POJ - 1007

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

CCCGGGGGGAAACATGAAGGGATCAGATTTATCGATGCATTTTTGGCCAA
TTTGGCCAAA
题解：找出各字符串中错误排序次数是多少，累计错误排序次数是利用反序累加，利用反序，就知道当前字符与后面字符是否大小排序，若不是就累加后面小于它的字符，所有字符串都累加错误次序后就按照error排序从小到大。第一次的时候WA是因为没看到如果错误次数相等就按照原来顺序。。。然后改了cmp函数，让a.error>=b.error就不会出现错误相同顺序不同！
#include <iostream>#include <algorithm>#include <cstdio>#include <string>using namespace std;struct DNA{    string str;    int error;};int getError(string str){    int sum = 0;    int count[4] = {0,0,0,0};    int len = str.length()-1;    for(int i = len; i >=0 ;i--){        char temp = str.at(i);        switch(temp){        case 'A':count[0]++;break;        case 'C':count[1]++,sum=sum+count[0];break;        case 'G':count[2]++,sum=sum+count[1]+count[0];break;        case 'T':sum=sum+count[0]+count[1]+count[2];break;        }    }    return sum;}bool cmp(DNA a,DNA b){    return a.error>=b.error?false:true;}int main(){    int n,m;    cin>>n>>m;    DNA d[105];    for(int i = 0; i < m ;i++){        string str;cin>>str;        d[i].str = str;d[i].error = getError(str);    }    sort(d,d+m,cmp);    for(int i = 0;i < m;i++){        cout<<d[i].str<<endl;    }    return 0;}