HDU 2489 Minimal Ratio Tree

Minimal Ratio Tree

Time Limit: 1000ms

Memory Limit: 32768KB

This problem will be judged on  HDU. Original ID: 2489
64-bit integer IO format: %I64d      Java class name: Maina

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

 

解题：枚举点集，对点集求最小生成树

 

 1 #include <bits/stdc++.h>

 2 using namespace std;

 3 const int INF = 0x3f3f3f3f;

 4 const int maxn = 20;

 5 int e[maxn][maxn],d[maxn],val[maxn],n,m;

 6 bool vis[maxn],us[maxn],ans[maxn];

 7 int prim(int s) {

 8     memset(d,0x3f,sizeof d);

 9     memset(us,false,sizeof us);

10     int sum = d[s] = 0;

11     while(true) {

12         int minV = INF,idx = -1;

13         for(int i = 0; i < n; ++i)

14             if(!us[i] && vis[i] && d[i] < minV) minV = d[idx = i];

15         if(minV == INF || idx == -1) break;

16         us[idx] = true;

17         sum += minV;

18         for(int i = 0; i < n; ++i) {

19             if(!us[i] && vis[i] && d[i] > e[idx][i])

20                 d[i] = e[idx][i];

21         }

22     }

23     return sum;

24 }

25 int wval,nval;

26 void dfs(int dep,int cnt,int nsum) {

27     if(cnt == m) {

28         int tmp = prim(dep - 1);

29         if(wval == INF || wval*nsum > tmp*nval) {

30             wval = tmp;

31             nval = nsum;

32             memcpy(ans,vis,sizeof vis);

33         }

34         return;

35     }

36     if(dep >= n) return;

37     vis[dep] = true;

38     dfs(dep+1,cnt+1,nsum+val[dep]);

39     vis[dep] = false;

40     dfs(dep+1,cnt,nsum);

41 }

42 int main() {

43     while(scanf("%d %d",&n,&m),n||m) {

44         for(int i = 0; i < n; ++i)

45             scanf("%d",&val[i]);

46         for(int i = 0; i < n; ++i)

47             for(int j = 0; j < n; ++j)

48                 scanf("%d",e[i]+j);

49         memset(vis,false,sizeof vis);

50         wval = INF;

51         dfs(0,0,0);

52         bool flag = false;

53         for(int i = 0; i < n; ++i)

54             if(ans[i]) {

55                 if(flag) putchar(' ');

56                 flag = true;

57                 printf("%d",i+1);

58             }

59         putchar('\n');

60     }

61     return 0;

62 }

63 

64 /*

65 3 2

66 30 20 10

67 0 6 2

68 6 0 3

69 2 3 0

70 

71 2 2

72 1 1

73 0 2

74 2 0

75 

76 0 0

77 */

View Code