poj 1007:DNA Sorting（水题，字符串逆序数排序）

DNA Sorting

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 80832		Accepted: 32533

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.

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 (0 < 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''. Since two strings can be equally sorted, then output them according to the orginal order.

Sample Input

10 6
AACATGAAGG
TTTTGGCCAA
TTTGGCCAAA
GATCAGATTT
CCCGGGGGGA
ATCGATGCAT

Sample Output

CCCGGGGGGA
AACATGAAGG
GATCAGATTT
ATCGATGCAT
TTTTGGCCAA
TTTGGCCAAA

Source

East Central North America 1998

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　　水题，字符串排序。

　　题意：给你m个长度为n的字符串，要求让你按照字符串的逆序数进行稳定排序。

　　　　所谓逆序数，就是给定一个排列和一个标准次序，如果这个排列中有两个元素与标准次序不同，则称这是一个逆序，这个排列的所有逆序总数称为这个排列的逆序数。

　　思路：定义一个结构体存储字符串和字符串的逆序数，在主程序里，输入字符串之后计算每一个字符串的逆序数。然后用sort进行排序，最后输出排序后的字符串。

　　代码：

 #include <iostream>
 #include <algorithm>
 using namespace std;
 struct Str{
     char s[];
     int n;  //逆序数个数
 };
 bool cmp(const Str &a,const Str &b)
 {
     if(a.n<b.n)
         return ;
     return ;
 }
 int main()
 {
     int n,m,i,j;
     Str str[];
     while(cin>>n>>m){
         for(i=;i<=m;i++)   //输入
             cin>>str[i].s;
         for(i=;i<=m;i++){
             int num=;
             int A=,C=,G=;
             for(j=n-;j>=;j--){
                 switch(str[i].s[j]){    //计算逆序数
                     case 'A':A++;break;
                     case 'C':C++;num+=A;break;
                     case 'G':G++;num+=A;num+=C;break;
                     case 'T':num+=A;num+=C;num+=G;break;
                     default:break;
                 }
             }
             str[i].n = num;
         }
         sort(str+,str+m+,cmp);
         for(i=;i<=m;i++)
             cout<<str[i].s<<endl;
     }
     return ;
 }

Freecode : www.cnblogs.com/yym2013