POJ_1007 DNA Sorting

DNA Sorting

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 72977 Accepted: 29150

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

大概题意，就是每个序列中，统计出非顺序数，既对应字母后面比它大的字母个数的总和，比如 DAABEC  = 4(D---AABC) + 1(E---C) = 5 ，一开始理解题目有点出入，卡了一阵子，最后发现原来是水题...

思路：对每个字母统计出后面出现的字母比它大的字母个数的总和，再进行排序，最后输出最大的哪个字符串。

具体代码如下：

#include<iostream>#include<string>using namespace std;struct dna{string str;int num;};struct dna tmp,dn[105];int main(){int n,m,i,j,k;cin >> n >> m;for(i = 1; i <= m; i++){cin >> dn[i].str;//计算逆序数dn[i].num = 0;j = 0;while(j < n - 1){while(j < n && dn[i].str[j] == 'A' )j++;for(k = j + 1; k < n; k++)if(dn[i].str[j] > dn[i].str[k])dn[i].num++;j++;}}//对dn的num进行排序for(i = 1; i < m; i++)for(j = i + 1; j <= m; j++){if(dn[i].num > dn[j].num){tmp.num = dn[i].num;tmp.str = dn[i].str;dn[i].num = dn[j].num;dn[i].str = dn[j].str;dn[j].num = tmp.num;dn[j].str = tmp.str;}}//输出排序后的dn的strfor(i = 1; i <= m; i++)cout << dn[i].str << endl;return 0;}