DNA Sorting

题目描述：

E - 05
Crawling in process...Crawling failedTime
Limit:2000MS    Memory Limit:65536KB   
64bit IO Format:%lld & %llu

SubmitStatus

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.

This problem contains multiple test cases!

The first line of a multiple input is an integer N, then a blank line followed by N input blocks. Each input block is in the format indicated in the problem description. There is a blank line between input blocks.

The output format consists of N output blocks. There is a blank line between output blocks.

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 (1 < 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''. If two or more strings are equally sorted, list them in the same order they are in the input file.

Sample Input

1

10 6

AACATGAAGG

TTTTGGCCAA

TTTGGCCAAA

GATCAGATTT

CCCGGGGGGA

ATCGATGCAT

Sample Output

CCCGGGGGGA

AACATGAAGG

GATCAGATTT

ATCGATGCAT

TTTTGGCCAA

TTTGGCCAAA

简述：

根据每一串字符串的逆序数来进行排序。

解题思路：

首先想到运用刚刚学的结构体来存储字符串然后运用冒泡排序，但发现并不可行，关键是vector容器的选择，另一个就是逆序数的计算。

解题细节：

1.因粗心大意，导致不能AC。

2.可以用sort进行排序，简单快捷.

3.当逆序数相等时，按原先固有位置排序。

代码：

#include <iostream>
#include <string>
#include <vector>
#include <algorithm>
using namespace std;
bool comp(const string &s1,const string &s2)//引用两个字符串
{
int i,j,k,m,n;
int sum1=0,sum2=0;
for(i=0;i<s1.size();i++)
{
for(j=i+1;j<s1.size();j++)
{
if(s1[i]>s1[j])
sum1++;
}
}
for(i=0;i<s2.size();i++)
{
for(j=i+1;j<s2.size();j++)
{
if(s2[i]>s2[j])
sum2++;
}
}
return sum1!=sum2?sum1<sum2:sum1<sum2;//如果两行字符串逆序数相同，那么按原先固有的位置排序
}
int main()
{
string s;
vector<string>v;//建立向量
int n,a,b;
cin>>n;
int i,j,k;
int p=0;//p来代表块数
for(i=0;i<n;i++)
{
cin.clear();
cin>>a>>b;
v.clear();//清空向量
p++;
for(j=0;j<b;j++)
{
cin>>s;
v.push_back(s);
}
sort(v.begin(),v.end(),comp);//采用sort排序
if(p!=1)
cout<<endl;//不是第一行的时候，产生新的一行
for(k=0;k<v.size();k++)
{
cout<<v[k]<<endl;
}
}
return 0;
}

心得：

解题不能过于着急，细心细致。