ZOJ Problem Set–1188 DNA Sorting

Time Limit: 2 Seconds      Memory Limit: 65536 KB

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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

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Source: East Central North America 1998

该题目中仅需要改变一下string的排序方式，然后使用STL容器就可以了。不过有一点，题目中说，如果按照规定的比较方法比较的结果是一样大的话，就按照输入的顺序输出，这一点为了节省了很多麻烦。

代码如下：

#include<iostream>
#include<set>
#include<string>
#include<algorithm>
#include<list>
using namespace std;
class DNA
{
public:
  string actg;
  DNA(string& s):actg(s){}
  bool operator<(const DNA& dna) const
  {
    return inversionNum() < dna.inversionNum();
  }
  int inversionNum() const
  {
    int bigToRight = 0;
    for(int i = 0; i < actg.length(); i++)
    {
      for(int j = i + 1; j < actg.length(); j++)
      {
        if(actg[i] > actg[j])
          bigToRight++;;
      }
    }
    return bigToRight;
  }
};
int main()
{
  int blocks;cin>>blocks;
  for(int block = 0; block < blocks; block++)
  {
    string state;
    if(block == 0)//仅在第一次的时候有如下两个空行
    {
      getline(cin,state);//用getline 吃掉输入blocks之后的回车
      getline(cin,state);//题目要求的空行
    }
    
    int n,m;
    cin>>n>>m;
    multiset<DNA> se;
    list<string> strList;
    for(int i = 0; i < m; i++)
    {
      string actg;cin>>actg;
      se.insert(DNA(actg));
    }
    for(multiset<DNA>::iterator it = se.begin(); it != se.end(); it++)
    {
      cout<<(*it).actg<<endl;
    }
    if(block != blocks - 1)//最后一次输出没有空行
    {
      cout<<endl;
    }
  }
  return 0;
}

完成之后，我就在想，如果按照题意中的比较方法比较的结果是相等的话，就按string默认的比较方式来排序的话，<重载的方法应该如下所示：

bool operator<(const DNA& dna) const
{
　　return inversionNum() < dna.inversionNum() ? true:actg<dna.actg;
}

但是真的这样做的话，程序是会出错的，老是说我的<重载有错误，这我就不知道错误在哪里了，这个有待研究一下，到底我这样写回导致怎么样的行为，如果要实现我的想法的话，<的重载又该怎么写呢？