poj 1007 DNA Sorting

DNA Sorting

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 76039		Accepted: 30490

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

Source

East Central North America 1998

 

题意：

      給m串长n的字符串，其价值为每字符比后面字符大的个数的和，按价值从小到大排序m串字符并输出

      我是直接用容器暴力的，也可以用其他排序

 

//Accepted    228K    32MS    C++    809B    2013-10-13 17:18:38
//简单的容器练习题.. 
#include<iostream>
#include<algorithm>
#include<string>
#include<map>
using namespace std;
map<string,int>M;
int n,m,a[105];
int get(string s) //得到价值
{
    int len=s.length();
    int sum=0;
    for(int i=0;i<len;i++)
        for(int j=i+1;j<len;j++)
            if(s[i]>s[j]) sum++;
    return sum;
}
int main(void)
{
    string s;
    while(cin>>m>>n)
    {
        for(int i=0;i<n;i++){
            cin>>s;
            M[s]=get(s);
            a[i]=M[s];
        }
        sort(a,a+n);
        for(int i=0;i<n;i++){
            //cout<<a[i]<<endl;
            for(map<string,int>::iterator it=M.begin();it!=M.end();it++){ //遍历n串
                if((it->second)==a[i])
                    cout<<it->first<<endl;
                //break;
            }
        }
    }
    return 0;
}