最小生成树+DFS

Minimal Ratio Tree

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

【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】

【Sample Output】

【题意】

给出一张n个点的图，图中的每一个结点以及每一条边都有其权值，要求从中选出m个点，找到m-1条边将其连接，使得边权值与点权值的比值达到最小。

【分析】

要使得比值最小，则点权值和尽可能地大同时边权值和尽可能地小。直接上考虑，边权值和尽可能小即对这m个点作最小生成树。

而题目给定的n不大，故可以用DFS搜出需要的m个点，然后对m个点进行最小生成树，中间注意判断和保存即可。

我用了一个dijkstra+优先队列的prim去找MST，这也是我第一次尝试使用STL的优先队列。

 #include<iostream>
 #include<cstdio>
 #include<cstring>
 #include<queue>

 using namespace std;

 bool flag[],outp[];
 int n,m,node[];
 int ma[][];
 double mi;

 typedef struct heaptyp
 {
     int num,key;
     friend bool operator < (heaptyp a,heaptyp b)
     {
         return a.num>b.num;
     }
 } heaptype;

 void prim(int s,int tot)
 {
     int i,j,now,ans;
     bool fla[];
     priority_queue<heaptype>heap;
     heaptype aaa;

     memset(fla,,sizeof(fla));
     fla[s]=true;
     ans=;
     for (i=;i<=n;i++)
     if (flag[i]&&ma[s][i])
     {
         heaptype temp;
         temp.num=ma[s][i];
         temp.key=i;
         heap.push(temp);
         aaa=heap.top();
     }

     for (j=;j<m;j++)
     {
         heaptype h=heap.top();
         heap.pop();
         aaa=heap.top();
         while (fla[h.key])
         {
             h=heap.top();
             heap.pop();
         }
         now=h.key;
         fla[now]=true;
         ans+=h.num;
         for (i=;i<=n;i++)
         if (flag[i]&&ma[now][i])
         if (!fla[i])
         {
             heaptype temp;
             temp.num=ma[now][i];
             temp.key=i;
             heap.push(temp);
             aaa=heap.top();
         }
     }
     double rat=(double)ans/tot;
     if (mi-rat>0.0000001)
     {
         mi=rat;
         for (int i=;i<=n;i++) outp[i]=fla[i];
     }
 }

 void dfs(int now,int last,int tot)
 {
      if (now==m)
      {
          int i;
          for (i=;i<=n;i++)
          if (flag[i]) break;
          prim(i,tot);
      }
      else
      {
          now++;
          for (int i=last+;i<=n-m+now;i++)
          {
              flag[i]=true;
              dfs(now,i,tot+node[i]);
              flag[i]=false;
          }
      }
 }

 int main()
 {
     scanf("%d%d",&n,&m);
     while (!(n==&&m==))
     {
         for (int i=;i<=n;i++) scanf("%d",&node[i]);
         for (int i=;i<=n;i++)
         for (int j=;j<=n;j++) scanf("%d",&ma[i][j]);

         mi=;
         memset(flag,,sizeof(flag));
         for (int i=;i<=n-m+;i++)
         {
             flag[i]=true;
             dfs(,i,node[i]);
             flag[i]=false;
         }

         int i;
         for (i=;i<=n;i++)
         if (outp[i])
         {
             printf("%d",i);
             break;
         }
         for (int j=i+;j<=n;j++)
         if (outp[j]) printf(" %d",j);
         printf("\n");
         scanf("%d%d",&n,&m);
     }

     return ;
 }