poj1007（串排序）

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http://poj.org/problem?id=1007

DNA Sorting

Time Limit: 1000MSMemory Limit: 10000KTotal Submissions: 66142Accepted: 26179

Description

One measure of ``unsortedness''in a sequence is the number of pairs of entries that are out oforder with respect to each other. For instance, in the lettersequence ``DAABEC'', this measure is 5, since D is greater thanfour letters to its right and E is greater than one letter to itsright. This measure is called the number of inversions in thesequence. The sequence ``AACEDGG'' has only one inversion (E andD)---it is nearly sorted---while the sequence ``ZWQM'' has 6inversions (it is as unsorted as can be---exactly the reverse ofsorted).

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, butrather in order of ``sortedness'', from ``most sorted'' to ``leastsorted''. All the strings are of the same length.

Input

The first line contains twointegers: a positive integer n (0 < n<= 50) giving the length of the strings; and apositive integer m (0 < m <= 100)giving the number of strings. These are followed by m lines, eachcontaining a string of length n.

Output

Output the list of inputstrings, arranged from ``most sorted'' to ``least sorted''. Sincetwo strings can be equally sorted, then output them according tothe orginal order.

Sample Input

10 6AACATGAAGGTTTTGGCCAATTTGGCCAAAGATCAGATTTCCCGGGGGGAATCGATGCAT

Sample Output

CCCGGGGGGAAACATGAAGGGATCAGATTTATCGATGCATTTTTGGCCAATTTGGCCAAA

Source

East Central North America 1998

题意：

输入m个长度为n的DNA序列，把他们按照逆序数从小到大稳定排序输出。

PS:“稳定排序”就是当序列中出现A1==A2时，排序前后A1与A2的相对位置不发生改变。

解题思路：

没难度，先求各个字符串的逆序数，再按逆序数对字符串快排，用qsort()函数。

虽然快排不是稳定的排序，但是只要在定义排序规则函数cmp做适当处理，a==b时返回0，即不处理a和b，就不会改变他们之间的相对位置了。这题纠结在怎么把比较完的字符串按照比较后的结果输出。网上看到的n[i]=n[i]*1000+i;输出时用 printf("%s\n",a[n[i]00]);真是大神啊。。

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
chara[101][101];
intn[101];
int cmp(constvoid*a,constvoid*b)
{
       return((*(int*)a-*(int*)b));
}
int main()
{
       intr,c,i,j,m;
       scanf("%d%d",&r,&c);
       for(i=0;i<c;i++)
       {
               scanf("%s",a[i]);//这边也十分值得借鉴
               for(j=0;j<r;j++)
                       for(m=j+1;m<r;m++)
                       {
                               if(a[i][j]>a[i][m])
                                       n[i]++;
                       }
                       n[i]=n[i]*1000+i;//妙啊
       }
       qsort(n,c,sizeof(n[0]),cmp);
       for(i=0;i<c;i++)
               printf("%s\n",a[n[i]%1000]);
       return 0;
}

代码2：个人来说比较喜欢这种。。。http://blog.csdn.net/woshixingaaa/article/details/5589070

#include<iostream>
#include<algorithm>
using namespace std;
struct f
{
int num;
char w[50];
}s[101];
bool cmp(struct f a,struct f b)
{
return a.num<b.num ;
}
int main()
{
int i,len,n,j,k;
cin>>len>>n;
for(i=0;i<n;i++)
{
  s[i].num =0;
  cin>>s[i].w;
  for(j=1;j<len;j++)
  {
   for(k=0;k<j;k++)
   {
    if(s[i].w[j]<s[i].w[k])
     s[i].num++;
   }
  }
}
sort(s,s+n,cmp);
for(i=0;i<n;i++)
  cout<<s[i].w<<endl;
return 0;
}