1007. DNA Sorting

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本文介绍了一种衡量DNA字符串序列“排序性”的方法，并提供了一个根据该排序性对字符串进行排列的C++程序实例。排序性的高低依据序列中逆序对的数量决定。

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

代码

#include<iostream>
#include <vector>
#include <string>
#include<map>
using namespace std;
int main() {
    int m, n;
    cin >> n >> m;
    string s;
    map<int, vector<string> > list;
    while (m--) {
        cin >> s;   
        int count = 0;
        for (int i = 0; i < s.size(); i++) {
            for (int j = i+1; j < s.size(); j++) {
                if (s[i] > s[j])
                    count++;
            }
        }
        list[count].push_back(s);
    }
    for (map<int, vector<string> >::iterator iter = list.begin(); iter != list.end(); iter++) {
        pair<int, vector<string> > temp = *iter;
        for (int i = 0; i < temp.second.size(); i++) {
            cout << temp.second[i] << endl;
        }
    }
    return 0;
}