POJ1007算法解析

POJ1007
DNA Sorting
Time Limit: 1000MS Memory Limit: 10000K
Total Submissions: 102642 Accepted: 41111

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 sequenceDAABEC”, 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 sequenceZWQM” 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'', frommost 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'' toleast 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

从网上找了一段非常精简的代码，非常的不错，用结构体定义每一行字符串

AC代码（G++）：

#include <iostream>#include <cstdlib>using namespace std;struct DNA{  string s;      int value;};int cmp(const DNA *a, const DNA *b)  {       return (a->value-b->value);       }   int main(){    int n,m;    while(cin>> m >>n)    {         DNA it[n];              for(int i=0;i!=n;i++)         {            cin >> it[i].s;            it[i].value = 0;            for(int j=0;j!=m;j++)                for(int k=j+1;k!=m;k++)                     if(it[i].s[j]>it[i].s[k])   it[i].value++;         }         qsort(it,n,sizeof(DNA), (int (*)(const void *, const void *))cmp);         for(int i=0;i!=n;i++)              cout << it[i].s <<endl ;                 }    return 0;   }