DNA Sorting

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 2203    Accepted Submission(s): 1075

Problem 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

这个题开始就没读懂,所以也不会做,后来学长讲完后才明白啥意思;

就是比如第一行数据AACATGAAGG     首先定义sum=0, A大于它后面字符的个数为0,sum+=0,第二个同样,第三个C大于它后面字符的个数为3,sum+=3。。。以此类推!

然后按每行数字从小到大输出字符串!!!

 #include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
struct as
{
char s[];
int m;
}aa[];
bool cmp(as s,as t)
{
return s.m < t.m;
}
int main()
{
int n,i,j,k,t,a,b,q;
scanf("%d",&n);
while(n--)
{
scanf("%d%d",&a,&b);
getchar();
for(k=;k<b;k++)
{ scanf("%s",aa[k].s);
{
for(t=;t<a;t++)
{
int sum=;
for(i=;i<a-;i++)
{
for(j=i+;j<a;j++)
if(aa[t].s[i]>aa[t].s[j])
sum++;
}
aa[t].m=sum;//将各组总数放在结构体中
} }
}
sort(aa, aa+b, cmp);//排序结构体
for(i=;i<b;i++)
printf("%s\n",aa[i].s);
}
return ;
}

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