poj 1007 (nyoj 160) DNA Sorting

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DNA Sorting

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 75164		Accepted: 30115

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

把每个序列的逆序数求出来然后根据逆序数稳定排序后升序输出，注意qsort中cmp函数的使用，当a == b时返回0则不处理两个数

#include<iostream>
#include<stdlib.h>
#include<algorithm>
#include<string>
using namespace std;
typedef struct NODE
{
int num;
string str;
}Node;

int comp(const void* a,const void* b)
{
Node* x=(Node*)a;
Node* y=(Node*)b;
return (x->num)-(y->num);
}
int main()
{
int n, m;
Node array[101];
while(cin>>n>>m)
{
int i;
for(i = 0; i < m; i++)
{
Node node;
cin>>node.str;
int j;
int count = 0;
int len = node.str.length();
for(j = 0; j < len - 1; j++)
{
int k;
if(node.str[j] == 'A')
continue;
for(k = j + 1; k < len; k++)
{
if(node.str[j] > node.str[k])
count ++;
}
}
node.num = count;
array[i] = node;
}
qsort(array, m, sizeof(Node), comp);
for(i = 0; i < m; i++)
cout<<array[i].str<<endl;

}
return 0;
}