【北大poj 】1007 DNA Sorting

北大poj1007

DNA Sorting

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 114198		Accepted: 45700

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

思路：按照题目意思，统计每个DNA的measure，按照从大到小稳定排序后输出。

代码实现：（G++提交）

#include 
#include 

using namespace std;

typedef struct B{
    int count; //DNA的measure
    int order; //该measure对应的串的下标
}B;

int cmp(B a , B b) //用来结构体从大到小排序
{
    return a.count < b.count;
}

int main() {
    int m,n; //长度和个数
    cin >> m >> n;
    char s[n][m+1];
    B b[n];
    int i = 0;
    while(i<n)
    {
        int num = 0; //measure
        cin >> s[i];
        s[i][m] = '\0';
        for(int j = 0; j<m-1; j++)
        {
            for(int k=j+1;k<m;k++)
            {
                if(s[i][j]-s[i][k]>0)
                    num++;
            }
        }
        b[i].count = num;
        b[i].order = i;
        i++;
    }
    sort(b,b+n,cmp); //结构体排序
    for(int i = 0;i<n;i++)
    {
        int tmp = b[i].order;
        cout << s[tmp] << endl;
    }


    return 0;
}

Memory	Time
716K	16MS