HDOJ 1379 DNA Sorting (sort 快排)

DNA Sorting

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 2249    Accepted Submission(s): 1107

Problem Description

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

110 6AACATGAAGGTTTTGGCCAATTTGGCCAAAGATCAGATTTCCCGGGGGGAATCGATGCAT

 

Sample Output

CCCGGGGGGAAACATGAAGGGATCAGATTTATCGATGCATTTTTGGCCAATTTGGCCAAA

注 - 此题为：HDOJ 1379 DNA Sorting  (sort 快排)

题意： 计算每一个字符串的等级 ，在进行排序  ，等级相同，按原顺序输出 

计算等级规则，例如：AACATGAAGG   ：

                  C 大于其后面的 3 个 A ，T 大于其后面的 5 个字符 ，第一个 G 大于其后面的 2 个 A，  字符串等级为  10

已AC代码：

#include<iostream>#include<cstring>#include<cstdio>#include<algorithm>using namespace std;struct node{char ch[120];int num; // 记录等级 int k;  // 记录顺序 }dna[120];int cmp(node a,node b){if(a.num==b.num)  //等级相同，按原顺序输出 return a.k<b.k; return a.num <b.num ;}int main(){int T;cin>>T;while(T--){int i,j,k,n,m;scanf("%d%d",&n,&m); //长度为 n,行数为 m for(i=0;i<m;++i){scanf("%s",dna[i].ch );dna[i].num=0;dna[i].k=i;for(j=0;j<n-1;++j)  // 计算等级 {for(k=j+1;k<n;++k){if(dna[i].ch[j] >dna[i].ch[k] )dna[i].num++;}}}sort(dna,dna+m,cmp);for(i=0;i<m;++i)cout<<dna[i].ch <<endl;}return 0;}