HDU 1379:DNA Sorting
DNA Sorting
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 2329 Accepted Submission(s): 1145
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.
1 10 6
AACATGAAGG
TTTTGGCCAA
TTTGGCCAAA
GATCAGATTT
CCCGGGGGGA
ATCGATGCAT
CCCGGGGGGA
AACATGAAGG
GATCAGATTT
ATCGATGCAT
TTTTGGCCAA
TTTGGCCAAA
你 离 开 了 , 我 的 世 界 里 只 剩 下 雨 。 。 。
#include <stdio.h>
#include <algorithm>
#include <iostream>
#include<string.h>
#include<math.h>
using namespace std;
struct DNA
{
char c[55];
int data;
} d[105];
int nixu(char *c)
{
int k=0;
int n=strlen(c);
for(int i=0; i<n; i++)
{
for(int j=i+1; j<n; j++)
if(c[i]>c[j])k++;
}
return k;
}
void paixu(int m)
{
for(int i=0; i<m; i++)
for(int j=0; j<m-i-1; j++)
if(d[j+1].data<d[j].data)
{
DNA a=d[j+1];
d[j+1]=d[j];
d[j]=a;
}
}
int main()
{
int t;
cin>>t;
while(t--)
{
int n,m;
cin>>n>>m;
getchar();
for(int i=0; i<m; i++)
{
scanf("%s",d[i].c);
d[i].data=nixu(d[i].c);
}
paixu(m);
for(int i=0; i<m; i++)
puts(d[i].c);
}
return 0;
}
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