【AKOJ】1023-DNA Sorting
DNA Sorting
Time Limit:1000MS Memory Limit:65536K
Total Submit:16 Accepted:9
原题链接
Description
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
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
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
Sample Input
10 6
AACATGAAGG
TTTTGGCCAA
TTTGGCCAAA
GATCAGATTT
CCCGGGGGGA
ATCGATGCAT
Sample Output
Sample Output
CCCGGGGGGA
AACATGAAGG
GATCAGATTT
ATCGATGCAT
TTTTGGCCAA
TTTGGCCAAA
Source
ahstu@ICPC01
#include
#include
struct st //定义字符串，与其数值的结构体
{
char ss[100];
int count;
}tt[100];
int co(char *ss) //计算字符串数值
{
int coun=0;
for(int i=0;ss[i]!='\0';i++)
{
for(int j=i+1;ss[j]!='\0';j++)
{
if(ss[i]>ss[j])
{coun++;}
}
}
return coun;
}
void swap(char *s,char *t) //字符串交换
{
char ch;
for(int i=0;i<100;i++)
{ch=t[i];t[i]=s[i];s[i]=ch;}
}
int main()
{
int m,n,i,temp;
scanf("%d%d",&m,&n);
for(i=0;itt[k+1].count)
{temp=tt[k].count;tt[k].count=tt[k+1].count;tt[k+1].count=temp;swap(tt[k].ss,tt[k+1].ss);} //数值交换的同时，字符串交换
}
}
for(i=0;i