E - 归并排序 求逆序数

Posted 2020-06-15

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


该题 利用数组的技巧 巧算逆序数  如函数f所示
运用结构体将 逆序数 与 字符串联系起来
运用algorithm中的stable_sort函数 排序 相同时，不改变原有序列！！

#include<iostream>
#include<algorithm>
using namespace std;
struct DNA{
    char str[55];
    int num;
}d[105];
bool cmp(DNA a,DNA b)  
{  
    return a.num<b.num;  
}  
int f(char s[],int n)
{
    int a[5]={0,0,0,0},m=0;
    for(int i=n-1;i>=0;i--){
        switch(s[i]){
        case ‘A‘:
            a[1]++;
            a[2]++;
            a[3]++;
            break;
        case ‘C‘:
            a[2]++;
            a[3]++;
            m+=a[1];
            break;
        case ‘G‘:
            a[3]++;
            m+=a[2];
            break;
        case ‘T‘:
            m+=a[3];
            break;
        }
    }
    return m;
}
int main() {
    int n,m;
    cin>>n>>m;
    for(int i=0;i<m;i++){
        for(int j=0;j<n;j++)cin>>d[i].str[j];
        d[i].num=f(d[i].str,n);
    }
    stable_sort(d,d+m,cmp);
    for(int i=0;i<m;i++) {
        for(int j=0;j<n;j++)cout<<d[i].str[j];
        cout<<endl;
    }
    //system("pause");
    return 0;
}