标签：acm   uva   brute force   backtracking   

题目如下：

Mapping the Swaps

Sorting an array can be done by swapping certain pairs of adjacent entriesin the array. This is the fundamental technique used in the well-knownbubble sort. If we list the identities of the pairs to be swapped, in thesequence they are to be swapped, we obtain what might be called a swap map.For example, suppose we wish to sort the array A whose elements are 3, 2,and 1 in that order. If the subscripts for this array are 1, 2, and 3,sorting the array can be accomplished by swapping A2 and A3, then swapping A1and A2, and finally swapping A2 and A3. If a pair is identified in a swapmap by indicating the subscript of the first element of the pair to beswapped, then this sorting process would be characterized with the swap map2 1 2.

It is instructive to note that there may be many ways in which swapping ofadjacent array entries can be used to sort an array. The previous array,containing 3 2 1, could also be sorted by swapping A1 and A2, then swappingA2 and A3, and finally swapping A1 and A2 again. The swap map that describesthis sorting sequence is 1 2 1.

For a given array, how many different swap maps exist? A little thoughtwill show that there are an infinite number of swap maps, since sequentialswapping of an arbitrary pair of elements will not change the order of the elements. Thus the swap map 1 1 1 2 1 will also leave our arrayelements in ascending order. But how many swap maps of minimum size willplace a given array in order? That is the question you are to answer in this problem.

Input

The input data will contain an arbitrary number of test cases, followed by asingle 0. Each test case will have a integer n that gives the size of anarray, and will be followed by the n integer values in the array.

Output

For each test case, print a message similar to those shown in thesample output below. In no test case will n be larger than 5.

Sample Input

2 9 7
2 12 50
3 3 2 1
3 9 1 5
0

Sample Output

There are 1 swap maps for input data set 1.
There are 0 swap maps for input data set 2.
There are 2 swap maps for input data set 3.
There are 1 swap maps for input data set 4.

AC的代码如下：

UVA Mapping the Swaps,布布扣,bubuko.com

UVA Mapping the Swaps

标签：acm   uva   brute force   backtracking   

原文地址：http://blog.csdn.net/u013840081/article/details/37830689