标签：acm uva brute force backtracking
题目如下：| Bandwidth |
Given a graph (V,E) where V is a set of nodes and E is a set of arcsin VxV, and an ordering on the elements in V, then the bandwidth of a node v is defined as the maximum distance in theordering between v and any node to which it is connected in thegraph. The bandwidth of the ordering is then defined as the maximum ofthe individual bandwidths. For example, consider the following graph:
This can be ordered in many ways, two of which are illustrated below:
For these orderings, the bandwidths of the nodes (in order) are 6, 6,1, 4, 1, 1, 6, 6 giving an ordering bandwidth of 6, and 5, 3, 1, 4, 3,5, 1, 4 giving an ordering bandwidth of 5.
Write a program that will find the ordering of a graph that minimisesthe bandwidth.
Input will consist of a series of graphs. Each graph will appear on aline by itself. The entire file will be terminated by a lineconsisting of a single #. For each graph, the input will consist ofa series of records separated by `;‘. Each record will consist of anode name (a single upper case character in the the range `A‘ to `Z‘),followed by a `:‘ and at least one of its neighbours. The graph willcontain no more than 8 nodes.
Output will consist of one line for each graph, listing the orderingof the nodes followed by an arrow (->) and the bandwidth for thatordering. All items must be separated from their neighbours by exactlyone space. If more than one ordering produces the same bandwidth, thenchoose the smallest in lexicographic ordering, that is the one thatwould appear first in an alphabetic listing.
A:FB;B:GC;D:GC;F:AGH;E:HD #
A B C F G D H E -> 3
给一个图，求出使图的带宽为最小值的节点的排列，带宽指每个节点的与相邻节点在排列中距离的最大值，图的带宽是节点带宽的最大值。由于节点最多只有八个，直接枚举全排列即可，如果节点数较多应该剪枝，但本题显然不需要。
AC的代码如下：
标签：acm uva brute force backtracking
原文地址：http://blog.csdn.net/u013840081/article/details/37931277
