zoj_1188 DNA Sorting

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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

10 6
AACATGAAGG
TTTTGGCCAA
TTTGGCCAAA
GATCAGATTT
CCCGGGGGGA
ATCGATGCAT

Sample Output

CCCGGGGGGA
AACATGAAGG
GATCAGATTT
ATCGATGCAT
TTTTGGCCAA
TTTGGCCAAA

分析：

本题是字符串排序问题，字符串内部按照ASCII大小排序，字符串之间按照倒位数量来排序。

倒位数量，就是该字符串中，每个字符与它右边的字符相比，逆序次数的总和。

排序时，如果两行字符串的倒位数量相同，那么按照原先输入时的顺序排序；输出时，每块之间需要空一行。

关键问题是：字符串内部如何排序？

首先要想到，要得到倒位数量，需要一个变量记录字符对比的次数，如果当前字符比右边字符大，该变量就加1，否则变量不变。

假设字符串一的倒位数量是c1，字符串二的倒位数量是c2.

那么字符串之间是如何比较的呢？

答案给出的比较方法中：return c1!=c2 ? c1<c2:c1<c2;

简化可得：return c1<c2;

另外还需要温习一下sort函数：

默认情况下为升序排列, sort(array,array+n);

也可以重载该函数，自定义排序方法cmp， sort(array,array+n,cmp);

例如降序排列：

bool cmp( int a, int b)

{

return a>b;

}

Answer：

#include <iostream>#include <string>#include <vector>#include <algorithm>using namespace std;bool comp(const string &s1, const string &s2){    int i,j,k;    int c1=0,c2=0;    //计算s1需要移动的次数c1    for(i=0;i<s1.size();i++)    {        for(j=i+1;j<s1.size();j++)        {            if(s1[i]>s1[j]) c1++;        }    }    //计算s1需要移动的次数c1    for(i=0;i<s2.size();i++)    {        for(j=i+1;j<s2.size();j++)        {            if(s2[i]>s2[j]) c2++;        }    }    //如果两行字符串移动的次数相同，那么按照原先固有的位置排序    return c1<c2;}int main(){    string s;    vector<string>v;    int n,a,b;//n为组数，a为字符串长度，b为字符串行数    int i,j,k;    int p=0;//组数从0开始    cin>>n;    for(i=0;i<n;i++)    {        cin.clear();        cin>>a>>b;        v.clear();        p++;        for(j=0;j<b;j++)        {            cin>>s;            v.push_back(s);        }        //按comp排序规则排序        sort(v.begin(),v.end(),comp);        if(p!=1) cout<<endl; //如果不是第一组，产生一个新空行        for(k=0;k<v.size();k++)        {            cout<<v[k]<<endl;        }    }}

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