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】
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

【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。

 

  1 #include<iostream>
  2 #include<cstdio>
  3 #include<cstring>
  4 #include<queue>
  5 
  6 using namespace std;
  7 
  8 bool flag[16],outp[16];
  9 int n,m,node[16];
 10 int ma[16][16];
 11 double mi;
 12 
 13 typedef struct heaptyp
 14 {
 15     int num,key;
 16     friend bool operator < (heaptyp a,heaptyp b)
 17     {
 18         return a.num>b.num;
 19     }
 20 } heaptype;
 21 
 22 void prim(int s,int tot)
 23 {
 24     int i,j,now,ans;
 25     bool fla[16];
 26     priority_queue<heaptype>heap;
 27     heaptype aaa;
 28     
 29     memset(fla,0,sizeof(fla));
 30     fla[s]=true;
 31     ans=0;
 32     for (i=1;i<=n;i++) 
 33     if (flag[i]&&ma[s][i])
 34     {
 35         heaptype temp;
 36         temp.num=ma[s][i];
 37         temp.key=i;
 38         heap.push(temp);
 39         aaa=heap.top();
 40     }
 41     
 42     for (j=1;j<m;j++)
 43     {
 44         heaptype h=heap.top();
 45         heap.pop();
 46         aaa=heap.top();
 47         while (fla[h.key])
 48         {
 49             h=heap.top();
 50             heap.pop();
 51         }
 52         now=h.key;
 53         fla[now]=true;
 54         ans+=h.num;
 55         for (i=1;i<=n;i++)
 56         if (flag[i]&&ma[now][i])
 57         if (!fla[i])
 58         {
 59             heaptype temp;
 60             temp.num=ma[now][i];
 61             temp.key=i;
 62             heap.push(temp);
 63             aaa=heap.top();
 64         }
 65     }
 66     double rat=(double)ans/tot;
 67     if (mi-rat>0.0000001)
 68     {
 69         mi=rat;
 70         for (int i=1;i<=n;i++) outp[i]=fla[i];
 71     }
 72 }
 73 
 74 void dfs(int now,int last,int tot)
 75 {
 76      if (now==m)
 77      {
 78          int i;
 79          for (i=1;i<=n;i++)
 80          if (flag[i]) break;
 81          prim(i,tot);
 82      }
 83      else 
 84      {
 85          now++;
 86          for (int i=last+1;i<=n-m+now;i++)
 87          {
 88              flag[i]=true;
 89              dfs(now,i,tot+node[i]);
 90              flag[i]=false;
 91          }
 92      }
 93 }
 94 
 95 int main()
 96 {
 97     scanf("%d%d",&n,&m);
 98     while (!(n==0&&m==0))
 99     {
100         for (int i=1;i<=n;i++) scanf("%d",&node[i]);
101         for (int i=1;i<=n;i++)
102         for (int j=1;j<=n;j++) scanf("%d",&ma[i][j]);
103         
104         mi=2147483647;
105         memset(flag,0,sizeof(flag));
106         for (int i=1;i<=n-m+1;i++)
107         {
108             flag[i]=true;
109             dfs(1,i,node[i]);
110             flag[i]=false;
111         }
112         
113         int i;
114         for (i=1;i<=n;i++) 
115         if (outp[i]) 
116         {
117             printf("%d",i);
118             break;
119         }
120         for (int j=i+1;j<=n;j++)
121         if (outp[j]) printf(" %d",j);
122         printf("\n");
123         scanf("%d%d",&n,&m);
124     }
125     
126     return 0;
127 }
View Code

 

posted @ 2014-09-06 20:57  辰帆  阅读(147)  评论(0)  编辑  收藏  举报