# POJ 1007 DNA Sorting C++版

POJ题解 2493浏览

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

Slyar:题意就是让你求逆序对，然后按照逆序对的大小将字符串输出，如果逆序对数目一样，则不要改变原来的顺序输出。

```#include
#include
#include

using namespace std;

struct dna
{
int pos;
int key;
string str;
};

/* sort比较函数 */
bool cmp(const dna &a, const dna &b)
{
if (a.key != b.key)
{
return a.key < b.key;
}
else
{
return a.pos < b.pos;
}
}

int main()
{
int n, m, count;
dna inv[110];
string str;
cin >> n >> m;

/* 求逆序数对的个数 */
for (int i = 0; i < m; i++)
{
cin >> str;
count = 0;
for (int j = 0; j < n - 1; j++)
{
for (int k = j + 1; k < n; k++)
{
if (str[j] > str[k]) count++;
}
}
/* 保存信息 */
inv[i].key = count;
inv[i].pos = i;
inv[i].str = str;
}

/* 按逆序数对大小/序号排序 */
sort(inv, inv + m, cmp);

/* 输出结果 */
for (int i = 0; i < m; i++)
{
cout << inv[i].str << endl;
}

//system("pause");
return 0;
}
```