DNA Sorting（排序）

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DNA Sorting

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 2182    Accepted Submission(s): 1062

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

 

思路：从小到大排列。。。不好说，看题；

代码：

 #include<stdio.h>
 #include<algorithm>
 using namespace std;
 typedef struct{
     int num;
     char str[];
 }node;
 int cmp(node a,node b){
     return a.num<b.num;
 }
 int search(int n,char *a){int flot=;
     for(int i=;i<n;++i){
         for(int j=i+;j<n;++j){
             if(a[j]-a[i]<)flot++;
         }
     }
     return flot;
 }
 int main(){
     int m,n,T;
     node dna[];
     scanf("%d",&T);
     while(T--){scanf("%d%d",&n,&m);
         int i=;
         for(int i=;i<m;++i){
             scanf("%s",dna[i].str);
             dna[i].num=search(n,dna[i].str);
         }
         sort(dna,dna+m,cmp);
         for(int i=;i<m;++i)printf("%s\n",dna[i].str);
     }
     return ;
 }

快排：

#include<stdio.h>
#include<algorithm>
#include<string.h>
using namespace std;
typedef struct{
    int num;
    char str[];
}node;
node dna[];
int search(int n,char *a){
    int flot=;
    for(int i=;i<n;++i){
        for(int j=i+;j<n;++j){
            if(a[j]-a[i]<)flot++;
        }
    }
    return flot;
}
void Swap(node &a, node &b){
    swap(a.num, b.num);
    char s[];
    strcpy(s, a.str);
    strcpy(a.str, b.str);
    strcpy(b.str, s);
}
int handle(int l, int r){
    int j = l - ;
    for(int i = l; i <= r; i++){
        if(dna[i].num <= dna[r].num){
            j++;
            Swap(dna[i], dna[j]);
        }
    }
    return j;
}
void quiksort(int l, int r){
    int m;
    if(l < r){
        m = handle(l, r);
        quiksort(l, m - );
        quiksort(m + , r);
    }
}
int main(){
    int m,n,T;

    scanf("%d",&T);
    while(T--){scanf("%d%d",&n,&m);
        int i=;
        for(int i=;i<m;++i){
            scanf("%s",dna[i].str);
            dna[i].num=search(n,dna[i].str);
        }
        quiksort(, m - );
        for(int i=;i<m;++i)printf("%s\n",dna[i].str);
    }
    return ;
}