poj 1007 DNA Sorting

DNA Sorting

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 61155		Accepted: 24162

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

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 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

10 6
AACATGAAGG
TTTTGGCCAA
TTTGGCCAAA
GATCAGATTT
CCCGGGGGGA
ATCGATGCAT

Sample Output

CCCGGGGGGA
AACATGAAGG
GATCAGATTT
ATCGATGCAT
TTTTGGCCAA
TTTGGCCAAA

 

先简单说一下题名意思。题名意思是说有一些字符串，这些字符串的字符不都是按先后顺序排的。这些字符串都对应一个值，这个值是倒序数的总和。例如：AACEDGG只有E和D是反的，所以这个值是1。GCA这个串的值就是3（GA、GC、CA都是反的）。

#include <stdio.h>
#include <string.h>
int Sum(char a[],int n)
{
  int k,j,sum=0;
  for (j=0;j<n;j++)
  {
      for (k=j+1;k<n;k++)
      {
         if (a[j]>a[k]) sum++; 
      } 
  }
  return sum;
}
int main(int argc, char *argv[])
{
  int n,m;
  int i,j,t;
  char a[128][60],c[110];
  int b[128];
  scanf("%d%d",&n,&m);
  for (i=0;i<m;i++)
  {
     scanf("%s",a[i]);  
     b[i]=Sum(a[i],n);   //求b[i]的值  
  }
  //冒泡  
for (i=0;i<m-1;i++)
  for (j=0;j<m-1-i;j++)  
  {
     if(b[j]>b[j+1])
     {
       t=b[j];      strcpy(c,a[j]);
       b[j]=b[j+1]; strcpy(a[j],a[j+1]);  
       b[j+1]=t;    strcpy(a[j+1],c);    //若前个b[i]大的话就交换 连同字符窜一起交换  
     }
  }
   for (i=0;i<m;i++)
   printf("%s\n",a[i]);
  return 0;
}

 

这道题错的主要原因是对于m，n没有区分开

数组开的不够大还是因为n，m混淆