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POJ 1007 DNA Sorting

2024-08-08 08:05:44   219人阅读

DNA Sorting
Time Limit: 1000MS Memory Limit: 10000K
Total Submissions: 85118 Accepted: 34274

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 6AACATGAAGGTTTTGGCCAATTTGGCCAAAGATCAGATTTCCCGGGGGGAATCGATGCAT

Sample Output

CCCGGGGGGAAACATGAAGGGATCAGATTTATCGATGCATTTTTGGCCAATTTGGCCAAA

解题思路：
这道题也是属于简单题，基本思路就是先算出每个DNA序列的逆序数，然后排个序输出（注意这里需要用稳定排序，我用了C++库的qsort，简单定义一下不处理‘==’的情况就可以了）。
求逆序数稍微有点技巧

代码：
 1 #include <iostream> 2 #include <algorithm> 3  4 /************************************************** 5  * POJ 1007 DNA Sorting 6  * author:樊列龙 7  * 2014-12-16 8  *************************************************/ 9 typedef struct DNA10 {11     int count;// 逆序数的个数12     char DNAStr[110];13 } DNA;14 15 int N;16 int SIZE;17 int countA,countC,countG;18 19 20 /** 计算DNA字符串的逆序数  */21 int getDNAInversionNumber(char* dnaStr)22 {23     int count = 0;24     countA = countC = countG = 0;25     for(int j = SIZE-1; j >= 0; j--)26     {27         switch(dnaStr[j])28         {29         case ‘A‘:30             countA++;31             break;32         case ‘C‘:33             countC++;34             count += countA;35             break;36         case ‘G‘:37             countG++;38             count += countA;39             count += countC;40             break;41         case ‘T‘:42             count += countA;43             count += countC;44             count += countG;45             break;46         }47     }48     return count;49 }50 51 int cmp(const void* a, const void* b)52 {53     DNA* DNA_A = (DNA*)a;54     DNA* DNA_B = (DNA*)b;55     return (DNA_A->count) - (DNA_B->count);56 }57 58 int main()59 {60     using namespace std;61     while(cin >> SIZE >> N)62     {63         DNA* dnas = new DNA[N];64         for(int i = 0; i < N; i++)65         {66             cin >> dnas[i].DNAStr;67             dnas[i].count = getDNAInversionNumber(dnas[i].DNAStr);68         }69 70         qsort(dnas, N, sizeof(DNA), cmp);71 72         for(int i = 0; i < N; i++)73         {74             cout << dnas[i].DNAStr << endl;75         }76     }77     return 0;78 }

 





POJ 1007 DNA Sorting


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