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 subgraph 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m1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。
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