poj 1007 DNA Sorting

DNA Sorting

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 95437		Accepted: 38399

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

题目大意：根据字符串的逆序数，从小到大输出字符串。
思路：将字符串和它的逆序数联系起来，通过对它的逆序数从小到大排序，对字符串排序。。。用结构体解决，用sort函数排序。

#include <iostream>
#include <algorithm>
#include <cstring>
using namespace std;

typedef struct{
    char s[51];
    int num;
}DNA;

//求一个字符串的逆序数
int InversionNumber(char *s, int n){
    int num = 0;
    for(int i = 0; i < n - 1; i++)
        for(int j = i + 1; j < n; j++){
            if(s[i] > s[j])
                num++;
        }
    return num;
}

bool cmp(DNA a, DNA b){
    return a.num < b.num;
}

int main(){
    int n, m;
    cin >> n >> m;
    DNA *a = new DNA [m];
    for(int i = 0; i < m; i++){
        cin >> a[i].s;
        a[i].num = InversionNumber(a[i].s, n);
    }
    sort(a, a + m, cmp);

    for(int i = 0; i < m; i++)
        cout << a[i].s << endl;
    return 0;
}