Solution of ZOJ 1188 DNA Sorting
One measure of ``unsortedness'' in a sequence is the number of pairs of entries that are out of order with respect to each other. For instance, in the letter sequence ``DAABEC'', this measure is 5, since D is greater than four letters to its right and E is greater than one letter to its right. This measure is called the number of inversions in the sequence. The sequence ``AACEDGG'' has only one inversion (E and D)--it is nearly sorted--while the sequence ``ZWQM'' has 6 inversions (it is as unsorted as can be--exactly the reverse of sorted).
You are responsible for cataloguing a sequence of DNA strings (sequences containing only the four letters A, C, G, and T). However, you want to catalog them, not in alphabetical order, but rather in order of ``sortedness'', from ``most sorted'' to ``least sorted''. All the strings are of the same length.
This problem contains multiple test cases!
The first line of a multiple input is an integer N, then a blank line followed by N input blocks. Each input block is in the format indicated in the problem description. There is a blank line between input blocks.
The output format consists of N output blocks. There is a blank line between output blocks.
Input
The first line contains two integers: a positive integer n (0 < n <= 50) giving the length of the strings; and a positive integer m (1 < m <= 100) giving the number of strings. These are followed by m lines, each containing a string of length n.
Output
Output the list of input strings, arranged from ``most sorted'' to ``least sorted''. If two or more strings are equally sorted, list them in the same order they are in the input file.
Sample Input
1
10 6
AACATGAAGG
TTTTGGCCAA
TTTGGCCAAA
GATCAGATTT
CCCGGGGGGA
ATCGATGCAT
Sample Output
CCCGGGGGGA
AACATGAAGG
GATCAGATTT
ATCGATGCAT
TTTTGGCCAA
TTTGGCCAAA
Source: East Central North America 1998
分析：此题关键点有二，一是容器的选择，二是逆序数的计算。下面给出我的实现代码，中间用到了multiset和functor：
#include <iostream> #include <cstdio> #include <string> #include <set> using namespace std; int num_inver( const string & str ) { int inv_num = 0; for( int i = 0; i < str.size() - 1; ++i ) { for( int j = i + 1; j < str.size(); ++j ) { if( str[i] > str[j] ) ++inv_num; } } return inv_num; } struct DNACompare { bool operator()( const string & str1, const string & str2 ) { return num_inver( str1 ) < num_inver( str2 ); } }; int main() { multiset<string, DNACompare> seqs; int num_blocks; scanf("%d", &num_blocks); while( num_blocks-- != 0 ) { string line; getline( cin, line ); int str_len, num_strs; scanf( "%d %d", &str_len, &num_strs ); for( int j = 0; j < num_strs; ++j ) { cin >> line; seqs.insert( line ); } multiset<string, DNACompare>::const_iterator iter; for( iter = seqs.begin(); iter != seqs.end(); ++iter ) printf("%s/n", iter->c_str()); seqs.clear(); if( num_blocks != 0 ) printf("/n"); } return 0; }